ENGN40: Dynamics and Vibrations - Brown University

Chapter 5

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Vibrations

5.1 Overview of Vibrations

5.1.1 Examples of practical vibration problems

Vibration is a continuous cyclic motion of a structure or a component.

Generally, engineers try to avoid vibrations, because vibrations have a number of unpleasant effects:

·Cyclic motion implies cyclic forces. Cyclic forces are very damaging to materials.

·Even modest levels of vibration can cause extreme discomfort;

·Vibrations generally lead to a loss of precision in controlling machinery.

Examples where vibration suppression is an issue include:

Structural vibrations.  Most buildings are mounted on top of special rubber pads, which are intended to isolate the building from ground vibrations.  The figure on the right shows vibration isolators being installed under the floor of a building during construction (from www.wilrep.com )

No vibrations course is complete without a mention of the Tacoma Narrows suspension bridge. This bridge, constructed in the s, was at the time the longest suspension bridge in the world. Because it was a new design, it suffered from an unforseen source of vibrations. In high wind, the roadway would exhibit violent torsional vibrations, as shown in the picture below.

You can watch newsreel footage of the vibration and even the final collapse at http://www.youtube.com/watch?v=HxTZ446tbzE  To the credit of the designers, the bridge survived for an amazingly long time before it finally failed.  It is thought that the vibrations were a form of self-excited vibration known as `flutter,’ or ‘galloping’  A similar form of vibration is known to occur in aircraft wings.  Interestingly, modern cable stayed bridges that also suffer from a new vibration problem: the cables are very lightly damped and can vibrate badly in high winds (this is a resonance problem, not flutter). You can find a detailed article on the subject at www.fhwa.dot.gov/bridge/pubs//chap3.cfm. Some bridge designs go as far as to incorporate active vibration suppression systems in their cables.

Vehicle suspension systems are familiar to everyone, but continue to evolve as engineers work to improve vehicle handling and ride (the figure above is from http://www.altairhyperworks.com. A radical new approach to suspension design emerged in when a research group led by Malcolm Smith at Cambridge University invented a new mechanical suspension element they called an ‘inerter’. This device can be thought of as a sort of generalized spring, but instead of exerting a force proportional to the relative displacement of its two ends, the inerter exerts a force that is proportional to the relative acceleration of its two ends. An actual realization is shown in the figure. You can find a detailed presentation on the theory behind the device at http://www-control.eng.cam.ac.uk/~mcs/lecture_j.pdf The device was adopted in secret by the McLaren Formula 1 racing team in (they called it the ‘J damper’, and a scandal erupted in Formula 1 racing when the Renault team managed to steal drawings for the device, but were unable to work out what it does. The patent for the device has now been licensed Penske and looks to become a standard element in formula 1 racing. It is only a matter of time before it appears on vehicles available to the rest of us.

Precision Machinery: The picture on the right shows one example of a precision instrument. It is essential to isolate electron microscopes from vibrations. A typical transmission electron microscope is designed to resolve features of materials down to atomic length scales. If the specimen vibrates by more than a few atomic spacings, it will be impossible to see! This is one reason that electron microscopes are always located in the basement – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ the basement of a building vibrates much less than the upper floors. Professor K.-S. Kim at Brown recently invented and patented a new vibration isolation system to support his atomic force microscope on the 7th floor of the Barus-Holley building – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ you can find the patent at United States Patent, Patent Number 7,543,791.

Here is another precision instrument that is very sensitive to vibrations.

The picture shows features of a typical hard disk drive. It is particularly important to prevent vibrations in the disk stack assembly and in the disk head positioner, since any relative motion between these two components will make it impossible to read data. The spinning disk stack assembly has some very interesting vibration characteristics (which fortunately for you, is beyond the scope of this course).

Vibrations are not always undesirable, however. On occasion, they can be put to good use. Examples of beneficial applications of vibrations include ultrasonic probes, both for medical application and for nondestructive testing. The picture shows a medical application of ultrasound: it is an image of someone’s colon. This type of instrument can resolve features down to a fraction of a millimeter, and is infinitely preferable to exploratory surgery. Ultrasound is also used to detect cracks in aircraft and structures.

Musical instruments and loudspeakers are a second example of systems which put vibrations to good use.  Finally, most mechanical clocks use vibrations to measure time. 

5.1.2 Vibration Measurement

When faced with a vibration problem, engineers generally start by making some measurements to try to isolate the cause of the problem. There are two common ways to measure vibrations:

1. An accelerometer is a small electro-mechanical device that gives an electrical signal proportional to its acceleration.   The picture shows a typical 3 axis MEMS accelerometer (you’ll use one in a project in this course).   MEMS accelerometers should be selected very carefully – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@  you can buy cheap accelerometers for less than $50, but these are usually meant just as sensors, not for making precision measurements.   For measurements you’ll need to select one that is specially designed for the frequency range you are interested in sensing.  The best accelerometers are expensive ‘inertial grade’ versions (suitable for so-called ‘inertial navigtation’ in which accelerations are integrated to determine position) which are often use Kalman filtering to fuse the accelerations with GPS measurements.

2.A displacement transducer is similar to an accelerometer, but gives an electrical signal proportional to its displacement.

Displacement transducers are generally preferable if you need to measure low frequency vibrations; accelerometers behave better at high frequencies.

5.1.3 Features of a Typical Vibration Response

The picture below shows a typical signal that you might record using an accelerometer or displacement transducer.

Important features of the response are

The signal is often (although not always) periodic: that is to say, it repeats itself at fixed intervals of time. Vibrations that do not repeat themselves in this way are said to be random. All the systems we consider in this course will exhibit periodic vibrations.

The PERIOD of the signal, T, is the time required for one complete cycle of oscillation, as shown in the picture.

 The FREQUENCY of the signal, f, is the number of cycles of oscillation per second. Cycles per second is often given the name Hertz: thus, a signal which repeats 100 times per second is said to oscillate at 100 Hertz.

 The ANGULAR FREQUENCY of the signal, ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ , is defined as ω=2πf MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jabg2 da9iaaikdacqaHapaCcaWGMbaaaa@3C71@ . We specify angular frequency in radians per second. Thus, a signal that oscillates at 100 Hz has angular frequency 200π MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdacaaIWa GaaGimaiabec8aWbaa@3A27@ radians per second.

Period, frequency and angular frequency are related by

f= 1 T ω=2πf= 2π T MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAgacqGH9a qpdaWcaaqaaiaaigdaaeaacaWGubaaaiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeM8a3jabg2da9i abgkdaYiabec8aWjaadAgacqGH9aqpdaWcaaqaaiaaikdacqaHapaC aeaacaWGubaaaaaa@@

The PEAK-TO-PEAK AMPLITUDE of the signal, A, is the difference between its maximum value and its minimum value, as shown in the picture

 The AMPLITUDE of the signal is generally taken to mean half its peak to peak amplitude. Engineers sometimes use amplitude as an abbreviation for peak to peak amplitude, however, so be careful.

 The ROOT MEAN SQUARE AMPLITUDE or RMS amplitude is defined as

σ= { 1 T ∫ 0 T [ y(t) ] 2 dt } 1/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeo8aZjabg2 da9maacmaabaWaaSaaaeaacaaIXaaabaGaamivaaaadaWdXbqaamaa dmaabaGaamyEaiaacIcacaWG0bGaaiykaaGaay5waiaaw2faamaaCa aaleqabaGaaGOmaaaaaeaacaaIWaaabaGaamivaaqdcqGHRiI8aOGa amizaiaadshaaiaawUhacaGL9baadaahaaWcbeqaaiaaigdacaGGVa GaaGOmaaaaaaa@4B36@

5.1.4 Harmonic Oscillations

 Harmonic oscillations are a particularly simple form of vibration response.  A conservative spring-mass system will exhibit harmonic motion – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@  if you have Java, Internet Explorer (or a browser plugin that allows you to run IE in another browser) you can run a Java Applet to visualize the motion.  You can find instructions for installing Java, the IE plugins, and giving permission for the Applet to run here. The address for the SHM simulator (cut and paste this into the Internet Explorer address bar)

http://www.brown.edu/Departments/Engineering/Courses/En4/java/shm.html

If the spring is perturbed from its static equilibrium position, it vibrates (press `start’ to watch the vibration). We will analyze the motion of the spring mass system soon. We will find that the displacement of the mass from its static equilibrium position, x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhaaaa@@ , has the form

x(t)=ΔXsin(ωt+ϕ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcqqHuoarcaWGybGaci4CaiaacMgacaGG UbGaaiikaiabeM8a3jaaykW7caWG0bGaey4kaSIaeqy1dyMaaiykaa aa@47FD@

Here, ΔX MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejaadI faaaa@387C@ is the amplitude of the displacement, ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ is the frequency of oscillations in radians per second, and ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@@ (in radians) is known as the `phase’ of the vibration. Vibrations of this form are said to be Harmonic.

Typical values for amplitude and frequency are listed in the table below

We can also express the displacement in terms of its period of oscillation T

x(t)=ΔXsin( 2π T t+ϕ ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcqqHuoarcaWGybGaci4CaiaacMgacaGG UbWaaeWaaeaadaWcaaqaaiaaikdacqaHapaCaeaacaWGubaaaiaads hacqGHRaWkcqaHvpGzaiaawIcacaGLPaaaaaa@@

The velocity v MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhaaaa@@ and acceleration a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggaaaa@371F@ of the mass follow as

v(t)=ΔVsin( ωt+ϕ ) V 0 =ωΔXcos( ωt+ϕ ) a(t)=−ΔAsin( ωt+ϕ )ΔA=ωΔV= ω 2 ΔX MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamODai aacIcacaWG0bGaaiykaiabg2da9iabfs5aejaadAfaciGGZbGaaiyA aiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0bGaey4kaSIaeqy1dy gacaGLOaGaayzkaaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGwbWaaSba aSqaaiaaicdaaeqaaOGaeyypa0JaeqyYdCNaeuiLdqKaamiwaiGaco gacaGGVbGaai4CamaabmaabaGaeqyYdCNaaGPaVlaadshacqGHRaWk cqaHvpGzaiaawIcacaGLPaaaaeaacaWGHbGaaiikaiaadshacaGGPa Gaeyypa0JaeyOeI0IaeuiLdqKaamyqaiGacohacaGGPbGaaiOBamaa bmaabaGaeqyYdCNaaGPaVlaadshacqGHRaWkcqaHvpGzaiaawIcaca GLPaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cq qHuoarcaWGbbGaeyypa0JaaGPaVlaaykW7caaMc8UaeqyYdCNaeuiL dqKaamOvaiabg2da9iabeM8a3naaCaaaleqabaGaaGOmaaaakiabfs 5aejaadIfaaaaa@B506@

Here, ΔV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejaadA faaaa@387A@ is the amplitude of the velocity, and ΔA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejaadg eaaaa@@ is the amplitude of the acceleration. Note the simple relationships between acceleration, velocity and displacement amplitudes.

Surprisingly, many complex engineering systems behave just like the spring mass system we are looking at here. To describe the behavior of the system, then, we need to know three things (in order of importance):

(1) The frequency (or period) of the vibrations

(2) The amplitude of the vibrations

(3) Occasionally, we might be interested in the phase, but this is rare.

So, our next problem is to find a way to calculate these three quantities for engineering systems.

We will do this in stages. First, we will analyze a number of freely vibrating, conservative systems. Second, we will examine free vibrations in a dissipative system, to show the influence of energy losses in a mechanical system. Finally, we will discuss the behavior of mechanical systems when they are subjected to oscillating forces.

5.2 Free vibration of conservative, single degree of freedom, linear systems.

First, we will explain what is meant by the title of this section.

Recall that a system is conservative if energy is conserved, i.e. potential energy + kinetic energy = constant during motion.

Free vibration means that no time varying external forces act on the system.

A system has one degree of freedom if its motion can be completely described by a single scalar variable. We’ll discuss this in a bit more detail later.

A system is said to be linear if its equation of motion is linear. We will see what this means shortly.

It turns out that all 1DOF, linear conservative systems behave in exactly the same way. By analyzing the motion of one representative system, we can learn about all others.

We will follow standard procedure, and use a spring-mass system as our representative example.

Problem: The figure shows a spring mass system. The spring has stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaaGimaaqabaaaaa@37F1@ . The mass is released with velocity v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaa WcbaGaaGimaaqabaaaaa@381A@ from position s 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohadaWgaa WcbaGaaGimaaqabaaaaa@@ at time t=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshacqGH9a qpcaaIWaaaaa@38F2@ . Find s(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohacaGGOa GaamiDaiaacMcaaaa@@ .

There is a standard approach to solving problems like this

(i) Get a differential equation for s using F=ma (or other methods to be discussed)

(ii) Solve the differential equation.

The picture shows a free body diagram for the mass.

Newton’s law of motion states that

F=ma⇒− F s i+(N−mg)j=m d 2 s d t 2 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcaWGTbGaaCyyaiabgkDiElabgkHiTiaadAeadaWgaaWcbaGaam4C aaqabaGccaWHPbGaey4kaSIaaiikaiaad6eacqGHsislcaWGTbGaam 4zaiaacMcacaWHQbGaeyypa0JaamyBamaalaaabaGaamizamaaCaaa leqabaGaaGOmaaaakiaadohaaeaacaWGKbGaamiDamaaCaaaleqaba GaaGOmaaaaaaGccaWHPbaaaa@4F8E@

The spring force is related to the length of the spring by F s =k(s− L 0 ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaam4CaaqabaGccqGH9aqpcaWGRbGaaiikaiaadohacqGHsisl caWGmbWaaSbaaSqaaiaaicdaaeqaaOGaaiykaaaa@3F28@ . The i component of the equation of motion and this equation then shows that

m d 2 s d t 2 +ks=k L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGaamizaiaa dshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaam4Aaiaadohacq GH9aqpcaWGRbGaamitamaaBaaaleaacaaIWaaabeaaaaa@435C@

This is our equation of motion for s.

Now, we need to solve this equation. We could, of course, use Matlab to do this – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ in fact here is the Matlab solution.

syms m k L0 s0 v0 real
syms v(t) s(t)
assume(k>0); assume(m>0);
diffeq = m*diff(s(t),t,2) + k*s(t) == k*L0;
v(t) = diff(s(t),t);
IC = [s(0)==s0, v(0)==v0];
s(t) = dsolve(diffeq,IC)

In practice we usually don’t need to use matlab (and of course in exams you won’t have access to matlab!)

5.2.1 Using tabulated solutions to solve equations of motion for vibration problems

Note that all vibrations problems have similar equations of motion.  Consequently, we can just solve the equation once, record the solution, and use it to solve any vibration problem we might be interested in.  The procedure to solve any vibration problem is:

1. Derive the equation of motion, using Newton’s laws (or sometimes you can use energy methods, as discussed in Section 5.3)

2. Do some algebra to arrange the equation of motion into a standard form

3. Look up the solution to this standard form in a table of solutions to vibration problems.

We have provided a table of standard solutions as a separate document that you can download and print for future reference.   Actually, this is exactly what MATLAB is doing when it solves a differential equation for you – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugGbabaaa aaaaaapeGaa83eGaaa@@  it is doing sophisticated pattern matching to look up the solution you want in a massive internal database.

We will illustrate the procedure using many examples.

5.2.2 Solution to the equation of motion for an undamped spring-mass system

We would like to solve

m d 2 s d t 2 +ks=k L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGaamizaiaa dshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaam4Aaiaadohacq GH9aqpcaWGRbGaamitamaaBaaaleaacaaIWaaabeaaaaa@435C@

with initial conditions  ds dt = v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaadohaaeaacaWGKbGaamiDaaaacqGH9aqpcaWG2bWaaSbaaSqa aiaaicdaaeqaaaaa@3CF3@ from position s 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohadaWgaa WcbaGaaGimaaqabaaaaa@@ at time t=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshacqGH9a qpcaaIWaaaaa@38F2@ .

We therefore consult our list of solutions to differential equations, and observe that it gives the solution to the following equation

1 ω n 2 d 2 x d t 2 +x=C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaamiEaiabg2da 9iaadoeaaaa@@

This is very similar to our equation, but not quite the same.  To make them identical, divide our equation through by k  

m k d 2 s d t 2 +s= L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam yBaaqaaiaadUgaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaam4CaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaki abgUcaRiaadohacqGH9aqpcaWGmbWaaSbaaSqaaiaaicdaaeqaaaaa @427C@

We see that if we define

m k = 1 ω n 2 ⇒ ω n = k m x=sC= L 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam yBaaqaaiaadUgaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaeqyYdC3a a0baaSqaaiaad6gaaeaacaaIYaaaaaaakiabgkDiElabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamiEaiabg2da9i aadohacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaam4qaiabg2da9i aadYeadaWgaaWcbaGaaGimaaqabaaaaa@7DFF@

then our equation is equivalent to the standard one.

HEALTH WARNING: it is important to note that this substitution only works if L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaaGimaaqabaaaaa@37F1@ is constant, so its time derivative is zero.

The solution for x is

x=C+ X 0 sin( ω n t+ϕ ) X 0 = ( x 0 −C) 2 + v 0 2 / ω n 2 ϕ= tan −1 ( ( x 0 −C) ω n v 0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai abg2da9iaadoeacqGHRaWkcaWGybWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbWaaeWaaeaacqaHjpWDdaWgaaWcbaGaamOBaa qabaGccaWG0bGaey4kaSIaeqy1dygacaGLOaGaayzkaaGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVdqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccqGH9aqpdaGcaaqaaiaacIcacaWG4bWaaSba aSqaaiaaicdaaeqaaOGaeyOeI0Iaam4qaiaacMcadaahaaWcbeqaai aaikdaaaGccqGHRaWkcaWG2bWaa0baaSqaaiaaicdaaeaacaaIYaaa aOGaai4laiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaeqaaO GaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlabew9aMjabg2da9iGacshacaGGHbGaaiOBamaaCa aaleqabaGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaacaGGOaGa amiEamaaBaaaleaacaaIWaaabeaakiabgkHiTiaadoeacaGGPaGaeq yYdC3aaSbaaSqaaiaad6gaaeqaaaGcbaGaamODamaaBaaaleaacaaI WaaabeaaaaaakiaawIcacaGLPaaaaaaa@84A4@

Here, x 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGimaaqabaaaaa@381C@ and v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaa WcbaGaaGimaaqabaaaaa@381A@ are the initial value of x and dx/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsgacaWG4b Gaai4laiaadsgacaWG0baaaa@3AB5@ its time derivative, which must be computed from the initial values of s and its time derivative

x 0 = s 0 v 0 = dx dt = ds dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGimaaqabaGccqGH9aqpcaWGZbWaaSbaaSqaaiaaicdaaeqa aOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadAhadaWgaaWcbaGa aGimaaqabaGccqGH9aqpdaWcaaqaaiaadsgacaWG4baabaGaamizai aadshaaaGaeyypa0ZaaSaaaeaacaWGKbGaam4CaaqaaiaadsgacaWG 0baaaaaa@5AC5@

When we present the solution, we have a choice of writing down the solution for x, and giving formulas for the various terms in the solution (this is what is usually done):

x= X 0 sin( ω n t+ϕ ) ω n = k m X 0 = ( s 0 − L 0 ) 2 + v 0 2 / ω n 2 ϕ= tan −1 ( ( s 0 − L 0 ) ω n v 0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai abg2da9iaadIfadaWgaaWcbaGaaGimaaqabaGcciGGZbGaaiyAaiaa c6gadaqadaqaaiabeM8a3naaBaaaleaacaWGUbaabeaakiaadshacq GHRaWkcqaHvpGzaiaawIcacaGLPaaacaaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8oabaGaeqyYdC3aaSbaaSqaaiaad6gaae qaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaadUgaaeaacaWGTbaaaaWc beaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaadIfadaWgaaWcbaGaaGimaaqabaGccqGH9aqpdaGc aaqaamaabmaabaGaam4CamaaBaaaleaacaaIWaaabeaakiabgkHiTi aadYeadaWgaaWcbaGaaGimaaqabaaakiaawIcacaGLPaaadaahaaWc beqaaiaaikdaaaGccqGHRaWkcaWG2bWaa0baaSqaaiaaicdaaeaaca aIYaaaaOGaai4laiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaa aeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlabew9aMjabg2da9iGacshacaGGHbGaaiOB amaaCaaaleqabaGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaaca GGOaGaam4CamaaBaaaleaacaaIWaaabeaakiabgkHiTiaadYeadaWg aaWcbaGaaGimaaqabaGccaGGPaGaeqyYdC3aaSbaaSqaaiaad6gaae qaaaGcbaGaamODamaaBaaaleaacaaIWaaabeaaaaaakiaawIcacaGL Paaaaaaa@@

Alternatively, we can express all the variables in the standard solution in terms of s

s= L 0 + ( s 0 − L 0 ) 2 + v 0 2 / ω n 2 sin( k m t+ tan −1 ( ( s 0 − L 0 ) ω n v 0 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohacqGH9a qpcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaOaaaeaadaqa daqaaiaadohadaWgaaWcbaGaaGimaaqabaGccqGHsislcaWGmbWaaS baaSqaaiaaicdaaeqaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaI YaaaaOGaey4kaSIaamODamaaDaaaleaacaaIWaaabaGaaGOmaaaaki aac+cacqaHjpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaaabeaakiGa cohacaGGPbGaaiOBamaabmaabaWaaOaaaeaadaWcaaqaaiaadUgaae aacaWGTbaaaaWcbeaakiaadshacqGHRaWkciGG0bGaaiyyaiaac6ga daahaaWcbeqaaiabgkHiTiaaigdaaaGcdaqadaqaamaalaaabaGaai ikaiaadohadaWgaaWcbaGaaGimaaqabaGccqGHsislcaWGmbWaaSba aSqaaiaaicdaaeqaaOGaaiykaiabeM8a3naaBaaaleaacaWGUbaabe aaaOqaaiaadAhadaWgaaWcbaGaaGimaaqabaaaaaGccaGLOaGaayzk aaaacaGLOaGaayzkaaaaaa@@

But this solution looks very messy (more like the Matlab solution).

Observe that:

The mass oscillates harmonically, as discussed in the preceding section;

The angular frequency of oscillation, ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ , is a characteristic property of the system, and is independent of the initial position or velocity of the mass. This is a very important observation, and we will expand upon it below. The characteristic frequency is known as the natural frequency of the system.

  Increasing the stiffness of the spring increases the natural frequency of the system;

  Increasing the mass reduces the natural frequency of the system.

5.2.3 Natural Frequencies and Mode Shapes.

We saw that the spring mass system described in the preceding section likes to vibrate at a characteristic frequency, known as its natural frequency. This turns out to be a property of all stable mechanical systems.

All stable, unforced, mechanical systems vibrate harmonically at certain discrete frequencies, known as natural frequencies of the system.

For the spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8hfGaaa@@ mass system, we found only one natural frequency. More complex systems have several natural frequencies. For example, the system of two masses shown below has two natural frequencies, given by

ω 1 = k m , ω 2 = 3k m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIXaaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaeqyYdC3aaSbaaSqaaiaaikdaaeqaaOGaey ypa0ZaaOaaaeaadaWcaaqaaiaaiodacaWGRbaabaGaamyBaaaaaSqa baaaaa@5A78@

A system with three masses would have three natural frequencies, and so on.

In general, a system with more than one natural frequency will not vibrate harmonically.

For example, suppose we start the two mass system vibrating, with initial conditions

x 1 = x o 1 d x 1 dt =0 x 2 = x o 2 d x 2 dt =0 }t=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaciaaeaqabe aacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0ZaaCbiaeaacaWG 4baaleqabaGaam4BaaaakmaaBaaaleaacaaIXaaabeaakiaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7daWcaaqaaiaadsgacaWG4bWaaSbaaS qaaiaaigdaaeqaaaGcbaGaamizaiaadshaaaGaeyypa0JaaGimaaqa aiaadIhadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpdaWfGaqaaiaadI haaSqabeaacaWGVbaaaOWaaSbaaSqaaiaaikdaaeqaaOGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVpaalaaabaGaamizaiaadIhadaWgaaWc baGaaGOmaaqabaaakeaacaWGKbGaamiDaaaacqGH9aqpcaaIWaaaai aaw2haaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadshacqGH9aqpcaaIWa aaaa@8E1F@

The response may be shown (see sect 5.5 if you want to know how) to be

x 1 = A 1 sin( ω 1 t+ ϕ 1 )+ A 2 sin( ω 2 t+ ϕ 2 ) x 2 = A 1 sin( ω 1 t+ ϕ 1 )− A 2 sin( ω 2 t+ ϕ 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamiEam aaBaaaleaacaaIXaaabeaakiabg2da9iaadgeadaWgaaWcbaGaaGym aaqabaGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3naaBaaale aacaaIXaaabeaakiaadshacqGHRaWkcqaHvpGzdaWgaaWcbaGaaGym aaqabaaakiaawIcacaGLPaaacaaMc8UaaGPaVlabgUcaRiaadgeada WgaaWcbaGaaGOmaaqabaGcciGGZbGaaiyAaiaac6gadaqadaqaaiab eM8a3naaBaaaleaacaaIYaaabeaakiaadshacqGHRaWkcqaHvpGzda WgaaWcbaGaaGOmaaqabaaakiaawIcacaGLPaaaaeaacaWG4bWaaSba aSqaaiaaikdaaeqaaOGaeyypa0JaamyqamaaBaaaleaacaaIXaaabe aakiGacohacaGGPbGaaiOBamaabmaabaGaeqyYdC3aaSbaaSqaaiaa igdaaeqaaOGaamiDaiabgUcaRiabew9aMnaaBaaaleaacaaIXaaabe aaaOGaayjkaiaawMcaaiaaykW7caaMc8UaeyOeI0IaamyqamaaBaaa leaacaaIYaaabeaakiGacohacaGGPbGaaiOBamaabmaabaGaeqyYdC 3aaSbaaSqaaiaaikdaaeqaaOGaamiDaiabgUcaRiabew9aMnaaBaaa leaacaaIYaaabeaaaOGaayjkaiaawMcaaaaaaa@79D2@

with

A 1 = 1 2 ( x 1 o + x 2 o ) A 2 = 1 2 ( x 1 o − x 2 o ) ϕ 1 = π 2 ϕ 2 = π 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyqam aaBaaaleaacaaIXaaabeaakiabg2da9maalaaabaGaaGymaaqaaiaa ikdaaaWaaeWaaeaadaWfGaqaaiaadIhadaWgaaWcbaGaaGymaaqaba aabeqaaiaad+gaaaGccqGHRaWkdaWfGaqaaiaadIhadaWgaaWcbaGa aGOmaaqabaaabeqaaiaad+gaaaaakiaawIcacaGLPaaacaaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaamyqamaaBaaaleaacaaIYaaabeaakiabg2da9maalaaa baGaaGymaaqaaiaaikdaaaWaaeWaaeaadaWfGaqaaiaadIhadaWgaa WcbaGaaGymaaqabaaabeqaaiaad+gaaaGccqGHsisldaWfGaqaaiaa dIhadaWgaaWcbaGaaGOmaaqabaaabeqaaiaad+gaaaaakiaawIcaca GLPaaaaeaacqaHvpGzdaWgaaWcbaGaaGymaaqabaGccqGH9aqpdaWc aaqaaiabec8aWbqaaiaaikdaaaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabew9aMnaa BaaaleaacaaIYaaabeaakiabg2da9maalaaabaGaeqiWdahabaGaaG Omaaaaaaaa@BC67@

In general, the vibration response will look complicated, and is not harmonic. The animation above shows a typical example (if you are using the pdf version of these notes the animation will not work)

However, if we choose the special initial conditions:

x 1 o = X 0 x 2 o = X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaxacabaGaam iEamaaBaaaleaacaaIXaaabeaaaeqabaGaam4Baaaakiabg2da9iaa dIfadaWgaaWcbaGaaGimaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8+aaCbiaeaacaWG4bWaaSbaaSqaaiaaikdaaeqaaaqabeaaca WGVbaaaOGaeyypa0JaamiwamaaBaaaleaacaaIWaaabeaaaaa@@

then the response is simply

x 1 = X 0 sin( ω 1 t+ ϕ 1 ) x 2 = X 0 sin( ω 1 t+ ϕ 1 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamiEam aaBaaaleaacaaIXaaabeaakiabg2da9iaadIfadaWgaaWcbaGaaGim aaqabaGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3naaBaaale aacaaIXaaabeaakiaadshacqGHRaWkcqaHvpGzdaWgaaWcbaGaaGym aaqabaaakiaawIcacaGLPaaacaaMc8UaaGPaVdqaaiaadIhadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpcaWGybWaaSbaaSqaaiaaicdaaeqa aOGaci4CaiaacMgacaGGUbWaaeWaaeaacqaHjpWDdaWgaaWcbaGaaG ymaaqabaGccaWG0bGaey4kaSIaeqy1dy2aaSbaaSqaaiaaigdaaeqa aaGccaGLOaGaayzkaaGaaGPaVlaaykW7aaaa@5D55@

i.e., both masses vibrate harmonically, at the first natural frequency, as shown in the animation to the right.

Similarly, if we choose

x 1 o = X 0 x 2 o =− X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaxacabaGaam iEamaaBaaaleaacaaIXaaabeaaaeqabaGaam4Baaaakiabg2da9iaa dIfadaWgaaWcbaGaaGimaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8+aaCbiaeaacaWG4bWaaSbaaSqaaiaaikdaaeqaaaqabeaaca WGVbaaaOGaeyypa0JaeyOeI0IaamiwamaaBaaaleaacaaIWaaabeaa aaa@@

then

x 1 = X 0 sin( ω 2 t+ ϕ 2 ) x 2 =− X 0 sin( ω 2 t+ ϕ 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamiEam aaBaaaleaacaaIXaaabeaakiabg2da9iaadIfadaWgaaWcbaGaaGim aaqabaGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3naaBaaale aacaaIYaaabeaakiaadshacqGHRaWkcqaHvpGzdaWgaaWcbaGaaGOm aaqabaaakiaawIcacaGLPaaacaaMc8UaaGPaVdqaaiaadIhadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpcqGHsislcaWGybWaaSbaaSqaaiaa icdaaeqaaOGaci4CaiaacMgacaGGUbWaaeWaaeaacqaHjpWDdaWgaa WcbaGaaGOmaaqabaGccaWG0bGaey4kaSIaeqy1dy2aaSbaaSqaaiaa ikdaaeqaaaGccaGLOaGaayzkaaGaaGPaVlaaykW7aaaa@5E46@

i.e., the system vibrates harmonically, at the second natural frequency.

The special initial displacements of a system that cause it to vibrate harmonically are called  `mode shapes’ for the system.

If a system has several natural frequencies, there is a corresponding mode of vibration for each natural frequency.

The natural frequencies are arguably the single most important property of any mechanical system. This is because, as we shall see, the natural frequencies coincide (almost) with the system’s resonant frequencies. That is to say, if you apply a time varying force to the system, and choose the frequency of the force to be equal to one of the natural frequencies, you will observe very large amplitude vibrations.

When designing a structure or component, you generally want to control its natural vibration frequencies very carefully. For example, if you wish to stop a system from vibrating, you need to make sure that all its natural frequencies are much greater than the expected frequency of any forces that are likely to act on the structure. If you are designing a vibration isolation platform, you generally want to make its natural frequency much lower than the vibration frequency of the floor that it will stand on. Design codes usually specify allowable ranges for natural frequencies of structures and components.

Once a prototype has been built, it is usual to measure the natural frequencies and mode shapes for a system. This is done by attaching a number of accelerometers to the system, and then hitting it with a hammer (this is usually a regular rubber tipped hammer, which might be instrumented to measure the impulse exerted by the hammer during the impact). By trial and error, one can find a spot to hit the device so as to excite each mode of vibration independent of any other. You can tell when you have found such a spot, because the whole system vibrates harmonically. The natural frequency and mode shape of each vibration mode is then determined from the accelerometer readings.

Impulse hammer tests can even be used on big structures like bridges or buildings – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ but you need a big hammer. In a recent test on a new cable stayed bridge in France, the bridge was excited by first attaching a barge to the center span with a high strength cable; then the cable was tightened to raise the barge part way out of the water; then, finally, the cable was released rapidly to set the bridge vibrating.

5.2.4 Calculating the number of degrees of freedom (and natural frequencies) of a system

When you analyze the behavior a system, it is helpful to know ahead of time how many vibration frequencies you will need to calculate. There are various ways to do this. Here are some rules that you can apply:

The number of degrees of freedom is equal to the number of independent coordinates required to describe the motion. This is only helpful if you can see by inspection how to describe your system. For the spring-mass system in the preceding section, we know that the mass can only move in one direction, and so specifying the length of the spring s will completely determine the motion of the system. The system therefore has one degree of freedom, and one vibration frequency. Section 5.6 provides several more examples where it is fairly obvious that the system has one degree of freedom.

For a 2D system, the number of degrees of freedom can be calculated from the equation

n=3r+2p− N c MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6gacqGH9a qpcaaIZaGaamOCaiabgUcaRiaaikdacaWGWbGaeyOeI0IaamOtamaa BaaaleaacaWGJbaabeaaaaa@3F4E@

where:

r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadkhaaaa@@ is the number of rigid bodies in the system

p is the number of particles in the system

N c MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6eadaWgaa WcbaGaam4yaaqabaaaaa@@ is the number of constraints (or, if you prefer, independent reaction forces) in the system.

To be able to apply this formula you need to know how many constraints appear in the problem. Constraints are imposed by things like rigid links, or contacts with rigid walls, which force the system to move in a particular way. The numbers of constraints associated with various types of 2D connections are listed in the table below. Notice that the number of constraints is always equal to the number of reaction forces you need to draw on an FBD to represent the joint

For a 3D system, the number of degrees of freedom can be calculated from the equation

n=6r+3p− N c MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6gacqGH9a qpcaaI2aGaamOCaiabgUcaRiaaiodacaWGWbGaeyOeI0IaamOtamaa BaaaleaacaWGJbaabeaaaaa@3F52@

where the symbols have the same meaning as for a 2D system. A table of various constraints for 3D problems is given below.

5.2.4 Calculating natural frequencies for 1DOF conservative systems

In light of the discussion in the preceding section, we clearly need some way to calculate natural frequencies for mechanical systems. We do not have time in this course to discuss more than the very simplest mechanical systems. We will therefore show you some tricks for calculating natural frequencies of 1DOF, conservative, systems. It is best to do this by means of examples.

Example 1: The spring-mass system revisited

Calculate the natural frequency of vibration for the system shown in the figure. Assume that the contact between the block and wedge is frictionless. The spring has stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaaGimaaqabaaaaa@37F1@

Our first objective is to get an equation of motion for s. We could do this by drawing a FBD, writing down Newton’s law, and looking at its components. However, for 1DOF systems it turns out that we can derive the EOM very quickly using the kinetic and potential energy of the system.

The potential energy and kinetic energy can be written down as:

V= 1 2 k ( s− L 0 ) 2 −mgssinαT= 1 2 m ( ds dt ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadUgadaqadaqaaiaadoha cqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaayzkaa WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamyBaiaadEgacaWGZbGa aGPaVlaaykW7ciGGZbGaaiyAaiaac6gacqaHXoqycaaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8Uaamivaiabg2da9maalaaabaGaaGymaaqaai aaikdaaaGaamyBamaabmaabaWaaSaaaeaacaWGKbGaam4Caaqaaiaa dsgacaWG0baaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaa a@6AAF@

(The second term in V is the gravitational potential energy – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ it is negative because the height of the mass decreases with increasing s). Now, note that since our system is conservative

T+V=constant ⇒ d dt ( T+V )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamivai abgUcaRiaadAfacqGH9aqpcaqGJbGaae4Baiaab6gacaqGZbGaaeiD aiaabggacaqGUbGaaeiDaaqaaiabgkDiEpaalaaabaGaamizaaqaai aadsgacaWG0baaamaabmaabaGaamivaiabgUcaRiaadAfaaiaawIca caGLPaaacqGH9aqpcaaIWaaaaaa@4C75@

Differentiate our expressions for T and V (use the chain rule) to see that

m ds dt d 2 s d t 2 +k(s− L 0 ) ds dt −mg ds dt sinα=0 ⇒ m k d 2 s d t 2 +s= L 0 + mg k sinα MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyBam aalaaabaGaamizaiaadohaaeaacaWGKbGaamiDaaaadaWcaaqaaiaa dsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGaamizaiaadshada ahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaam4AaiaacIcacaWGZbGa eyOeI0IaamitamaaBaaaleaacaaIWaaabeaakiaacMcadaWcaaqaai aadsgacaWGZbaabaGaamizaiaadshaaaGaeyOeI0IaamyBaiaadEga daWcaaqaaiaadsgacaWGZbaabaGaamizaiaadshaaaGaci4CaiaacM gacaGGUbGaeqySdeMaeyypa0JaaGimaaqaaiabgkDiEpaalaaabaGa amyBaaqaaiaadUgaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYa aaaOGaam4CaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaa kiabgUcaRiaadohacqGH9aqpcaWGmbWaaSbaaSqaaiaaicdaaeqaaO Gaey4kaSYaaSaaaeaacaWGTbGaam4zaaqaaiaadUgaaaGaci4Caiaa cMgacaGGUbGaeqySdegaaaa@6F17@

Finally, we must turn this equation of motion into one of the standard solutions to vibration equations.

Our equation looks very similar to

1 ω n 2 d 2 x d t 2 +x=C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaamiEaiabg2da 9iaadoeaaaa@@

By comparing this with our equation we see that the natural frequency of vibration is

ω n = 2k 3m (rad/s)      = 1 2π 2k 3m ( Hz ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaa ikdacaWGRbaabaGaaG4maiaad2gaaaaaleqaaOGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caGGOaGaaeOCaiaabggacaqG KbGaae4laiaabohacaqGPaaabaGaaeiiaiaabccacaqGGaGaaeiiai aabccacaqG9aWaaSaaaeaacaqGXaaabaGaaeOmaiabec8aWbaadaGc aaqaamaalaaabaGaaGOmaiaadUgaaeaacaaIZaGaamyBaaaaaSqaba GccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVpaabmaabaGaaeisaiaabQhaaiaawIcacaGLPaaaaaaa@8F4A@

Summary of procedure for calculating natural frequencies:

(1)  Describe the motion of the system, using a single scalar variable (In the example, we chose to describe motion using the distance s);

(2) Write down the potential energy V and kinetic energy T of the system in terms of the scalar variable;

(3) Use d dt ( T+V )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaaqaaiaadsgacaWG0baaamaabmaabaGaamivaiabgUcaRiaadAfa aiaawIcacaGLPaaacqGH9aqpcaaIWaaaaa@3EF3@ to get an equation of motion for your scalar variable;

(4) Arrange the equation of motion in standard form;

(5) Read off the natural frequency by comparing your equation to the standard form.

Example 2: A nonlinear system.

We will illustrate the procedure with a second example, which will demonstrate another useful trick.

Find the natural frequency of vibration for a pendulum, shown in the figure.

We will idealize the mass as a particle, to keep things simple.

We will follow the steps outlined earlier:

(1) We describe the motion using the angle θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@

(2) We write down T and V:

V=−mgLcosθ T= 1 2 m ( L dθ dt ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamOvai abg2da9iabgkHiTiaad2gacaWGNbGaamitaiGacogacaGGVbGaai4C aiabeI7aXbqaaiaadsfacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYa aaaiaad2gadaqadaqaaiaadYeadaWcaaqaaiaadsgacqaH4oqCaeaa caWGKbGaamiDaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaa aaaaa@4C72@

(if you don’t see the formula for the kinetic energy, you can write down the position vector of the mass as r=Lsinθi−Lcosθj MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhacqGH9a qpcaWGmbGaci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiabgkHiTiaa dYeaciGGJbGaai4BaiaacohacqaH4oqCcaWHQbaaaa@45C6@ , differentiate to find the velocity: v=Lcosθ dθ dt i+Lsinθ dθ dt j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAhacqGH9a qpcaWGmbGaci4yaiaac+gacaGGZbGaeqiUde3aaSaaaeaacaWGKbGa eqiUdehabaGaamizaiaadshaaaGaaCyAaiabgUcaRiaadYeaciGGZb GaaiyAaiaac6gacqaH4oqCdaWcaaqaaiaadsgacqaH4oqCaeaacaWG KbGaamiDaaaacaWHQbaaaa@4EE1@ , and then compute T=m(v⋅v)/2=m L 2 ( dθ dt ) 2 ( sin 2 θ+ cos 2 θ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpcaWGTbGaaiikaiaahAhacqGHflY1caWH2bGaaiykaiaac+cacaaI YaGaeyypa0JaamyBaiaadYeadaahaaWcbeqaaiaaikdaaaGcdaqada qaamaalaaabaGaamizaiabeI7aXbqaaiaadsgacaWG0baaaaGaayjk aiaawMcaamaaCaaaleqabaGaaGOmaaaakiaacIcaciGGZbGaaiyAai aac6gadaahaaWcbeqaaiaaikdaaaGccqaH4oqCcqGHRaWkciGGJbGa ai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccqaH4oqCcaGGPaaaaa@581C@ and use a trig identity. You can also use the circular motion formulas, if you prefer).

(3) Differentiate with respect to time:

m L 2 d 2 θ d t 2 dθ dt +mgLsinθ dθ dt =0 ⇒ L g d 2 θ d t 2 +sinθ=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyBai aadYeadaahaaWcbeqaaiaaikdaaaGcdaWcaaqaaiaadsgadaahaaWc beqaaiaaikdaaaGccqaH4oqCaeaacaWGKbGaamiDamaaCaaaleqaba GaaGOmaaaaaaGcdaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiD aaaacqGHRaWkcaWGTbGaam4zaiaadYeaciGGZbGaaiyAaiaac6gacq aH4oqCdaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaacqGH 9aqpcaaIWaaabaGaeyO0H49aaSaaaeaacaWGmbaabaGaam4zaaaada WcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccqaH4oqCaeaacaWG KbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkciGGZbGaai yAaiaac6gacqaH4oqCcqGH9aqpcaaIWaaaaaa@643F@

(4) Arrange the EOM into standard form. Houston, we have a problem. There is no way this equation can be arranged into standard form. This is because the equation is nonlinear ( sinθ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacohacaGGPb GaaiOBaiabeI7aXbaa@3AC7@ is a nonlinear function of θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@ ). There is, however, a way to deal with this problem. We will show what needs to be done, summarizing the general steps as we go along.

(i) Find the static equilibrium configuration(s) for the system.

If the system is in static equilibrium, it does not move. We can find values of θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@ for which the system is in static equilibrium by setting all time derivatives of θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@ in the equation of motion to zero, and then solving the equation. Here,

sin θ o =0⇒ θ 0 =0,π,2π... MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacohacaGGPb GaaiOBaiabeI7aXnaaBaaaleaacaWGVbaabeaakiabg2da9iaaicda caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl abgkDiElaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeI7aXnaaBaaaleaaca aIWaaabeaakiabg2da9iaaicdacaGGSaGaeqiWdaNaaiilaiaaikda cqaHapaCcaGGUaGaaiOlaiaac6caaaa@6AFC@

Here, we have used θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaaaaa@38D5@ to denote the special values of θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@ for which the system happens to be in static equilibrium. Note that θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaaaaa@38D5@ is always a constant.

(ii) Assume that the system vibrates with small amplitude about a static equilibrium configuration of interest.

To do this, we let θ= θ 0 +x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXjabg2 da9iabeI7aXnaaBaaaleaacaaIWaaabeaakiabgUcaRiaadIhaaaa@3D7A@ , where x<<1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacqGH8a apcqGH8aapcaaIXaaaaa@39F9@ .

Here, x represents a small change in angle from an equilibrium configuration.. Note that x will vary with time as the system vibrates. Instead of solving for θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@ , we will solve for x. Before going on, make sure that you are comfortable with the physical significance of both x and θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaaaaa@38D5@ .

(iii) Linearize the equation of motion, by expanding all nonlinear terms as Taylor Maclaurin series about the equilibrium configuration.

We substitute for θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXbaa@37EF@ in the equation of motion, to see that

L g d 2 x d t 2 +sin( θ 0 +x)=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam itaaqaaiaadEgaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamiEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaki abgUcaRiGacohacaGGPbGaaiOBaiaacIcacqaH4oqCdaWgaaWcbaGa aGimaaqabaGccqGHRaWkcaWG4bGaaiykaiabg2da9iaaicdaaaa@491C@

(Recall that θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaaaaa@38D5@ is constant, so its time derivatives vanish)

Now, recall the Taylor-Maclaurin series expansion of a function f(x)has the form

f( x 0 +x)=f( x 0 )+x f ′ ( x 0 )+ 1 2 x 2 f ″ ( x 0 )+... MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAgacaGGOa GaamiEamaaBaaaleaacaaIWaaabeaakiabgUcaRiaadIhacaGGPaGa eyypa0JaamOzaiaacIcacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaai ykaiabgUcaRiaadIhaceWGMbGbauaacaGGOaGaamiEamaaBaaaleaa caaIWaaabeaakiaacMcacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYa aaaiaadIhadaahaaWcbeqaaiaaikdaaaGcceWGMbGbayaacaGGOaGa amiEamaaBaaaleaacaaIWaaabeaakiaacMcacqGHRaWkcaGGUaGaai Olaiaac6caaaa@532B@

where

f ′ ( x 0 )≡ df dx | x= x 0 f ″ ( x 0 )≡ d 2 f d x 2 | x= x 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiqadAgagaqbai aacIcacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaaiykaiabggMi6oaa eiaabaWaaSaaaeaacaWGKbGaamOzaaqaaiaadsgacaWG4baaaaGaay jcSdWaaSbaaSqaaiaadIhacqGH9aqpcaWG4bWaaSbaaWqaaiaaicda aeqaaaWcbeaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlqadAgagaGbaiaacIcacaWG4bWaaSbaaSqaaiaaicda aeqaaOGaaiykaiabggMi6oaaeiaabaWaaSaaaeaacaWGKbWaaWbaaS qabeaacaaIYaaaaOGaamOzaaqaaiaadsgacaWG4bWaaWbaaSqabeaa caaIYaaaaaaaaOGaayjcSdWaaSbaaSqaaiaadIhacqGH9aqpcaWG4b WaaSbaaWqaaiaaicdaaeqaaaWcbeaaaaa@7C3A@

Apply this to the nonlinear term in our equation of motion

sin( θ 0 +x )=sin θ 0 +xcos θ 0 − 1 2 x 2 sin θ 0 +... MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacohacaGGPb GaaiOBamaabmaabaGaeqiUde3aaSbaaSqaaiaaicdaaeqaaOGaey4k aSIaamiEaaGaayjkaiaawMcaaiabg2da9iGacohacaGGPbGaaiOBai abeI7aXnaaBaaaleaacaaIWaaabeaakiabgUcaRiaadIhaciGGJbGa ai4BaiaacohacqaH4oqCdaWgaaWcbaGaaGimaaqabaGccqGHsislda WcaaqaaiaaigdaaeaacaaIYaaaaiaadIhadaahaaWcbeqaaiaaikda aaGcciGGZbGaaiyAaiaac6gacqaH4oqCdaWgaaWcbaGaaGimaaqaba GccqGHRaWkcaGGUaGaaiOlaiaac6caaaa@59D5@

Now, since x<<1, we can assume that x n <

sin( θ 0 +x )≈sin θ 0 +xcos θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacohacaGGPb GaaiOBamaabmaabaGaeqiUde3aaSbaaSqaaiaaicdaaeqaaOGaey4k aSIaamiEaaGaayjkaiaawMcaaiabgIKi7kGacohacaGGPbGaaiOBai abeI7aXnaaBaaaleaacaaIWaaabeaakiabgUcaRiaadIhaciGGJbGa ai4BaiaacohacqaH4oqCdaWgaaWcbaGaaGimaaqabaaaaa@4D9C@

Finally, we can substitute back into our equation of motion, to obtain

L g d 2 x d t 2 +cos θ 0 x=−sin θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam itaaqaaiaadEgaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamiEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaki abgUcaRiGacogacaGGVbGaai4CaiabeI7aXnaaBaaaleaacaaIWaaa beaakiaadIhacqGH9aqpcqGHsislciGGZbGaaiyAaiaac6gacqaH4o qCdaWgaaWcbaGaaGimaaqabaaaaa@4C83@

(iv) Compare the linear equation with the standard form to deduce the natural frequency.

We can do this for each equilibrium configuration.

θ 0 =0,2π,4π...⇒ L g d 2 x d t 2 +x=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaakiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7 caaIYaGaeqiWdaNaaiilaiaaykW7caaMc8UaaGinaiabec8aWjaac6 cacaGGUaGaaiOlaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlabgkDiElaaykW7caaMc8+aaSaaaeaacaWGmbaabaGaam4zaaaaca aMc8+aaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqa aiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRiaadI hacqGH9aqpcaaIWaaaaa@649F@

whence

ω n = g L (rad/sec) f n = 1 2π g L (Hz) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaa dEgaaeaacaWGmbaaaaWcbeaakiaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaacIcacaqGYbGaaeyyaiaabsgacaqGVaGaae4CaiaabwgacaqG JbGaaeykaaqaaiaadAgadaWgaaWcbaGaamOBaaqabaGccqGH9aqpda WcaaqaaiaaigdaaeaacaaIYaGaeqiWdahaamaakaaabaWaaSaaaeaa caWGNbaabaGaamitaaaaaSqabaGccaaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaacIcacaqGibGaaeOEaiaabMcaaaaa@852D@

Note that all these values of θ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaaaaa@38D5@ really represent the same configuration: the mass is hanging below the pivot. We have rediscovered the well-known expression for the natural frequency of a freely swinging pendulum.

Next, try the remaining static equilibrium configuration

θ 0 =π,3π,5π...⇒ L g d 2 x d t 2 −x=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaakiabg2da9iabec8aWjaacYcacaaMc8UaaGPa VlaaiodacqaHapaCcaGGSaGaaGPaVlaaykW7caaI1aGaeqiWdaNaai Olaiaac6cacaGGUaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaeyO0H4TaaGPaVlaaykW7daWcaaqaaiaadYeaaeaacaWGNbaaai aaykW7daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baa baGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyOeI0Iaam iEaiabg2da9iaaicdaaaa@65AF@

If we look up this equation in our list of standard solutions, we find it does not have a harmonic solution. Instead, the solution is

x(t)= 1 2 ( x 0 + v 0 α ) e −αt + 1 2 ( x 0 − v 0 α ) e αt α= g L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9maalaaabaGaaGymaaqaaiaaikda aaWaaeWaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaS aaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaaGcbaGaeqySdegaaaGa ayjkaiaawMcaaiaadwgadaahaaWcbeqaaiabgkHiTiabeg7aHjaads haaaGccqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYaaaamaabmaabaGa amiEamaaBaaaleaacaaIWaaabeaakiabgkHiTmaalaaabaGaamODam aaBaaaleaacaaIWaaabeaaaOqaaiabeg7aHbaaaiaawIcacaGLPaaa caWGLbWaaWbaaSqabeaacqaHXoqycaWG0baaaaGcbaGaeqySdeMaey ypa0ZaaOaaaeaadaWcaaqaaiaadEgaaeaacaWGmbaaaaWcbeaaaaaa @5B60@

where x 0 =x(t=0) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGimaaqabaGccqGH9aqpcaWG4bGaaiikaiaadshacqGH9aqp caaIWaGaaiykaaaa@3E3B@ and v 0 = dx dt | t=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaa WcbaGaaGimaaqabaGccqGH9aqpdaabcaqaamaalaaabaGaamizaiaa dIhaaeaacaWGKbGaamiDaaaaaiaawIa7amaaBaaaleaacaWG0bGaey ypa0JaaGimaaqabaaaaa@417D@

Thus, except for some rather special initial conditions, x increases without bound as time increases. This is a characteristic of an unstable mechanical system.

If we visualize the system with θ 0 =π MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIWaaabeaakiabg2da9iabec8aWbaa@3BA2@ , we can see what is happening. This equilibrium configuration has the pendulum upside down!

No wonder the equation is predicting an instability…

Here is a question to think about. Our solution predicts that both x and dx/dtbecome infinitely large. We know that a real pendulum would never rotate with infinite angular velocity. What has gone wrong?

Example 3: We will look at one more nonlinear system, to make sure that you are comfortable with this procedure. Calculate the resonant frequency of small oscillations about the equilibrium configuration θ=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXjabg2 da9iaaicdaaaa@39AF@  for the system shown. The spring has stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaaGimaaqabaaaaa@37F1@ .

We follow the same procedure as before.

The potential and kinetic energies of the system are

V= 1 2 k ( Lsinθ ) 2 + 1 2 mgLcosθ T= 1 2 m L 2 3 ( dθ dt ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamOvai abg2da9maalaaabaGaaGymaaqaaiaaikdaaaGaam4AamaabmaabaGa amitaiGacohacaGGPbGaaiOBaiabeI7aXbGaayjkaiaawMcaamaaCa aaleqabaGaaGOmaaaakiabgUcaRmaalaaabaGaaGymaaqaaiaaikda aaGaamyBaiaadEgacaWGmbGaci4yaiaac+gacaGGZbGaeqiUdehaba Gaamivaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaSaaaeaa caWGTbGaamitamaaCaaaleqabaGaaGOmaaaaaOqaaiaaiodaaaWaae WaaeaadaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaaaiaa wIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaaa@5A00@

Hence

d dt (T+V)= m L 2 3 d 2 θ d t 2 dθ dt +k L 2 sinθcosθ dθ dt − 1 2 mgLsinθ dθ dt =0 ⇒ m L 2 3 d 2 θ d t 2 +( k L 2 cosθ− mgL 2 )sinθ=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaaSaaae aacaWGKbaabaGaamizaiaadshaaaGaaiikaiaadsfacqGHRaWkcaWG wbGaaiykaiabg2da9maalaaabaGaamyBaiaadYeadaahaaWcbeqaai aaikdaaaaakeaacaaIZaaaamaalaaabaGaamizamaaCaaaleqabaGa aGOmaaaakiabeI7aXbqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYa aaaaaakmaalaaabaGaamizaiabeI7aXbqaaiaadsgacaWG0baaaiab gUcaRiaadUgacaWGmbWaaWbaaSqabeaacaaIYaaaaOGaci4CaiaacM gacaGGUbGaeqiUdeNaci4yaiaac+gacaGGZbGaeqiUde3aaSaaaeaa caWGKbGaeqiUdehabaGaamizaiaadshaaaGaeyOeI0YaaSaaaeaaca aIXaaabaGaaGOmaaaacaWGTbGaam4zaiaadYeaciGGZbGaaiyAaiaa c6gacqaH4oqCdaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaa aacqGH9aqpcaaIWaaabaGaeyO0H49aaSaaaeaacaWGTbGaamitamaa CaaaleqabaGaaGOmaaaaaOqaaiaaiodaaaWaaSaaaeaacaWGKbWaaW baaSqabeaacaaIYaaaaOGaeqiUdehabaGaamizaiaadshadaahaaWc beqaaiaaikdaaaaaaOGaey4kaSYaaeWaaeaacaWGRbGaamitamaaCa aaleqabaGaaGOmaaaakiGacogacaGGVbGaai4CaiabeI7aXjabgkHi TmaalaaabaGaamyBaiaadEgacaWGmbaabaGaaGOmaaaaaiaawIcaca GLPaaaciGGZbGaaiyAaiaac6gacqaH4oqCcqGH9aqpcaaIWaaaaaa@8E90@

Once again, we have found a nonlinear equation of motion. This time we know what to do. We are told to find natural frequency of oscillation about θ=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXjabg2 da9iaaicdaaaa@39AF@ , so we don’t need to solve for the equilibrium configurations this time. We set θ=0+x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXjabg2 da9iaaicdacqGHRaWkcaWG4baaaa@3B8E@ , with x<<1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacqGH8a apcqGH8aapcaaIXaaaaa@39F9@ and substitute back into the equation of motion:

m L 2 3 d 2 x d t 2 +( k L 2 cosx− mgL 2 )sinx=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam yBaiaadYeadaahaaWcbeqaaiaaikdaaaaakeaacaaIZaaaamaalaaa baGaamizamaaCaaaleqabaGaaGOmaaaakiaadIhaaeaacaWGKbGaam iDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaqadaqaaiaadUga caWGmbWaaWbaaSqabeaacaaIYaaaaOGaci4yaiaac+gacaGGZbGaam iEaiabgkHiTmaalaaabaGaamyBaiaadEgacaWGmbaabaGaaGOmaaaa aiaawIcacaGLPaaaciGGZbGaaiyAaiaac6gacaWG4bGaeyypa0JaaG imaaaa@@

Now, expand all the nonlinear terms (it is OK to do them one at a time and then multiply everything out. You can always throw away all powers of x greater than one as you do so)

cosx≈1sinx≈x ⇒ m L 2 3 d 2 x d t 2 +( k L 2 − mgL 2 )x=0 ⇒ m 3k( 1−mg/2kL ) d 2 x d t 2 +x=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaci4yai aac+gacaGGZbGaamiEaiabgIKi7kaaigdacaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlGacohacaGGPbGaaiOBaiaadIhacqGHijYU caWG4baabaGaeyO0H49aaSaaaeaacaWGTbGaamitamaaCaaaleqaba GaaGOmaaaaaOqaaiaaiodaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaa caaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYa aaaaaakiabgUcaRmaabmaabaGaam4AaiaadYeadaahaaWcbeqaaiaa ikdaaaGccqGHsisldaWcaaqaaiaad2gacaWGNbGaamitaaqaaiaaik daaaaacaGLOaGaayzkaaGaamiEaiabg2da9iaaicdaaeaacqGHshI3 daWcaaqaaiaad2gaaeaacaaIZaGaam4AamaabmaabaGaaGymaiabgk HiTiaad2gacaWGNbGaai4laiaaikdacaWGRbGaamitaaGaayjkaiaa wMcaaaaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4b aabaGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIa amiEaiabg2da9iaaicdaaaaa@@

We now have an equation in standard form, and can read off the natural frequency

ω n = 3k m ( 1− mg 2kL ) (rad/sec) f n = 1 2π 3k m ( 1− mg 2kL ) (Hz) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaa iodacaWGRbaabaGaamyBaaaadaqadaqaaiaaigdacqGHsisldaWcaa qaaiaad2gacaWGNbaabaGaaGOmaiaadUgacaWGmbaaaaGaayjkaiaa wMcaaaWcbeaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caqGOaGaaeOCaiaabggacaqGKbGaae4laiaabohaca qGLbGaae4yaiaabMcaaeaacaWGMbWaaSbaaSqaaiaad6gaaeqaaOGa eyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaiabec8aWbaadaGcaaqaam aalaaabaGaaG4maiaadUgaaeaacaWGTbaaamaabmaabaGaaGymaiab gkHiTmaalaaabaGaamyBaiaadEgaaeaacaaIYaGaam4AaiaadYeaaa aacaGLOaGaayzkaaaaleqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caGGOaGaaeisaiaabQhacaqGPaaaaa a@852F@

Question: what happens for mg>2kL MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gacaWGNb GaeyOpa4JaaGOmaiaadUgacaWGmbaaaa@3B9C@ ?

Example 3: A system with a rigid body (the KE of a rigid body will be defined in the next section of the course – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieqajugybabaaa aaaaaapeGaa83eGaaa@@ just live with it for now!).

Calculate the natural frequency of vibration for the system shown in the figure. Assume that the cylinder rolls without slip on the wedge. The spring has stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaaGimaaqabaaaaa@37F1@

Our first objective is to get an equation of motion for s. We do this by writing down the potential and kinetic energies of the system in terms of s.

The potential energy is easy:

V= 1 2 k ( s− L 0 ) 2 −mgssinα MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadUgadaqadaqaaiaadoha cqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaayzkaa WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamyBaiaadEgacaWGZbGa aGPaVlaaykW7ciGGZbGaaiyAaiaac6gacqaHXoqyaaa@4C03@

The first term represents the energy in the spring, while second term accounts for the gravitational potential energy.

The kinetic energy is slightly more tricky. Note that the magnitude of the angular velocity of the disk is related to the magnitude of its translational velocity by

Rω= ds dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadkfacqaHjp WDcqGH9aqpdaWcaaqaaiaadsgacaWGZbaabaGaamizaiaadshaaaaa aa@3DB6@

Thus, the combined rotational and translational kinetic energy follows as

T= 1 2 m R 2 2 ω 2 + 1 2 m ( ds dt ) 2 = 1 2 3m 2 ( ds dt ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamivai abg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaSaaaeaacaWGTbGa amOuamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaeqyYdC3aaW baaSqabeaacaaIYaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOm aaaacaWGTbWaaeWaaeaadaWcaaqaaiaadsgacaWGZbaabaGaamizai aadshaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGcbaGa aGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaaig daaeaacaaIYaaaaiaaykW7caaMc8+aaSaaaeaacaaIZaGaamyBaaqa aiaaikdaaaGaaGPaVpaabmaabaWaaSaaaeaacaWGKbGaam4Caaqaai aadsgacaWG0baaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaa aaaa@613D@

Now, note that since our system is conservative

T+V=constant ⇒ d dt ( T+V )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamivai abgUcaRiaadAfacqGH9aqpcaqGJbGaae4Baiaab6gacaqGZbGaaeiD aiaabggacaqGUbGaaeiDaaqaaiabgkDiEpaalaaabaGaamizaaqaai aadsgacaWG0baaamaabmaabaGaamivaiabgUcaRiaadAfaaiaawIca caGLPaaacqGH9aqpcaaIWaaaaaa@4C75@

Differentiate our expressions for T and V to see that

3m 2 d 2 s d t 2 ds dt +k(s− L 0 ) ds dt −mg ds dt sinα=0 ⇒ 3m 2k d 2 s d t 2 +s= L 0 + mg k sinα MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaaSaaae aacaaIZaGaamyBaaqaaiaaikdaaaWaaSaaaeaacaWGKbWaaWbaaSqa beaacaaIYaaaaOGaam4CaaqaaiaadsgacaWG0bWaaWbaaSqabeaaca aIYaaaaaaakmaalaaabaGaamizaiaadohaaeaacaWGKbGaamiDaaaa cqGHRaWkcaWGRbGaaiikaiaadohacqGHsislcaWGmbWaaSbaaSqaai aaicdaaeqaaOGaaiykamaalaaabaGaamizaiaadohaaeaacaWGKbGa amiDaaaacqGHsislcaWGTbGaam4zamaalaaabaGaamizaiaadohaae aacaWGKbGaamiDaaaaciGGZbGaaiyAaiaac6gacqaHXoqycqGH9aqp caaIWaaabaGaeyO0H49aaSaaaeaacaaIZaGaamyBaaqaaiaaikdaca WGRbaaamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadoha aeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkca WGZbGaeyypa0JaamitamaaBaaaleaacaaIWaaabeaakiabgUcaRmaa laaabaGaamyBaiaadEgaaeaacaWGRbaaaiGacohacaGGPbGaaiOBai abeg7aHbaaaa@@

The last equation is almost in one of the standard forms given on the handout, except that the right hand side is not zero. There is a trick to dealing with this problem – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ simply subtract the constant right hand side from s, and call the result x. (This only works if the right hand side is a constant, of course). Thus let

x=s− L 0 − mg k sinα MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacqGH9a qpcaWGZbGaeyOeI0IaamitamaaBaaaleaacaaIWaaabeaakiabgkHi TmaalaaabaGaamyBaiaadEgaaeaacaWGRbaaaiGacohacaGGPbGaai OBaiabeg7aHbaa@@

and substitute into the equation of motion:

3m 2k d 2 x d t 2 +x+ L 0 + mg k sinα= L 0 + mg k sinα ⇒ 3m 2k d 2 x d t 2 +x=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaaSaaae aacaaIZaGaamyBaaqaaiaaikdacaWGRbaaamaalaaabaGaamizamaa CaaaleqabaGaaGOmaaaakiaadIhaaeaacaWGKbGaamiDamaaCaaale qabaGaaGOmaaaaaaGccqGHRaWkcaWG4bGaey4kaSIaamitamaaBaaa leaacaaIWaaabeaakiabgUcaRmaalaaabaGaamyBaiaadEgaaeaaca WGRbaaaiGacohacaGGPbGaaiOBaiabeg7aHjabg2da9iaadYeadaWg aaWcbaGaaGimaaqabaGccqGHRaWkdaWcaaqaaiaad2gacaWGNbaaba Gaam4AaaaaciGGZbGaaiyAaiaac6gacqaHXoqyaeaacqGHshI3daWc aaqaaiaaiodacaWGTbaabaGaaGOmaiaadUgaaaWaaSaaaeaacaWGKb WaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWaaWba aSqabeaacaaIYaaaaaaakiabgUcaRiaadIhacqGH9aqpcaaIWaaaaa a@@

This is now in the form

1 ω n 2 d 2 x d t 2 +x=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaamiEaiabg2da 9iaaicdaaaa@@

and by comparing this with our equation we see that the natural frequency of vibration is

ω n = 2k 3m (rad/s)      = 1 2π 2k 3m ( Hz ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaa ikdacaWGRbaabaGaaG4maiaad2gaaaaaleqaaOGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caGGOaGaaeOCaiaabggacaqG KbGaae4laiaabohacaqGPaaabaGaaeiiaiaabccacaqGGaGaaeiiai aabccacaqG9aWaaSaaaeaacaqGXaaabaGaaeOmaiabec8aWbaadaGc aaqaamaalaaabaGaaGOmaiaadUgaaeaacaaIZaGaamyBaaaaaSqaba GccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVpaabmaabaGaaeisaiaabQhaaiaawIcacaGLPaaaaaaa@8F4A@

5.3 Free vibration of a damped, single degree of freedom, linear spring mass system.

We analyzed vibration of several conservative systems in the preceding section. In each case, we found that if the system was set in motion, it continued to move indefinitely. This is counter to our everyday experience. Usually, if you start something vibrating, it will vibrate with a progressively decreasing amplitude and eventually stop moving.

The reason our simple models predict the wrong behavior is that we neglected energy dissipation. In this section, we explore the influence of energy dissipation on free vibration of a spring-mass system. As before, although we model a very simple system, the behavior we predict turns out to be representative of a wide range of real engineering systems.

5.3.1 Vibration of a damped spring-mass system

The spring mass dashpot system shown is released with velocity u 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwhadaWgaa WcbaGaaGimaaqabaaaaa@@ from position s 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohadaWgaa WcbaGaaGimaaqabaaaaa@@ at time t=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshacqGH9a qpcaaIWaaaaa@38F2@ . Find s(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohacaGGOa GaamiDaiaacMcaaaa@@ .

Once again, we follow the standard approach to solving problems like this

(i) Get a differential equation for s using F=ma

(ii) Solve the differential equation.

You may have forgotten what a dashpot (or damper) does. Suppose we apply a force F to a dashpot, as shown in the figure. We would observe that the dashpot stretched at a rate proportional to the force

F=c dL dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeacqGH9a qpcaWGJbWaaSaaaeaacaWGKbGaamitaaqaaiaadsgacaWG0baaaaaa @3C9E@

One can buy dampers (the shock absorbers in your car contain dampers): a damper generally consists of a plunger inside an oil filled cylinder, which dissipates energy by churning the oil. Thus, it is possible to make a spring-mass-damper system that looks very much like the one in the picture. More generally, however, the spring mass system is used to represent a complex mechanical system. In this case, the damper represents the combined effects of all the various mechanisms for dissipating energy in the system, including friction, air resistance, deformation losses, and so on.

To proceed, we draw a free body diagram, showing the forces exerted by the spring and damper on the mass.

Newton’s law then states that

k(s− L 0 )+c ds dt =ma=m d 2 s d t 2 ⇒ m k d 2 s d t 2 + c k ds dt +s= L 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaam4Aai aacIcacaWGZbGaeyOeI0IaamitamaaBaaaleaacaaIWaaabeaakiaa cMcacqGHRaWkcaWGJbWaaSaaaeaacaWGKbGaam4Caaqaaiaadsgaca WG0baaaiabg2da9iaad2gacaWGHbGaeyypa0JaamyBamaalaaabaGa amizamaaCaaaleqabaGaaGOmaaaakiaadohaaeaacaWGKbGaamiDam aaCaaaleqabaGaaGOmaaaaaaaakeaacqGHshI3daWcaaqaaiaad2ga aeaacaWGRbaaamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaaki aadohaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGH RaWkdaWcaaqaaiaadogaaeaacaWGRbaaamaalaaabaGaamizaiaado haaeaacaWGKbGaamiDaaaacqGHRaWkcaaMc8Uaam4Caiabg2da9iaa dYeadaWgaaWcbaGaaGimaaqabaaaaaa@@

This is our equation of motion for s.

Now, we check our list of solutions to differential equations, and see that we have a solution to:

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaaaa@4E43@

We can get our equation into this form by setting

s=x ω n = k m ς= c 2 km C= L 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohacqGH9a qpcaWG4bGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlabeM8a3naaBaaaleaacaWGUbaabeaaki abg2da9maakaaabaWaaSaaaeaacaWGRbaabaGaamyBaaaaaSqabaGc caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua eqOWdyLaeyypa0ZaaSaaaeaacaWGJbaabaGaaGOmamaakaaabaGaam 4Aaiaad2gaaSqabaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadoeacqGH9aqpcaWGmb WaaSbaaSqaaiaaicdaaeqaaaaa@80BE@

As before, ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ is known as the natural frequency of the system. We have discovered a new parameter, ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ , which is called the damping coefficient. It plays a very important role, as we shall see below.

Now, we can write down the solution for x:

Overdamped System ς>1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg6 da+iaaigdaaaa@39A1@

x(t)=C+exp(−ς ω n t){ v 0 +(ς ω n + ω d )( x 0 −C) 2 ω d exp( ω d t)− v 0 +(ς ω n − ω d )( x 0 −C) 2 ω d exp(− ω d t) } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSIaciyzaiaacIhacaGG WbGaaiikaiabgkHiTiabek8awjabeM8a3naaBaaaleaacaWGUbaabe aakiaadshacaGGPaWaaiWaaeaadaWcaaqaaiaadAhadaWgaaWcbaGa aGimaaqabaGccqGHRaWkcaGGOaGaeqOWdyLaeqyYdC3aaSbaaSqaai aad6gaaeqaaOGaey4kaSIaeqyYdC3aaSbaaSqaaiaadsgaaeqaaOGa aiykaiaacIcacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0Iaam 4qaiaacMcaaeaacaaIYaGaeqyYdC3aaSbaaSqaaiaadsgaaeqaaaaa kiGacwgacaGG4bGaaiiCaiaacIcacqaHjpWDdaWgaaWcbaGaamizaa qabaGccaWG0bGaaiykaiabgkHiTmaalaaabaGaamODamaaBaaaleaa caaIWaaabeaakiabgUcaRiaacIcacqaHcpGvcqaHjpWDdaWgaaWcba GaamOBaaqabaGccqGHsislcqaHjpWDdaWgaaWcbaGaamizaaqabaGc caGGPaGaaiikaiaadIhadaWgaaWcbaGaaGimaaqabaGccqGHsislca WGdbGaaiykaaqaaiaaikdacqaHjpWDdaWgaaWcbaGaamizaaqabaaa aOGaciyzaiaacIhacaGGWbGaaiikaiabgkHiTiabeM8a3naaBaaale aacaWGKbaabeaakiaadshacaGGPaaacaGL7bGaayzFaaaaaa@85B9@

where ω d = ω n ς 2 −1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGKbaabeaakiabg2da9iabeM8a3naaBaaaleaacaWGUbaa beaakmaakaaabaGaeqOWdy1aaWbaaSqabeaacaaIYaaaaOGaeyOeI0 IaaGymaaWcbeaaaaa@417C@

Critically Damped System ς=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9iaaigdaaaa@399F@

x(t)=C+{ ( x 0 −C)+[ v 0 + ω n ( x 0 −C) ]t }exp(− ω n t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSYaaiWaaeaacaGGOaGa amiEamaaBaaaleaacaaIWaaabeaakiabgkHiTiaadoeacaGGPaGaey 4kaSYaamWaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIa eqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaaiikaiaadIhadaWgaaWcba GaaGimaaqabaGccqGHsislcaWGdbGaaiykaaGaay5waiaaw2faaiaa dshaaiaawUhacaGL9baaciGGLbGaaiiEaiaacchacaGGOaGaeyOeI0 IaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaamiDaiaacMcaaaa@5AFF@

Underdamped System ς<1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabgY da8iaaigdaaaa@399D@

x(t)=C+exp(−ς ω n t){ ( x 0 −C)cos ω d t+ v 0 +ς ω n ( x 0 −C) ω d sin ω d t } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSIaciyzaiaacIhacaGG WbGaaiikaiabgkHiTiabek8awjabeM8a3naaBaaaleaacaWGUbaabe aakiaadshacaGGPaWaaiWaaeaacaGGOaGaamiEamaaBaaaleaacaaI WaaabeaakiabgkHiTiaadoeacaGGPaGaci4yaiaac+gacaGGZbGaeq yYdC3aaSbaaSqaaiaadsgaaeqaaOGaamiDaiabgUcaRmaalaaabaGa amODamaaBaaaleaacaaIWaaabeaakiabgUcaRiabek8awjabeM8a3n aaBaaaleaacaWGUbaabeaakiaacIcacaWG4bWaaSbaaSqaaiaaicda aeqaaOGaeyOeI0Iaam4qaiaacMcaaeaacqaHjpWDcaWLa8+aaSbaaS qaaiaadsgaaeqaaaaakiGacohacaGGPbGaaiOBaiabeM8a3naaBaaa leaacaWGKbaabeaakiaadshaaiaawUhacaGL9baaaaa@6D57@

where ω d = ω n 1− ς 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGKbaabeaakiabg2da9iabeM8a3naaBaaaleaacaWGUbaa beaakmaakaaabaGaaGymaiabgkHiTiabek8awnaaCaaaleqabaGaaG Omaaaaaeqaaaaa@@ is known as the damped natural frequency of the system.

In all the preceding equations, x 0 , v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGimaaqabaGccaGGSaGaaGPaVlaaykW7caWG2bWaaSbaaSqa aiaaicdaaeqaaaaa@3DCD@ are the values of x and its time derivative at time t=0.

These expressions are rather too complicated to visualize what the system is doing for any given set of parameters. – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ if you have Java, Internet Explorer (or a browser plugin that allows you to run IE in another browser) you can run a Java Applet to visualize the motion. You can find instructions for installing Java, the IE plugins, and giving permission for the Applet to run here. The address for the free vibration simulator (cut and paste this into the Internet Explorer address bar) is

http://www.brown.edu/Departments/Engineering/Courses/En4/java/free.html

You can use the sliders to set the values of either m, k, and c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ (in this case the program will calculate the values of ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ and ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ for you, and display the results), or alternatively, you can set the values of ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ and ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ directly. You can also choose values for the initial conditions x 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGimaaqabaaaaa@381C@ and v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaa WcbaGaaGimaaqabaaaaa@381A@ . When you press `start,’ the applet will animate the behavior of the system, and will draw a graph of the position of the mass as a function of time. You can also choose to display the phase plane, which shows the velocity of the mass as a function of its position, if you wish. You can stop the animation at any time, change the parameters, and plot a new graph on top of the first to see what has changed. If you press `reset’, all your graphs will be cleared, and you can start again.

Try the following tests to familiarize yourself with the behavior of the system

Set the dashpot coefficient c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ to a low value, so that the damping coefficient ς<1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabgY da8iaaigdaaaa@399D@ . Make sure the graph is set to display position versus time, and press `start.’ You should see the system vibrate. The vibration looks very similar to the behavior of the conservative system we analyzed in the preceding section, except that the amplitude decays with time. Note that the system vibrates at a frequency very slightly lower than the natural frequency of the system.

Keeping the value of c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ fixed, vary the values of spring constant and mass to see what happens to the frequency of vibration and also to the rate of decay of vibration. Is the behavior consistent with the solutions given above?

Keep the values of k and m fixed, and vary c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ . You should see that, as you increase c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ , the vibration dies away more and more quickly. What happens to the frequency of oscillations as c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ is increased? Is this behavior consistent with the predictions of the theory?

Now, set the damping coefficient (not the dashpot coefficient this time) to ς=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9iaaigdaaaa@399F@ . For this value, the system no longer vibrates; instead, the mass smoothly returns to its equilibrium position x=0. If you need to design a system that returns to its equilibrium position in the shortest possible time, then it is customary to select system parameters so that ς=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9iaaigdaaaa@399F@ . A system of this kind is said to be critically damped.

Set ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ to a value greater than 1. Under these conditions, the system decays more slowly towards its equilibrium configuration.

Keeping ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ >1, experiment with the effects of changing the stiffness of the spring and the value of the mass. Can you explain what is happening mathematically, using the equations of motion and their solution?

Finally, you might like to look at the behavior of the system on its phase plane. In this course, we will not make much use of the phase plane, but it is a powerful tool for visualizing the behavior of nonlinear systems. By looking at the patterns traced by the system on the phase plane, you can often work out what it is doing. For example, if the trajectory encircles the origin, then the system is vibrating. If the trajectory approaches the origin, the system is decaying to its equilibrium configuration.

We now know the effects of energy dissipation on a vibrating system. One important conclusion is that if the energy dissipation is low, the system will vibrate. Furthermore, the frequency of vibration is very close to that of an undamped system. Consequently, if you want to predict the frequency of vibration of a system, you can simplify the calculation by neglecting damping.

5.3.2 Using Free Vibrations to Measure Properties of a System

We will describe one very important application of the results developed in the preceding section.

It often happens that we need to measure the dynamical properties of an engineering system. For example, we might want to measure the natural frequency and damping coefficient for a structure after it has been built, to make sure that design predictions were correct, and to use in future models of the system.

You can use the free vibration response to do this, as follows. First, you instrument your design by attaching accelerometers to appropriate points. You then use an impulse hammer to excite a particular mode of vibration, as discussed in Section 5.1.3. You use your accelerometer readings to determine the displacement at the point where the structure was excited: the results will be a graph similar to the one shown below.

We then identify a nice looking peak, and call the time there t 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshadaWgaa WcbaGaaGimaaqabaaaaa@@ , as shown.

The following quantities are then measured from the graph:

1. The period of oscillation. The period of oscillation was defined in Section 5.1.2: it is the time between two peaks, as shown. Since the signal is (supposedly) periodic, it is often best to estimate T as follows

T= t n − t 0 n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpdaWcaaqaaiaadshadaWgaaWcbaGaamOBaaqabaGccqGHsislcaWG 0bWaaSbaaSqaaiaaicdaaeqaaaGcbaGaamOBaaaaaaa@3E13@

where t n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshadaWgaa WcbaGaamOBaaqabaaaaa@@ is the time at which the nth peak occurs, as shown in the picture.

2. The Logarithmic Decrement. This is a new quantity, defined as follows

δ=log( x( t n ) x( t n+1 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabg2 da9iGacYgacaGGVbGaai4zamaabmaabaWaaSaaaeaacaWG4bGaaiik aiaadshadaWgaaWcbaGaamOBaaqabaGccaGGPaaabaGaamiEaiaacI cacaWG0bWaaSbaaSqaaiaad6gacqGHRaWkcaaIXaaabeaakiaacMca aaaacaGLOaGaayzkaaaaaa@47DA@

where x( t n ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDamaaBaaaleaacaWGUbaabeaakiaacMcaaaa@3AB1@ is the displacement at the nth peak, as shown. In principle, you should be able to pick any two neighboring peaks, and calculate δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKbaa@37DE@ . You should get the same answer, whichever peaks you choose. It is often more accurate to estimate δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKbaa@37DE@ using the following formula

δ= 1 n log( x( t 0 ) x( t n ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabg2 da9maalaaabaGaaGymaaqaaiaad6gaaaGaciiBaiaac+gacaGGNbWa aeWaaeaadaWcaaqaaiaadIhacaGGOaGaamiDamaaBaaaleaacaaIWa aabeaakiaacMcaaeaacaWG4bGaaiikaiaadshadaWgaaWcbaGaamOB aaqabaGccaGGPaaaaaGaayjkaiaawMcaaaaa@47C2@

This expression should give the same answer as the earlier definition.

Now, it turns out that we can deduce ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ and ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@  from T and δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKbaa@37DE@ , as follows.

ς= δ 4 π 2 + δ 2 ω n = 4 π 2 + δ 2 T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9maalaaabaGaeqiTdqgabaWaaOaaaeaacaaI0aGaeqiWda3aaWba aSqabeaacaaIYaaaaOGaey4kaSIaeqiTdq2aaWbaaSqabeaacaaIYa aaaaqabaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeM8a3n aaBaaaleaacaWGUbaabeaakiabg2da9maalaaabaWaaOaaaeaacaaI 0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGaey4kaSIaeqiTdq2aaW baaSqabeaacaaIYaaaaaqabaaakeaacaWGubaaaaaa@617D@

Why does this work? Let us calculate T and δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKbaa@37DE@ using the exact solution to the equation of motion for a damped spring-mass system. Recall that, for an underdamped system, the solution has the form

x(t)=exp(−ς ω n t){ x 0 cos ω d t+ v 0 +ς ω n x n ω d sin ω d t } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpciGGLbGaaiiEaiaacchacaGGOaGaeyOe I0IaeqOWdyLaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaamiDaiaacM cadaGadaqaaiaadIhadaWgaaWcbaGaaGimaaqabaGcciGGJbGaai4B aiaacohacqaHjpWDdaWgaaWcbaGaamizaaqabaGccaWG0bGaey4kaS YaaSaaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaeqOW dyLaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaamiEamaaBaaaleaaca WGUbaabeaaaOqaaiabeM8a3naaBaaaleaacaWGKbaabeaaaaGcciGG ZbGaaiyAaiaac6gacqaHjpWDdaWgaaWcbaGaamizaaqabaGccaWG0b aacaGL7bGaayzFaaaaaa@@

where ω d = ω n 1− ς 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGKbaabeaakiabg2da9iabeM8a3naaBaaaleaacaWGUbaa beaakmaakaaabaGaaGymaiabgkHiTiabek8awnaaCaaaleqabaGaaG Omaaaaaeqaaaaa@@ . Hence, the period of oscillation is

T= 2π ω d = 2π ω n 1− ς 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpdaWcaaqaaiaaikdacqaHapaCaeaacqaHjpWDdaWgaaWcbaGaamiz aaqabaaaaOGaeyypa0ZaaSaaaeaacaaIYaGaeqiWdahabaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOWaaOaaaeaacaaIXaGaeyOeI0IaeqOW dy1aaWbaaSqabeaacaaIYaaaaaqabaaaaaaa@@

Similarly,

δ=log exp(−ς ω n t n ){ x 0 cos ω d t n + v 0 +ς ω n x n v 0 sin ω d t n } exp(−ς ω n ( t n +T)){ x 0 cos ω d ( t n +T)+ v 0 +ς ω n x n v 0 sin ω d ( t n +T) } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabg2 da9iGacYgacaGGVbGaai4zamaalaaabaGaciyzaiaacIhacaGGWbGa aiikaiabgkHiTiabek8awjabeM8a3naaBaaaleaacaWGUbaabeaaki aadshadaWgaaWcbaGaamOBaaqabaGccaGGPaWaaiWaaeaacaWG4bWa aSbaaSqaaiaaicdaaeqaaOGaci4yaiaac+gacaGGZbGaeqyYdC3aaS baaSqaaiaadsgaaeqaaOGaamiDamaaBaaaleaacaWGUbaabeaakiab gUcaRmaalaaabaGaamODamaaBaaaleaacaaIWaaabeaakiabgUcaRi abek8awjabeM8a3naaBaaaleaacaWGUbaabeaakiaadIhadaWgaaWc baGaamOBaaqabaaakeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaaaaki GacohacaGGPbGaaiOBaiabeM8a3naaBaaaleaacaWGKbaabeaakiaa dshadaWgaaWcbaGaamOBaaqabaaakiaawUhacaGL9baaaeaaciGGLb GaaiiEaiaacchacaGGOaGaeyOeI0IaeqOWdyLaeqyYdC3aaSbaaSqa aiaad6gaaeqaaOGaaiikaiaadshadaWgaaWcbaGaamOBaaqabaGccq GHRaWkcaWGubGaaiykaiaacMcadaGadaqaaiaadIhadaWgaaWcbaGa aGimaaqabaGcciGGJbGaai4BaiaacohacqaHjpWDdaWgaaWcbaGaam izaaqabaGccaGGOaGaamiDamaaBaaaleaacaWGUbaabeaakiabgUca RiaadsfacaGGPaGaey4kaSYaaSaaaeaacaWG2bWaaSbaaSqaaiaaic daaeqaaOGaey4kaSIaeqOWdyLaeqyYdC3aaSbaaSqaaiaad6gaaeqa aOGaamiEamaaBaaaleaacaWGUbaabeaaaOqaaiaadAhadaWgaaWcba GaaGimaaqabaaaaOGaci4CaiaacMgacaGGUbGaeqyYdC3aaSbaaSqa aiaadsgaaeqaaOGaaiikaiaadshadaWgaaWcbaGaamOBaaqabaGccq GHRaWkcaWGubGaaiykaaGaay5Eaiaaw2haaaaaaaa@9D5C@

where we have noted that t n+1 = t n +T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshadaWgaa WcbaGaamOBaiabgUcaRiaaigdaaeqaaOGaeyypa0JaamiDamaaBaaa leaacaWGUbaabeaakiabgUcaRiaadsfaaaa@3EDB@ .

Fortunately, this horrendous equation can be simplified greatly:  substitute for T in terms of ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@  and ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ , then cancel everything you possibly can to see that

δ= 2πς 1− ς 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabg2 da9maalaaabaGaaGOmaiabec8aWjabek8awbqaamaakaaabaGaaGym aiabgkHiTiabek8awnaaCaaaleqabaGaaGOmaaaaaeqaaaaaaaa@@

Finally, we can solve for ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@  and ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@  to see that:

ς= δ 4 π 2 + δ 2 ω n = 4 π 2 + δ 2 T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9maalaaabaGaeqiTdqgabaWaaOaaaeaacaaI0aGaeqiWda3aaWba aSqabeaacaaIYaaaaOGaey4kaSIaeqiTdq2aaWbaaSqabeaacaaIYa aaaaqabaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeM8a3n aaBaaaleaacaWGUbaabeaakiabg2da9maalaaabaWaaOaaaeaacaaI 0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGaey4kaSIaeqiTdq2aaW baaSqabeaacaaIYaaaaaqabaaakeaacaWGubaaaaaa@617D@

as promised.

Note that this procedure can never give us values for k, m or c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ .  However, if we wanted to find these, we could perform a static test on the structure.  If we measure the deflection d under a static load F, then we know that

k= F d MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadUgacqGH9a qpdaWcaaqaaiaadAeaaeaacaWGKbaaaaaa@39F3@

Once k had been found, m and c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@  are easily deduced from the relations

ω n = k m ς= c 2 km MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlabek8awjabg2da9maalaaabaGaam 4yaaqaaiaaikdadaGcaaqaaiaadUgacaWGTbaaleqaaaaaaaa@5CE3@

5.4 Forced vibration of damped, single degree of freedom, linear spring mass systems.

Finally, we solve the most important vibration problems of all. In engineering practice, we are almost invariably interested in predicting the response of a structure or mechanical system to external forcing. For example, we may need to predict the response of a bridge or tall building to wind loading, earthquakes, or ground vibrations due to traffic. Another typical problem you are likely to encounter is to isolate a sensitive system from vibrations. For example, the suspension of your car is designed to isolate a sensitive system (you) from bumps in the road. Electron microscopes are another example of sensitive instruments that must be isolated from vibrations. Electron microscopes are designed to resolve features a few nanometers in size. If the specimen vibrates with amplitude of only a few nanometers, it will be impossible to see! Great care is taken to isolate this kind of instrument from vibrations. That is one reason they are almost always in the basement of a building: the basement vibrates much less than the floors above.

We will again use a spring-mass system as a model of a real engineering system. As before, the spring-mass system can be thought of as representing a single mode of vibration in a real system, whose natural frequency and damping coefficient coincide with that of our spring-mass system.

We will consider three types of forcing applied to the spring-mass system, as shown below:

External Forcing models the behavior of a system which has a time varying force acting on it. An example might be an offshore structure subjected to wave loading.

Base Excitation models the behavior of a vibration isolation system. The base of the spring is given a prescribed motion, causing the mass to vibrate. This system can be used to model a vehicle suspension system, or the earthquake response of a structure.

Rotor Excitation models the effect of a rotating machine mounted on a flexible floor. The crank with length Y 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGzbWaaSbaaS qaaiaaicdaaeqaaaaa@373A@ and mass m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWgaa WcbaGaaGimaaqabaaaaa@@ rotates at constant angular velocity, causing the mass m to vibrate.

Of course, vibrating systems can be excited in other ways as well, but the equations of motion will always reduce to one of the three cases we consider here.

Notice that in each case, we will restrict our analysis to harmonic excitation. For example, the external force applied to the first system is given by

F(t)= F 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeacaGGOa GaamiDaiaacMcacqGH9aqpcaWGgbWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaeqyYdCNaaGPaVlaadshaaaa@@

The force varies harmonically, with amplitude F 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaaGimaaqabaaaaa@37EA@ and frequency ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ . Similarly, the base motion for the second system is

y(t)= Y 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGzbWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaeqyYdCNaaGPaVlaadshaaaa@@

and the distance between the small mass m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWgaa WcbaGaaGimaaqabaaaaa@@ and the large mass m for the third system has the same form.

We assume that at time t=0, the initial position and velocity of each system is

x= x 0 dx dt = v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacqGH9a qpcaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8+aaSaaaeaacaWGKbGaamiEaaqaaiaadsgacaWG0baaaiabg2da 9iaadAhadaWgaaWcbaGaaGimaaqabaaaaa@536C@

In each case, we wish to calculate the displacement of the mass x from its static equilibrium configuration, as a function of time t. It is of particular interest to determine the influence of forcing amplitude and frequency on the motion of the mass.

We follow the same approach to analyze each system: we set up, and solve the equation of motion.

5.4.1 Equations of Motion for Forced Spring Mass Systems

Equation of Motion for External Forcing

We have no problem setting up and solving equations of motion by now. First draw a free body diagram for the system, as show on the right

Newton’s law of motion gives

m d 2 s d t 2 =F(t)−k(s− L 0 )−c ds dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGaamizaiaa dshadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaamOraiaacIcaca WG0bGaaiykaiabgkHiTiaadUgacaGGOaGaam4CaiabgkHiTiaadYea daWgaaWcbaGaaGimaaqabaGccaGGPaGaeyOeI0Iaam4yamaalaaaba GaamizaiaadohaaeaacaWGKbGaamiDaaaaaaa@4D8B@

Rearrange and susbstitute for F(t)

m k d 2 s d t 2 + c k ds dt +s= L 0 + 1 k F 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam yBaaqaaiaadUgaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaam4CaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaki abgUcaRmaalaaabaGaam4yaaqaaiaadUgaaaWaaSaaaeaacaWGKbGa am4CaaqaaiaadsgacaWG0baaaiabgUcaRiaadohacqGH9aqpcaWGmb WaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaSaaaeaacaaIXaaabaGa am4AaaaacaWGgbWaaSbaaSqaaiaaicdaaeqaaOGaci4CaiaacMgaca GGUbGaeqyYdCNaaGPaVlaadshaaaa@54A3@

Check out our list of solutions to standard ODEs. We find that if we set

ω n = k m ,ς= c 2 km ,K= 1 k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdyLaeyypa0ZaaSaa aeaacaWGJbaabaGaaGOmamaakaaabaGaam4Aaiaad2gaaSqabaaaaO GaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaadUeacqGH9aqpdaWcaaqaaiaaigdaaeaacaWGRb aaaaaa@@ ,

our equation can be reduced to the form

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C+K F 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaiabgUcaRiaadUeacaWGgbWaaSbaaSqaaiaaicdaae qaaOGaci4CaiaacMgacaGGUbGaeqyYdCNaaGPaVlaadshaaaa@58D9@

which is on the list.

The (horrible) solution to this equation is given in the list of solutions. We will discuss the solution later, after we have analyzed the other two systems.

Equation of Motion for Base Excitation

Exactly the same approach works for this system. The free body diagram is shown in the figure. Note that the force in the spring is now k(x-y) because the length of the spring is L 0 +x−y MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaaGimaaqabaGccqGHRaWkcaWG4bGaeyOeI0IaamyEaaaa@3BC5@ . Similarly, the rate of change of length of the dashpot is d(x-y)/dt.

Newton’s second law then tells us that

m d 2 s d t 2 =−k(s−y− L 0 )−c( ds dt − dy dt ) ⇒ m k d 2 s d t 2 + c k ds dt +s= L 0 +y+ c k dy dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyBam aalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadohaaeaacaWG KbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcqGHsislca WGRbGaaiikaiaadohacqGHsislcaWG5bGaeyOeI0IaamitamaaBaaa leaacaaIWaaabeaakiaacMcacqGHsislcaWGJbWaaeWaaeaadaWcaa qaaiaadsgacaWGZbaabaGaamizaiaadshaaaGaeyOeI0YaaSaaaeaa caWGKbGaamyEaaqaaiaadsgacaWG0baaaaGaayjkaiaawMcaaaqaai abgkDiEpaalaaabaGaamyBaaqaaiaadUgaaaWaaSaaaeaacaWGKbWa aWbaaSqabeaacaaIYaaaaOGaam4CaaqaaiaadsgacaWG0bWaaWbaaS qabeaacaaIYaaaaaaakiabgUcaRmaalaaabaGaam4yaaqaaiaadUga aaWaaSaaaeaacaWGKbGaam4CaaqaaiaadsgacaWG0baaaiabgUcaRi aadohacqGH9aqpcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIa amyEaiabgUcaRmaalaaabaGaam4yaaqaaiaadUgaaaWaaSaaaeaaca WGKbGaamyEaaqaaiaadsgacaWG0baaaaaaaa@@

Make the following substitutions

ω n = k m ,ς= c 2 km ,K=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdyLaeyypa0ZaaSaa aeaacaWGJbaabaGaaGOmamaakaaabaGaam4Aaiaad2gaaSqabaaaaO GaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaadUeacqGH9aqpcaaIXaaaaa@@

and the equation reduces to the standard form

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C+K( y+ 2ς ω n dy dt ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaiabgUcaRiaadUeadaqadaqaaiaadMhacqGHRaWkda WcaaqaaiaaikdacqaHcpGvaeaacqaHjpWDdaWgaaWcbaGaamOBaaqa baaaaOWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG0baaaaGaay jkaiaawMcaaaaa@5C9E@

Given the initial conditions

x= x 0 dx dt = v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacqGH9a qpcaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVpaalaaabaGaamizaiaadIhaaeaaca WGKbGaamiDaaaacqGH9aqpcaWG2bWaaSbaaSqaaiaaicdaaeqaaaaa @4BB5@

and the base motion

y(t)= Y 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGzbWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaeqyYdCNaaGPaVlaadshaaaa@@

we can look up the solution in our handy list of solutions to ODEs.

Equation of motion for Rotor Excitation

Finally, we will derive the equation of motion for the third case. Free body diagrams are shown in the figure for both the rotor and the mass

Note that the horizontal acceleration of the mass m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWgaa WcbaGaaGimaaqabaaaaa@@ is

a= d 2 d t 2 (s+y)= d 2 s d t 2 + d 2 y d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggacqGH9a qpdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaaakeaacaWGKbGa amiDamaaCaaaleqabaGaaGOmaaaaaaGccaGGOaGaam4CaiabgUcaRi aadMhacaGGPaGaeyypa0ZaaSaaaeaacaWGKbWaaWbaaSqabeaacaaI YaaaaOGaam4CaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaa aakiabgUcaRmaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaa dMhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaaaaa@4E6D@

Hence, applying Newton’s second law in the horizontal direction for both masses:

m d 2 s d t 2 =H−k(s− L 0 )−c ds dt m 0 ( d 2 s d t 2 + d 2 y d t 2 )=−H MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyBam aalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadohaaeaacaWG KbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcaWGibGaey OeI0Iaam4AaiaacIcacaWGZbGaeyOeI0IaamitamaaBaaaleaacaaI WaaabeaakiaacMcacqGHsislcaWGJbWaaSaaaeaacaWGKbGaam4Caa qaaiaadsgacaWG0baaaaqaaiaad2gadaWgaaWcbaGaaGimaaqabaGc daqadaqaamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaado haaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWk daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5baabaGaam izaiaadshadaahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzkaaGa eyypa0JaeyOeI0Iaamisaaaaaa@5DC7@

Add these two equations to eliminate H and rearrange

m+ m 0 k d 2 s d t 2 + c k ds dt +s= L 0 − m 0 k d 2 y d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam yBaiabgUcaRiaad2gadaWgaaWcbaGaaGimaaqabaaakeaacaWGRbaa amaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadohaaeaaca WGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaWcaaqa aiaadogaaeaacaWGRbaaamaalaaabaGaamizaiaadohaaeaacaWGKb GaamiDaaaacqGHRaWkcaWGZbGaeyypa0JaamitamaaBaaaleaacaaI WaaabeaakiabgkHiTmaalaaabaGaamyBamaaBaaaleaacaaIWaaabe aaaOqaaiaadUgaaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamyEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaaa a@556A@

To arrange this into standard form, make the following substitutions

ω n = k (m+ m 0 ) ς= c 2 k(m+ m 0 ) K= m 0 m+ m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaaiikaiaad2gacqGHRaWkcaWGTbWaaSbaaSqaaiaaicdaaeqaaO GaaiykaaaaaSqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlabek8awjabg2da9maalaaabaGaam4yaaqaai aaikdadaGcaaqaaiaadUgacaGGOaGaamyBaiabgUcaRiaad2gadaWg aaWcbaGaaGimaaqabaGccaGGPaaaleqaaaaakiaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaadUeacqGH9aqpdaWcaaqaaiaad2ga daWgaaWcbaGaaGimaaqabaaakeaacaWGTbGaey4kaSIaamyBamaaBa aaleaacaaIWaaabeaaaaaaaa@67FA@

whereupon the equation of motion reduces to

1 ω n 2 d 2 s d t 2 + 2ς ω n ds dt +s= L 0 − K ω n 2 d 2 y d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadohaaeaacaWGKbGaamiDaaaacqGHRaWkcaWGZbGa eyypa0JaamitamaaBaaaleaacaaIWaaabeaakiabgkHiTmaalaaaba Gaam4saaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGc daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5baabaGaam izaiaadshadaahaaWcbeqaaiaaikdaaaaaaaaa@5A62@

Finally, look at the picture to convince yourself that if the crank rotates with angular velocity ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ , then

y(t)= Y 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGzbWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaeqyYdCNaaGPaVlaadshaaaa@@

where Y 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMfadaWgaa WcbaGaaGimaaqabaaaaa@37FD@ is the length of the crank.

The solution can once again be found in the list of solutions to ODEs.

5.4.2 Definition of Transient and Steady State Response.

If you have looked at the list of solutions to the equations of motion we derived in the preceding section, you will have discovered that they look horrible. Unless you have a great deal of experience with visualizing equations, it is extremely difficult to work out what the equations are telling us.

If you have Java, Internet Explorer (or a browser plugin that allows you to run IE in another browser) you can run a Java Applet to visualize the motion. You can find instructions for installing Java, the IE plugins, and giving permission for the Applet to run here. The address for the free vibration simulator (cut and paste this into the Internet Explorer address bar) is

http://www.brown.edu/Departments/Engineering/Courses/En4/java/forced.html

The applet simply calculates the solution to the equations of motion using the formulae given in the list of solutions, and plots graphs showing features of the motion. You can use the sliders to set various parameters in the system, including the type of forcing, its amplitude and frequency; spring constant, damping coefficient and mass; as well as the position and velocity of the mass at time t=0. Note that you can control the properties of the spring-mass system in two ways: you can either set values for k, m and c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ using the sliders, or you can set ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ , K and ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ instead.

We will use the applet to demonstrate a number of important features of forced vibrations, including the following:

The steady state response of a forced, damped, spring mass system is independent of the initial conditions.

To convince yourself of this, run the applet (click on `start’ and let the system run for a while). Now, press `stop’; change the initial position of the mass, and press `start’ again.

You will see that, after a while, the solution with the new initial conditions is exactly the same as it was before. Change the type of forcing, and repeat this test. You can change the initial velocity too, if you wish.

We call the behavior of the system as time gets very large the `steady state’ response; and as you see, it is independent of the initial position and velocity of the mass.

The behavior of the system while it is approaching the steady state is called the `transient’ response.The transient response depends on everything…

Now, reduce the damping coefficient and repeat the test. You will find that the system takes longer to reach steady state. Thus, the length of time to reach steady state depends on the properties of the system (and also the initial conditions).

The observation that the system always settles to a steady state has two important consequences. Firstly, we rarely know the initial conditions for a real engineering system (who knows what the position and velocity of a bridge is at time t=0?) . Now we know this doesn’t matter – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ the response is not sensitive to the initial conditions. Secondly, if we aren’t interested in the transient response, it turns out we can greatly simplify the horrible solutions to our equations of motion.

When analyzing forced vibrations, we (almost) always neglect the transient response of the system, and calculate only the steady state behavior.

If you look at the solutions to the equations of motion we calculated in the preceding sections, you will see that each solution has the form

x(t)= x h (t)+ x p (t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWG4bWaaSbaaSqaaiaadIgaaeqaaOGa aiikaiaadshacaGGPaGaey4kaSIaamiEamaaBaaaleaacaWGWbaabe aakiaacIcacaWG0bGaaiykaaaa@445C@

The term x h (t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaamiAaaqabaGccaGGOaGaamiDaiaacMcaaaa@3AAB@ accounts for the transient response, and is always zero for large time. The second term gives the steady state response of the system.

Following standard convention, we will list only the steady state solutions below. You should bear in mind, however, that the steady state is only part of the solution, and is only valid if the time is large enough that the transient term can be neglected.

5.4.3 Summary of Steady-State Response of Forced Spring Mass Systems.

This section summarizes all the formulas you will need to solve problems involving forced vibrations.

Solution for External Forcing

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Equation of Motion

1 ω n 2 d 2 s d t 2 + 2ς ω n ds dt +s=C+KF(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadohaaeaacaWGKbGaamiDaaaacqGHRaWkcaWGZbGa eyypa0Jaam4qaiabgUcaRiaadUeacaWGgbGaaiikaiaadshacaGGPa aaaa@@

with

ω n = k m ,ς= c 2 km ,K= 1 k C= L 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdyLaeyypa0ZaaSaa aeaacaWGJbaabaGaaGOmamaakaaabaGaam4Aaiaad2gaaSqabaaaaO GaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaadUeacqGH9aqpdaWcaaqaaiaaigdaaeaacaWGRb aaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaadoeacqGH9aqpcaWGmbWaaSbaaSqaaiaaicdaaeqaaa aa@74D1@

Steady State Solution:

s(t)=C+ X 0 sin( ωt+ϕ ) X 0 =K F 0 M(ω/ ω n ,ζ) M(ω/ ω n ,ζ)= 1 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ςω/ ω n 1− ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaam4Cai aacIcacaWG0bGaaiykaiabg2da9iaadoeacqGHRaWkcaWGybWaaSba aSqaaiaaicdaaeqaaOGaci4CaiaacMgacaGGUbWaaeWaaeaacqaHjp WDcaaMc8UaamiDaiabgUcaRiabew9aMbGaayjkaiaawMcaaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaamiwamaaBaaaleaacaaIWaaabeaakiabg2da9iaadUeacaWG gbWaaSbaaSqaaiaaicdaaeqaaOGaamytaiaacIcacqaHjpWDcaGGVa GaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaaiilaiabeA7a6jaacMca aeaacaWGnbGaaiikaiabeM8a3jaac+cacqaHjpWDdaWgaaWcbaGaam OBaaqabaGccaGGSaGaeqOTdONaaiykaiabg2da9maalaaabaGaaGym aaqaamaacmaabaWaaeWaaeaacaaIXaGaeyOeI0IaeqyYdC3aaWbaaS qabeaacaaIYaaaaOGaai4laiabeM8a3naaDaaaleaacaWGUbaabaGa aGOmaaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgU caRmaabmaabaGaaGOmaiabek8awjabeM8a3jaac+cacqaHjpWDdaWg aaWcbaGaamOBaaqabaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaik daaaaakiaawUhacaGL9baadaahaaWcbeqaaiaaigdacaGGVaGaaGOm aaaaaaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqaHvpGzcqGH9a qpciGG0bGaaiyyaiaac6gadaahaaWcbeqaaiabgkHiTiaaigdaaaGc daWcaaqaaiabgkHiTiaaikdacqaHcpGvcqaHjpWDcaGGVaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaaGcbaGaaGymaiabgkHiTiabeM8a3naa CaaaleqabaGaaGOmaaaakiaac+cacqaHjpWDdaqhaaWcbaGaamOBaa qaaiaaikdaaaaaaaaaaa@E15B@

Here, the function M is called the ‘magnification’ for the system.M and ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@@ are graphed below, as a function of ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA5@

(a)                                                                             (b)

Steady state vibration of a force spring-mass system (a) Magnification (b) phase.

Solution for Base Excitation

Equation of Motion

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=K( y+ 2ς ω n dy dt ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4samaabmaabaGaamyEaiabgUcaRmaalaaabaGaaGOmai abek8awbqaaiabeM8a3naaBaaaleaacaWGUbaabeaaaaGcdaWcaaqa aiaadsgacaWG5baabaGaamizaiaadshaaaaacaGLOaGaayzkaaaaaa@5AF4@

with

ω n = k m ,ς= λ 2 km ,K=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdyLaeyypa0ZaaSaa aeaacqaH7oaBaeaacaaIYaWaaOaaaeaacaWGRbGaamyBaaWcbeaaaa GccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8Uaam4saiabg2da9iaaigdaaaa@@

Steady State solution

x(t)= X 0 sin( ωt+ϕ ) X 0 =K Y 0 M(ω/ ω n ,ζ) M= { 1+ ( 2ςω/ ω n ) 2 } 1/2 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ς ω 3 / ω n 3 1−(1−4 ς 2 ) ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9iaadIfadaWgaaWcbaGaaGimaaqa baGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0b Gaey4kaSIaeqy1dygacaGLOaGaayzkaaGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGybWaaSbaaSqaaiaaic daaeqaaOGaeyypa0Jaam4saiaadMfadaWgaaWcbaGaaGimaaqabaGc caWGnbGaaiikaiabeM8a3jaac+cacqaHjpWDdaWgaaWcbaGaamOBaa qabaGccaGGSaGaeqOTdONaaiykaaqaaiaad2eacqGH9aqpdaWcaaqa amaacmaabaGaaGymaiabgUcaRmaabmaabaGaaGOmaiabek8awjabeM 8a3jaac+cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaakiaawIcacaGL PaaadaahaaWcbeqaaiaaikdaaaaakiaawUhacaGL9baadaahaaWcbe qaaiaaigdacaGGVaGaaGOmaaaaaOqaamaacmaabaWaaeWaaeaacaaI XaGaeyOeI0IaeqyYdC3aaWbaaSqabeaacaaIYaaaaOGaai4laiabeM 8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaOGaayjkaiaawMcaamaa CaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaGaaGOmaiabek8awj abeM8a3jaac+cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaakiaawIca caGLPaaadaahaaWcbeqaaiaaikdaaaaakiaawUhacaGL9baadaahaa WcbeqaaiaaigdacaGGVaGaaGOmaaaaaaGccaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeqy1dy Maeyypa0JaciiDaiaacggacaGGUbWaaWbaaSqabeaacqGHsislcaaI XaaaaOWaaSaaaeaacqGHsislcaaIYaGaeqOWdyLaeqyYdC3aaWbaaS qabeaacaaIZaaaaOGaai4laiabeM8a3naaDaaaleaacaWGUbaabaGa aG4maaaaaOqaaiaaigdacqGHsislcaGGOaGaaGymaiabgkHiTiaais dacqaHcpGvdaahaaWcbeqaaiaaikdaaaGccaGGPaGaeqyYdC3aaWba aSqabeaacaaIYaaaaOGaai4laiabeM8a3naaDaaaleaacaWGUbaaba GaaGOmaaaaaaaaaaa@C19E@

The expressions for M MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eaaaa@370B@  and ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@@  are graphed below, as a function of ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA5@

(a)                                                                        (b)

Steady state vibration of a base excited spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugGbabaaa aaaaaapeGaa8hfGaaa@@ mass system (a) Amplitude and (b) phase

Solution for Rotor Excitation

Equation of Motion

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=− K ω n 2 d 2 y d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0JaeyOeI0YaaSaaaeaacaWGlbaabaGaeqyYdC3aa0baaSqaai aad6gaaeaacaaIYaaaaaaakmaalaaabaGaamizamaaCaaaleqabaGa aGOmaaaakiaadMhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaa aaaaaaaa@58B0@

with

ω n = k (m+ m 0 ) ς= λ 2 k(m+ m 0 ) K= m 0 m+ m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaaiikaiaad2gacqGHRaWkcaWGTbWaaSbaaSqaaiaaicdaaeqaaO GaaiykaaaaaSqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlabek8awjabg2da9maalaaabaGaeq4UdWgaba GaaGOmamaakaaabaGaam4AaiaacIcacaWGTbGaey4kaSIaamyBamaa BaaaleaacaaIWaaabeaakiaacMcaaSqabaaaaOGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8Uaam4saiabg2da9maalaaabaGaamyB amaaBaaaleaacaaIWaaabeaaaOqaaiaad2gacqGHRaWkcaWGTbWaaS baaSqaaiaaicdaaeqaaaaaaaa@68C6@

Steady state solution

x(t)= X 0 sin( ωt+ϕ ) X 0 =K Y 0 M(ω/ ω n ,ζ) M= ω 2 / ω n 2 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ςω/ ω n 1− ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9iaadIfadaWgaaWcbaGaaGimaaqa baGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0b Gaey4kaSIaeqy1dygacaGLOaGaayzkaaGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaamiwamaaBaaaleaacaaIWaaabeaakiabg2da9iaadUeacaWG zbWaaSbaaSqaaiaaicdaaeqaaOGaamytaiaacIcacqaHjpWDcaGGVa GaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaaiilaiabeA7a6jaacMca aeaacaWGnbGaeyypa0ZaaSaaaeaacqaHjpWDdaahaaWcbeqaaiaaik daaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaaGc baWaaiWaaeaadaqadaqaaiaaigdacqGHsislcqaHjpWDdaahaaWcbe qaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaI YaaaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaS YaaeWaaeaacaaIYaGaeqOWdyLaeqyYdCNaai4laiabeM8a3naaBaaa leaacaWGUbaabeaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaa aaaOGaay5Eaiaaw2haamaaCaaaleqabaGaaGymaiaac+cacaaIYaaa aaaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7cqaHvpGzcqGH9aqpciGG0bGaaiyyaiaac6ga daahaaWcbeqaaiabgkHiTiaaigdaaaGcdaWcaaqaaiabgkHiTiaaik dacqaHcpGvcqaHjpWDcaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaa aaaakeaacaaIXaGaeyOeI0IaeqyYdC3aaWbaaSqabeaacaaIYaaaaO Gaai4laiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaaaaaa@B666@

The expressions for X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FC@  and ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@@  are graphed below, as a function of ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA5@

Steady state vibration of a rotor excited spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugGbabaaa aaaaaapeGaa8hfGaaa@@ mass system (a) Amplitude (b) Phase

5.4.4 Features of the Steady State Response of Spring Mass Systems to Forced Vibrations.

Now, we will discuss the implications of the results in the preceding section.

The steady state response is always harmonic, and has the same frequency as that of the forcing.

To see this mathematically, note that in each case the solution has the form x(t)= X 0 sin(ωt+ϕ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGybWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaaiikaiabeM8a3jaaykW7caWG0bGaey4kaS Iaeqy1dyMaaiykaaaa@@ . Recall that ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ defines the frequency of the force, the frequency of base excitation, or the rotor angular velocity. Thus, the frequency of vibration is determined by the forcing, not by the properties of the spring-mass system. This is unlike the free vibration response.

You can also check this out using our applet. To switch off the transient solution, click on the checkbox labeled `show transient’. Then, try running the applet with different values for k, m and c MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadogaaaa@@ , as well as different forcing frequencies, to see what happens. As long as you have switched off the transient solution, the response will always be harmonic.

The amplitude of vibration is strongly dependent on the frequency of excitation, and on the properties of the spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8hfGaaa@@ mass system.

To see this mathematically, note that the solution has the form x(t)= X 0 sin(ωt+ϕ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGybWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaaiikaiabeM8a3jaaykW7caWG0bGaey4kaS Iaeqy1dyMaaiykaaaa@@ . Observe that X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FC@ is the amplitude of vibration, and look at the preceding section to find out how the amplitude of vibration varies with frequency, the natural frequency of the system, the damping factor, and the amplitude of the forcing. The formulae for X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FC@ are quite complicated, but you will learn a great deal if you are able to sketch graphs of X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FC@ as a function of ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA5@ for various values of ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ .

You can also use our applet to study the influence of forcing frequency, the natural frequency of the system, and the damping coefficient. If you plot position-v-time curves, make sure you switch off the transient solution to show clearly the steady state behavior. Note also that if you click on the `amplitude – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ v- frequency’ radio button just below the graphs, you will see a graph showing the steady state amplitude of vibration as a function of forcing frequency. The current frequency of excitation is marked as a square dot on the curve (if you don’t see the square dot, it means the frequency of excitation is too high to fit on the scale – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ if you lower the excitation frequency and press `start’ again you should see the dot appear). You can change the properties of the spring mass system (or the natural frequency and damping coefficient) and draw new amplitude-v-frequency curves to see how the response of the system has changed.

Try the following tests

(i) Keeping the natural frequency fixed (or k and m fixed), plot ampltude-v-frequency graphs for various values of damping coefficient (or the dashpot coefficient). What happens to the maximum amplitude of vibration as damping is reduced?

(ii) Keep the damping coefficient fixed at around 0.1. Plot graphs of amplitude-v-frequency for various values of the natural frequency of the system. How does the maximum vibration amplitude change as natural frequency is varied? What about the frequency at which the maximum occurs?

(iii) Keep the dashpot coefficient fixed at a lowish value. Plot graphs of amplitude-v-frequency for various values of spring stiffness and mass. Can you reconcile the behavior you observe with the results of test (ii)?

(iv) Try changing the type of forcing to base excitation and rotor excitation. Can you see any differences in the amplitude-v-frequency curves for different types of forcing?

(v) Set the damping coefficient to a low value (below 0.1). Keep the natural frequency fixed. Run the program for different excitation frequencies. Watch what the system is doing. Observe the behavior when the excitation frequency coincides with the natural frequency of the system. Try this test for each type of excitation.

If the forcing frequency is close to the natural frequency of the system, and the system is lightly damped, huge vibration amplitudes may occur. This phenomenon is known as resonance.

If you ran the tests in the preceding section, you will have seen the system resonate. Note that the system resonates at a very similar frequency for each type of forcing.

As a general rule, engineers try to avoid resonance like the plague. Resonance is bad vibrations. Large amplitude vibrations imply large forces; and large forces cause material failure. There are exceptions to this rule, of course. Musical instruments, for example, are supposed to resonate, so as to amplify sound. Musicians who play string, wind and brass instruments spend years training their lips or bowing arm to excite just the right vibration modes in their instruments to make them sound perfect. Resonance is a good thing in energy harvesting systems, and many instruments, such as MEMS gyroscopes, and atomic force microscopes, work by measuring how an external stimulus of some sort (rotation, or a surface force) changes the resonant frequency of a system.

There is a phase lag between the forcing and the system response, which depends on the frequency of excitation and the properties of the spring-mass system.

The response of the system is x(t)= X 0 sin(ωt+ϕ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGybWaaSbaaSqaaiaaicdaaeqaaOGa ci4CaiaacMgacaGGUbGaaiikaiabeM8a3jaaykW7caWG0bGaey4kaS Iaeqy1dyMaaiykaaaa@@ . Expressions for ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@@ are given in the preceding section. Note that the phase lag is always negative.

You can use the applet to examine the physical significance of the phase lag. Note that you can have the program plot a graph of phase-v-frequency for you, if you wish.

It is rather unusual to be particularly interested in the phase of the vibration, so we will not discuss it in detail here.

5.4.5 Engineering implications of vibration behavior

The solutions listed in the preceding sections give us general guidelines for engineering a system to avoid (or create!) vibrations.

Preventing a system from vibrating: Suppose that we need to stop a structure or component from vibrating – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ e.g. to stop a tall building from swaying. Structures are always deformable to some extent – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ this is represented qualitatively by the spring in a spring-mass system. They always have mass – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ this is represented by the mass of the block. Finally, the damper represents energy dissipation. Forces acting on a system generally fluctuate with time. They probably aren’t perfectly harmonic, but they usually do have a fairly well defined frequency (visualize waves on the ocean, for example, or wind gusts. Many vibrations are man-made, in which case their frequency is known – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ for example vehicles traveling on a road tend to induce vibrations with a frequency of about 2Hz, corresponding to the bounce of the car on its suspension).

So how do we stop the system from vibrating? We know that its motion is given by

x(t)= X 0 sin( ωt+ϕ ) X 0 = K F 0 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ςω/ ω n 1− ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9iaadIfadaWgaaWcbaGaaGimaaqa baGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0b Gaey4kaSIaeqy1dygacaGLOaGaayzkaaaabaGaamiwamaaBaaaleaa caaIWaaabeaakiabg2da9maalaaabaGaam4saiaadAeadaWgaaWcba GaaGimaaqabaaakeaadaGadaqaamaabmaabaGaaGymaiabgkHiTiab eM8a3naaCaaaleqabaGaaGOmaaaakiaac+cacqaHjpWDdaqhaaWcba GaamOBaaqaaiaaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaa ikdaaaGccqGHRaWkdaqadaqaaiaaikdacqaHcpGvcqaHjpWDcaGGVa GaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaGccaGLOaGaayzkaaWaaWba aSqabeaacaaIYaaaaaGccaGL7bGaayzFaaWaaWbaaSqabeaacaaIXa Gaai4laiaaikdaaaaaaOGaaGPaVlaaykW7aeaacaaMc8Uaeqy1dyMa eyypa0JaciiDaiaacggacaGGUbWaaWbaaSqabeaacqGHsislcaaIXa aaaOWaaSaaaeaacqGHsislcaaIYaGaeqOWdyLaeqyYdCNaai4laiab eM8a3naaBaaaleaacaWGUbaabeaaaOqaaiaaigdacqGHsislcqaHjp WDdaahaaWcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaa d6gaaeaacaaIYaaaaaaaaaaa@85F6@

ω n = k m ,ς= c 2 km ,K= 1 k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdyLaeyypa0ZaaSaa aeaacaWGJbaabaGaaGOmamaakaaabaGaam4Aaiaad2gaaSqabaaaaO GaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaadUeacqGH9aqpdaWcaaqaaiaaigdaaeaacaWGRb aaaaaa@@

To minimize vibrations, we must design the system to make the vibration amplitude X 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FD@ as small as possible. The formula for X 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FD@ is a bit scary, which is why we plot graphs of the solution. The graphs show that we will observe vibrations with large amplitudes if (i) The frequency ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA6@ is close to 1; and (ii) the damping ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeA7a6baa@37F7@ is small. At first sight, it looks like we could minimize vibrations by making ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA6@ very large. This is true in principle, and can be done in some designs, e.g. if the force acts on a very localized area of the structure, and will only excite a single vibration mode. For most systems, this approach will not work, however. This is because real components generally have a very large number of natural frequencies of vibration, corresponding to different vibration modes. We could design the system so that ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA6@ is large for the mode with the lowest frequency – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ and perhaps some others – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ but there will always be other modes with higher frequencies, which will have smaller values of ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA6@ . There is a risk that one of these will be close to resonance. Consequently, we generally design the system so that ω/ ω n <<1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH8aapcqGH8aapcaaI Xaaaaa@3E73@ for the mode with the lowest natural frequency. In fact, design codes usually specify the minimum allowable value of ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3BA6@ for vibration critical components. This will guarantee that ω/ ω n <<1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH8aapcqGH8aapcaaI Xaaaaa@3E73@ for all modes, and hence the vibration amplitude X 0 →K F 0 = F 0 /k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccqGHsgIRcaWGlbGaamOramaaBaaaleaacaaI Waaabeaakiabg2da9iaadAeadaWgaaWcbaGaaGimaaqabaGccaGGVa Gaam4Aaaaa@40E3@ . This tells us that the best approach to avoid vibrations is to make the structure as stiff as possible. This will make the natural frequency large, and will also make F 0 /k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaaGimaaqabaGccaGGVaGaam4Aaaaa@@ small.

Designing a suspension or vibration isolation system. Suspensions, and vibration isolation systems, are examples of base excited systems. In this case, the system really consists of a mass (the vehicle, or the isolation table) on a spring (the shock absorber or vibration isolation pad). We expect that the base will vibrate with some characteristic frequency ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ . Our goal is to design the system to minimize the vibration of the mass.

Our vibration solution predicts that the mass vibrates with displacement

x(t)= X 0 sin( ωt+ϕ ) X 0 = K Y 0 { 1+ ( 2ςω/ ω n ) 2 } 1/2 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ς ω 3 / ω n 3 1−(1−4 ς 2 ) ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9iaadIfadaWgaaWcbaGaaGimaaqa baGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0b Gaey4kaSIaeqy1dygacaGLOaGaayzkaaaabaGaamiwamaaBaaaleaa caaIWaaabeaakiabg2da9maalaaabaGaam4saiaadMfadaWgaaWcba GaaGimaaqabaGcdaGadaqaaiaaigdacqGHRaWkdaqadaqaaiaaikda cqaHcpGvcqaHjpWDcaGGVaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaa GccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGL7bGaayzF aaWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaaakeaadaGadaqaam aabmaabaGaaGymaiabgkHiTiabeM8a3naaCaaaleqabaGaaGOmaaaa kiaac+cacqaHjpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaaakiaawI cacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqadaqaaiaa ikdacqaHcpGvcqaHjpWDcaGGVaGaeqyYdC3aaSbaaSqaaiaad6gaae qaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGL7bGa ayzFaaWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaaaaOGaaGPaVd qaaiaaykW7cqaHvpGzcqGH9aqpciGG0bGaaiyyaiaac6gadaahaaWc beqaaiabgkHiTiaaigdaaaGcdaWcaaqaaiabgkHiTiaaikdacqaHcp GvcqaHjpWDdaahaaWcbeqaaiaaiodaaaGccaGGVaGaeqyYdC3aa0ba aSqaaiaad6gaaeaacaaIZaaaaaGcbaGaaGymaiabgkHiTiaacIcaca aIXaGaeyOeI0IaaGinaiabek8awnaaCaaaleqabaGaaGOmaaaakiaa cMcacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0 baaSqaaiaad6gaaeaacaaIYaaaaaaaaaaa@9D09@

ω n = k m ,ς= c 2 km ,K=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdyLaeyypa0ZaaSaa aeaacaWGJbaabaGaaGOmamaakaaabaGaam4Aaiaad2gaaSqabaaaaO GaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaadUeacqGH9aqpcaaIXaaaaa@@

Again, the graph is helpful to understand how the vibration amplitude X 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaaaaa@37FD@ varies with system parameters.

Clearly, we can minimize the vibration amplitude of the mass by making ω/ ω n >>1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH+aGpcqGH+aGpcaaI Xaaaaa@3E7B@ .We can do this by making the spring stiffness as small as possible (use a soft spring), and making the mass large. It also helps to make the damping ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeA7a6baa@37F7@ small. This is counter-intuitive – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ people often think that the energy dissipated by the shock absorbers in their suspensions that makes them work. There are some disadvantages to making the damping too small, however. For one thing, if the system is lightly damped, and is disturbed somehow, the subsequent transient vibrations will take a very long time to die out. In addition, there is always a risk that the frequency of base excitation is lower than we expect – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ if the system is lightly damped, a potentially damaging resonance may occur.

Suspension design involves a bit more than simply minimizing the vibration of the mass, of course – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ the car will handle poorly if the wheels begin to leave the ground. A very soft suspension generally has poor handling, so the engineers must trade off handling against vibration isolation.

5.4.6 Using Forced Vibration Response to Measure Properties of a System.

We often measure the natural frequency and damping coefficient for a mode of vibration in a structure or component, by measuring the forced vibration response of the system.

Here is how this is done. We find some way to apply a harmonic excitation to the system (base excitation might work; or you can apply a force using some kind of actuator, or you could deliberately mount an unbalanced rotor on the system).

Then, we mount accelerometers on our system, and use them to measure the displacement of the structure, at the point where it is being excited, as a function of frequency.

We then plot a graph, which usually looks something like the picture on the right. We read off the maximum response X max MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaciyBaiaacggacaGG4baabeaaaaa@3A16@ , and draw a horizontal line at amplitude X max / 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaciyBaiaacggacaGG4baabeaakiaac+cadaGcaaqaaiaaikda aSqabaaaaa@3BAA@ . Finally, we measure the frequencies ω 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIXaaabeaaaaa@38ED@ , ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIYaaabeaaaaa@38EE@ and ω max MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaaciGGTbGaaiyyaiaacIhaaeqaaaaa@3B06@ as shown in the picture.

We define the bandwidth of the response Δω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejabeM 8a3baa@396C@ as

Δω= ω 2 − ω 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejabeM 8a3jabg2da9iabeM8a3naaBaaaleaacaaIYaaabeaakiabgkHiTiab eM8a3naaBaaaleaacaaIXaaabeaaaaa@40D2@

Like the logarithmic decrement, the bandwidth of the forced harmonic response is a measure of the damping in a system.

It turns out that we can estimate the natural frequency of the system and its damping coefficient using the following formulae

ς≈ Δω 2 ω max ω n ≈ ω max MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabgI Ki7oaalaaabaGaeuiLdqKaeqyYdChabaGaaGOmaiabeM8a3naaBaaa leaaciGGTbGaaiyyaiaacIhaaeqaaaaakiaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeM8a3n aaBaaaleaacaWGUbaabeaakiabgIKi7kabeM8a3naaBaaaleaaciGG TbGaaiyyaiaacIhaaeqaaaaa@679F@

The formulae are accurate for small ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ - say ς<0.2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabgY da8iaaicdacaGGUaGaaGOmaaaa@3B0A@ .

To understand the origin of these formulae, recall that the amplitude of vibration due to external forcing is given by

X 0 = K F 0 ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccqGH9aqpdaWcaaqaaiaadUeacaWGgbWaaSba aSqaaiaaicdaaeqaaaGcbaWaaOaaaeaadaqadaqaaiaaigdacqGHsi slcqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0ba aSqaaiaad6gaaeaacaaIYaaaaaGccaGLOaGaayzkaaWaaWbaaSqabe aacaaIYaaaaOGaey4kaSYaaeWaaeaacaaIYaGaeqOWdyLaeqyYdCNa ai4laiabeM8a3naaBaaaleaacaWGUbaabeaaaOGaayjkaiaawMcaam aaCaaaleqabaGaaGOmaaaaaeqaaaaaaaa@522C@

We can find the frequency at which the amplitude is a maximum by differentiating with respect to ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ , setting the derivative equal to zero and solving the resulting equation for frequency. It turns out that the maximum amplitude occurs at a frequency

ω max = ω n 1−2 ς 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyypa0JaeqyYdC3aaSba aSqaaiaad6gaaeqaaOWaaOaaaeaacaaIXaGaeyOeI0IaaGOmaiabek 8awnaaCaaaleqabaGaaGOmaaaaaeqaaaaa@440E@

For small ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ , we see that

ω max ≈ ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyisISRaeqyYdC3aaSba aSqaaiaad6gaaeqaaaaa@3FAD@

Next, to get an expression relating the bandwidth Δω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejabeM 8a3baa@396C@ to ς MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awbaa@37DE@ , we first calculate the frequencies ω 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIXaaabeaaaaa@38ED@ and ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIYaaabeaaaaa@38EE@ . Note that the maximum amplitude of vibration can be calculated by setting ω= ω max MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jabg2 da9iabeM8a3naaBaaaleaaciGGTbGaaiyyaiaacIhaaeqaaaaa@3DD9@ , which gives

X max = K F 0 2ς 1− ς 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaciyBaiaacggacaGG4baabeaakiabg2da9maalaaabaGaam4s aiaadAeadaWgaaWcbaGaaGimaaqabaaakeaacaaIYaGaeqOWdy1aaO aaaeaacaaIXaGaeyOeI0IaeqOWdy1aaWbaaSqabeaacaaIYaaaaaqa baaaaaaa@@

Now, at the two frequencies of interest, we know X 0 = X max / 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccqGH9aqpcaWGybWaaSbaaSqaaiGac2gacaGG HbGaaiiEaaqabaGccaGGVaWaaOaaaeaacaaIYaaaleqaaaaa@3E7D@ , so that ω 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIXaaabeaaaaa@38ED@ and ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIYaaabeaaaaa@38EE@ must be solutions of the equation

K F 0 2ς 1− ς 2 1 2 = K F 0 ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω ω n ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam 4saiaadAeadaWgaaWcbaGaaGimaaqabaaakeaacaaIYaGaeqOWdy1a aOaaaeaacaaIXaGaeyOeI0IaeqOWdy1aaWbaaSqabeaacaaIYaaaaa qabaaaaOWaaSaaaeaacaaIXaaabaWaaOaaaeaacaaIYaaaleqaaaaa kiabg2da9maalaaabaGaam4saiaadAeadaWgaaWcbaGaaGimaaqaba aakeaadaGcaaqaamaabmaabaGaaGymaiabgkHiTiabeM8a3naaCaaa leqabaGaaGOmaaaakiaac+cacqaHjpWDdaqhaaWcbaGaamOBaaqaai aaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH RaWkdaqadaqaaiaaikdacqaHcpGvcqaHjpWDcqaHjpWDdaWgaaWcba GaamOBaaqabaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaa beaaaaaaaa@5AA4@

Rearrange this equation to see that

ω 4 −2 ω 2 ω n 2 (1−2 ς 2 )+ ω n 4 −8 ς 2 ω n 4 (1− ς 2 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaCa aaleqabaGaaGinaaaakiabgkHiTiaaikdacqaHjpWDdaahaaWcbeqa aiaaikdaaaGccqaHjpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaGcca GGOaGaaGymaiabgkHiTiaaikdacqaHcpGvdaahaaWcbeqaaiaaikda aaGccaGGPaGaey4kaSIaeqyYdC3aa0baaSqaaiaad6gaaeaacaaI0a aaaOGaeyOeI0IaaGioaiabek8awnaaCaaaleqabaGaaGOmaaaakiab eM8a3naaDaaaleaacaWGUbaabaGaaGinaaaakiaacIcacaaIXaGaey OeI0IaeqOWdy1aaWbaaSqabeaacaaIYaaaaOGaaiykaiabg2da9iaa icdaaaa@5B58@

This is a quadratic equation for ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaCa aaleqabaGaaGOmaaaaaaa@38EF@ and has solutions

ω 1 = { ω n 2 (1−2 ς 2 )−2 ω n 2 ς 1− ς 2 } 1/2 ω 2 = { ω n 2 (1−2 ς 2 )+2 ω n 2 ς 1− ς 2 } 1/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC 3aa0baaSqaaiaaigdaaeaaaaGccqGH9aqpdaGadaqaaiabeM8a3naa DaaaleaacaWGUbaabaGaaGOmaaaakiaacIcacaaIXaGaeyOeI0IaaG Omaiabek8awnaaCaaaleqabaGaaGOmaaaakiaacMcacqGHsislcaaI YaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaOGaeqOWdy1aaO aaaeaacaaIXaGaeyOeI0IaeqOWdy1aaWbaaSqabeaacaaIYaaaaaqa baaakiaawUhacaGL9baadaahaaWcbeqaaiaaigdacaGGVaGaaGOmaa aaaOqaaiabeM8a3naaDaaaleaacaaIYaaabaaaaOGaeyypa0ZaaiWa aeaacqaHjpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaGccaGGOaGaaG ymaiabgkHiTiaaikdacqaHcpGvdaahaaWcbeqaaiaaikdaaaGccaGG PaGaey4kaSIaaGOmaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaa aakiabek8awnaakaaabaGaaGymaiabgkHiTiabek8awnaaCaaaleqa baGaaGOmaaaaaeqaaaGccaGL7bGaayzFaaWaaWbaaSqabeaacaaIXa Gaai4laiaaikdaaaaaaaa@718C@

Expand both expressions in a Taylor series about ς=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9iaaicdaaaa@399E@ to see that

ω 1 ≈ ω n (1−ς) ω 2 ≈ ω n (1+ς) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC 3aaSbaaSqaaiaaigdaaeqaaOGaeyisISRaeqyYdC3aaSbaaSqaaiaa d6gaaeqaaOGaaiikaiaaigdacqGHsislcqaHcpGvcaGGPaaabaGaeq yYdC3aaSbaaSqaaiaaikdaaeqaaOGaeyisISRaeqyYdC3aaSbaaSqa aiaad6gaaeqaaOGaaiikaiaaigdacqGHRaWkcqaHcpGvcaGGPaaaaa a@4E4C@

so, finally, we confirm that

Δω= ω 2 − ω 1 =2ς ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejabeM 8a3jabg2da9iabeM8a3naaBaaaleaacaaIYaaabeaakiabgkHiTiab eM8a3naaBaaaleaacaaIXaaabeaakiabg2da9iaaikdacqaHcpGvcq aHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@472F@

5.4.7 Example Problems in Forced Vibrations

Example 1: A structure is idealized as a damped spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8hfGaaa@@ mass system with stiffness 10 kN/m; mass 2Mg; and dashpot coefficient 2 kNs/m. It is subjected to a harmonic force of amplitude 500N at frequency 0.5Hz. Calculate the steady state amplitude of vibration.

Start by calculating the properties of the system:

ω n = k m =2.23 rad/sς= c 2 km =0.224K= 1 k = 1   m/N MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccqGH9aqpcaaIYaGaaiOlaiaaikdacaaIZa GaaeiiaiaabkhacaqGHbGaaeizaiaab+cacaqGZbGaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqOWdy Laeyypa0ZaaSaaaeaacaWGJbaabaGaaGOmamaakaaabaGaam4Aaiaa d2gaaSqabaaaaOGaeyypa0JaaGimaiaac6cacaaIYaGaaGOmaiaais dacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaadUeacqGH9aqpdaWcaaqaaiaaigdaaeaacaWGRbaaaiabg2da9m aalaaabaGaaGymaaqaaiaaigdacaaIWaGaaGimaiaaicdacaaIWaaa aiaabccacaqGGaGaaeyBaiaab+cacaqGobaaaa@774B@

Now, the list of solutions to forced vibration problems gives

x(t)= X 0 sin( ωt+ϕ ) X 0 = K F 0 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ςω/ ω n 1− ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9iaadIfadaWgaaWcbaGaaGimaaqa baGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0b Gaey4kaSIaeqy1dygacaGLOaGaayzkaaaabaGaamiwamaaBaaaleaa caaIWaaabeaakiabg2da9maalaaabaGaam4saiaadAeadaWgaaWcba GaaGimaaqabaaakeaadaGadaqaamaabmaabaGaaGymaiabgkHiTiab eM8a3naaCaaaleqabaGaaGOmaaaakiaac+cacqaHjpWDdaqhaaWcba GaamOBaaqaaiaaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaa ikdaaaGccqGHRaWkdaqadaqaaiaaikdacqaHcpGvcqaHjpWDcaGGVa GaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaGccaGLOaGaayzkaaWaaWba aSqabeaacaaIYaaaaaGccaGL7bGaayzFaaWaaWbaaSqabeaacaaIXa Gaai4laiaaikdaaaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua eqy1dyMaeyypa0JaciiDaiaacggacaGGUbWaaWbaaSqabeaacqGHsi slcaaIXaaaaOWaaSaaaeaacqGHsislcaaIYaGaeqOWdyLaeqyYdCNa ai4laiabeM8a3naaBaaaleaacaWGUbaabeaaaOqaaiaaigdacqGHsi slcqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0ba aSqaaiaad6gaaeaacaaIYaaaaaaaaaaa@A1BB@

For the present problem:

ω=0.5×2πrad/s⇒ω/ ω n =π/2.23=1.41 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jabg2 da9iaaicdacaGGUaGaaGynaiabgEna0kaaikdacqaHapaCcaaMc8Ua aeOCaiaabggacaqGKbGaae4laiaabohacaaMc8UaaGPaVlaaykW7cq GHshI3caaMc8UaaGPaVlabeM8a3jaab+cacqaHjpWDdaWgaaWcbaGa aeOBaaqabaGccqGH9aqpcqaHapaCcaGGVaGaaGOmaiaac6cacaaIYa GaaG4maiabg2da9iaaigdacaGGUaGaaGinaiaaigdaaaa@5F8A@

Substituting numbers into the expression for the vibration amplitude shows that

X 0 =43mm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccqGH9aqpcaaI0aGaaG4maiaaykW7caaMc8Ua aeyBaiaab2gaaaa@3F7D@

Example 2: A car and its suspension system are idealized as a damped spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8hfGaaa@@ mass system, with natural frequency 0.5Hz and damping coefficient 0.2. Suppose the car drives at speed V over a road with sinusoidal roughness. Assume the roughness wavelength is 10m, and its amplitude is 20cm. At what speed does the maximum amplitude of vibration occur, and what is the corresponding vibration amplitude?

Let s denote the distance traveled by the car, and let L denote the wavelength of the roughness and H the roughness amplitude. Then, the height of the wheel above the mean road height may be expressed as

y=Hsin( 2πs L ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacqGH9a qpcaWGibGaci4CaiaacMgacaGGUbWaaeWaaeaadaWcaaqaaiaaikda cqaHapaCcaWGZbaabaGaamitaaaaaiaawIcacaGLPaaaaaa@41BD@

Noting that s=Vt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadohacqGH9a qpcaWGwbGaamiDaaaa@3A0B@ , we have that

y(t)=Hsin( 2πV L t ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGibGaci4CaiaacMgacaGGUbWaaeWa aeaadaWcaaqaaiaaikdacqaHapaCcaWGwbaabaGaamitaaaacaWG0b aacaGLOaGaayzkaaaaaa@44EB@

i.e., the wheel oscillates vertically with harmonic motion, at frequency ω=2πV/L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jabg2 da9iaaikdacqaHapaCcaWGwbGaai4laiaadYeaaaa@3DE4@ .

Now, the suspension has been idealized as a spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8hfGaaa@@ mass system subjected to base excitation. The steady state vibration is

x(t)= X 0 sin( ωt+ϕ ) X 0 =K Y 0 M M= { 1+ ( 2ςω/ ω n ) 2 } 1/2 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 ϕ= tan −1 −2ς ω 3 / ω n 3 1−(1−4 ς 2 ) ω 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaiqabaGaamiEai aacIcacaWG0bGaaiykaiabg2da9iaadIfadaWgaaWcbaGaaGimaaqa baGcciGGZbGaaiyAaiaac6gadaqadaqaaiabeM8a3jaaykW7caWG0b Gaey4kaSIaeqy1dygacaGLOaGaayzkaaGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca WGybWaaSbaaSqaaiaaicdaaeqaaOGaeyypa0Jaam4saiaadMfadaWg aaWcbaGaaGimaaqabaGccaWGnbaabaGaamytaiabg2da9maalaaaba WaaiWaaeaacaaIXaGaey4kaSYaaeWaaeaacaaIYaGaeqOWdyLaeqyY dCNaai4laiabeM8a3naaBaaaleaacaWGUbaabeaaaOGaayjkaiaawM caamaaCaaaleqabaGaaGOmaaaaaOGaay5Eaiaaw2haamaaCaaaleqa baGaaGymaiaac+cacaaIYaaaaaGcbaWaaiWaaeaadaqadaqaaiaaig dacqGHsislcqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGGVaGaeqyY dC3aa0baaSqaaiaad6gaaeaacaaIYaaaaaGccaGLOaGaayzkaaWaaW baaSqabeaacaaIYaaaaOGaey4kaSYaaeWaaeaacaaIYaGaeqOWdyLa eqyYdCNaai4laiabeM8a3naaBaaaleaacaWGUbaabeaaaOGaayjkai aawMcaamaaCaaaleqabaGaaGOmaaaaaOGaay5Eaiaaw2haamaaCaaa leqabaGaaGymaiaac+cacaaIYaaaaaaakiaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqaHvpGz cqGH9aqpciGG0bGaaiyyaiaac6gadaahaaWcbeqaaiabgkHiTiaaig daaaGcdaWcaaqaaiabgkHiTiaaikdacqaHcpGvcqaHjpWDdaahaaWc beqaaiaaiodaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaaca aIZaaaaaGcbaGaaGymaiabgkHiTiaacIcacaaIXaGaeyOeI0IaaGin aiabek8awnaaCaaaleqabaGaaGOmaaaakiaacMcacqaHjpWDdaahaa WcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaa caaIYaaaaaaaaaaa@BD03@

For light damping, the maximum amplitude of vibration occurs at around the natural frequency. Therefore, the critical speed follows from

ω= 2πV L = ω n ⇒V= ω n L/2π=5m/s=18 km/hr MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqyYdC Naeyypa0ZaaSaaaeaacaaIYaGaeqiWdaNaamOvaaqaaiaadYeaaaGa eyypa0JaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaGcbaGaeyO0H4Taam Ovaiabg2da9iabeM8a3naaBaaaleaacaWGUbaabeaakiaadYeacaGG VaGaaGOmaiabec8aWjabg2da9iaaiwdacaaMc8UaaeyBaiaab+caca qGZbGaaeypaiaabgdacaqG4aGaaeiiaiaabUgacaqGTbGaae4laiaa bIgacaqGYbaaaaa@599F@

Note that K=1 for base excitation, so that the amplitude of vibration at ω/ ω n =1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH9aqpcaaIXaaaaa@3D70@ is approximately

X 0 ≈ Y 0 2ς =20/0.4=50cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccqGHijYUdaWcaaqaaiaadMfadaWgaaWcbaGa aGimaaqabaaakeaacaaIYaGaeqOWdyfaaiabg2da9iaaikdacaaIWa Gaai4laiaaicdacaGGUaGaaGinaiabg2da9iaaiwdacaaIWaGaaGPa VlaabogacaqGTbaaaa@492F@

Note that at this speed, the suspension system is making the vibration worse. The amplitude of the car’s vibration is greater than the roughness of the road. Suspensions work best if they are excited at frequencies well above their resonant frequencies.

Example 3: The suspension system discussed in the preceding problem has the following specifications. For the roadway described in the preceding section, the amplitude of vibration may not exceed 35cm at any speed. At 55 miles per hour, the amplitude of vibration must be less than 10cm. The car weighs lb. Select values for the spring stiffness and the dashpot coefficient.

We must first determine values for ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeA7a6baa@37F7@ and ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaaaaa@@ that will satisfy the design specifications. To this end:

(i)The specification requires that

X 0 Y 0 < 35 20 =1.75 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam iwamaaBaaaleaacaaIWaaabeaaaOqaaiaadMfadaWgaaWcbaGaaGim aaqabaaaaOGaeyipaWZaaSaaaeaacaaIZaGaaGynaaqaaiaaikdaca aIWaaaaiabg2da9iaaigdacaGGUaGaaG4naiaaiwdaaaa@41DD@

for any value of ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHjpWDaaa@@ (remember ω=2πV/L MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHjpWDcqGH9a qpcaaIYaGaeqiWdaNaamOvaiaac+cacaWGmbaaaa@3D21@ ). Recall that X 0 =K Y 0 M(ω/ ω n ,ζ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGybWaaSbaaS qaaiaaicdaaeqaaOGaeyypa0Jaam4saiaadMfadaWgaaWcbaGaaGim aaqabaGccaWGnbGaaiikaiabeM8a3jaac+cacqaHjpWDdaWgaaWcba GaamOBaaqabaGccaGGSaGaeqOTdONaaiykaaaa@44F5@ and that K=1 for a base excited spring — MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8hfGaaa@@ mass system. This tells us that the magnification M= X 0 / Y 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGnbGaeyypa0 JaamiwamaaBaaaleaacaaIWaaabeaakiaac+cacaWGzbWaaSbaaSqa aiaaicdaaeqaaaaa@3B92@ has to be below 1.75 for any frequency. The graph shows that if ς>0.4 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg6 da+iaaicdacaGGUaGaaGinaaaa@3B10@ , the magnification never exceeds 1.75. We also see that smaller values of ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEaaa@@ make the suspension more effective (M is smaller) at high frequencies. So ς=0.4 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9iaaicdacaGGUaGaaGinaaaa@3B0E@ is a good choice.

If you prefer not to use the graph, you can use the approximation M max ≈1/(2ζ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGnbWaaSbaaS qaaiGac2gacaGGHbGaaiiEaaqabaGccqGHijYUcaaIXaGaai4laiaa cIcacaaIYaGaeqOTdONaaiykaaaa@@ which suggests that ζ>1/(2×1.75) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEcqGH+a GpcaaIXaGaai4laiaacIcacaaIYaGaey41aqRaaGymaiaac6cacaaI 3aGaaGynaiaacMcaaaa@40C2@ which gives ζ≈0.3 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEcqGHij YUcaaIWaGaaiOlaiaaiodaaaa@3B0D@ - but the approximation is not very accurate for such large values of ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEaaa@@ (to get a better estimate you’d have to maximize the formula for magnification with respect to ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHjpWDaaa@@ but that’s very messy).

(ii)Now, the frequency of base excitation at 55mph is

ω= 2πV L = 2π×0.447×55 10 =15.45  rad/s MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jabg2 da9maalaaabaGaaGOmaiabec8aWjaadAfaaeaacaWGmbaaaiabg2da 9maalaaabaGaaGOmaiabec8aWjabgEna0kaaicdacaGGUaGaaGinai aaisdacaaI3aGaey41aqRaaGynaiaaiwdaaeaacaaIXaGaaGimaaaa cqGH9aqpcaaIXaGaaGynaiaac6cacaaI0aGaaGynaiaabccacaqGGa GaaeOCaiaabggacaqGKbGaae4laiaabohaaaa@55F7@

We must choose system parameters so that, at this excitation frequency, X 0 / Y 0 <10/20=1/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccaGGVaGaamywamaaBaaaleaacaaIWaaabeaa kiabgYda8iaaigdacaaIWaGaai4laiaaikdacaaIWaGaeyypa0JaaG ymaiaac+cacaaIYaaaaa@@ .

This tells us that M must be less than ½ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa8xVaaaa@37AC@ when ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHjpWDaaa@@ is 15.45 rad/s or greater. We already know that ζ=0.4 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEcqGH9a qpcaaIWaGaaiOlaiaaisdaaaa@3A63@ , and following the curve for this value of ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEaaa@@ we see that M<1/2 if ω/ ω n >2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jaac+ cacqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH+aGpcaaIYaaaaa@3D73@ . Therefore, we must pick ω n <ω/2=7.7  rad/s MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabgYda8iabeM8a3jaac+cacaaIYaGaeyyp a0JaaG4naiaac6cacaaI3aGaaeiiaiaabccacaqGYbGaaeyyaiaabs gacaqGVaGaae4Caaaa@@ .

Again, if you prefer not to use the graph, you can also solve

M= { 1+ ( 2ςω/ ω n ) 2 } 1/2 { ( 1− ω 2 / ω n 2 ) 2 + ( 2ςω/ ω n ) 2 } 1/2 < 1 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eacqGH9a qpdaWcaaqaamaacmaabaGaaGymaiabgUcaRmaabmaabaGaaGOmaiab ek8awjabeM8a3jaac+cacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaki aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakiaawUhacaGL9baa daahaaWcbeqaaiaaigdacaGGVaGaaGOmaaaaaOqaamaacmaabaWaae WaaeaacaaIXaGaeyOeI0IaeqyYdC3aaWbaaSqabeaacaaIYaaaaOGa ai4laiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaOGaayjkai aawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaGaaGOm aiabek8awjabeM8a3jaac+cacqaHjpWDdaWgaaWcbaGaamOBaaqaba aakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakiaawUhacaGL 9baadaahaaWcbeqaaiaaigdacaGGVaGaaGOmaaaaaaGccqGH8aapda WcaaqaaiaaigdaaeaacaaIYaaaaaaa@663F@

for ω/ ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHjpWDcaGGVa GaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaa@3AE2@ , but this is a pain, and the graph is accurate enough for a design estimate.

Finally, we can compute properties of the system. We have that

ω n = k m ⇒7.7= k 0.44× ⇒k=78  kN/m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGUbaabeaakiabg2da9maakaaabaWaaSaaaeaacaWGRbaa baGaamyBaaaaaSqabaGccqGHshI3caaI3aGaaiOlaiaaiEdacqGH9a qpdaGcaaqaamaalaaabaGaam4AaaqaaiaaicdacaGGUaGaaGinaiaa isdacqGHxdaTcaaIZaGaaGimaiaaicdacaaIWaaaaaWcbeaakiabgk DiElaadUgacqGH9aqpcaaI3aGaaGioaiaabccacaqGGaGaae4Aaiaa b6eacaqGVaGaaeyBaaaa@556F@

Similarly

ς= λ 2 mk ⇒λ=2×0.4× ×.44× =8kNs/m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabek8awjabg2 da9maalaaabaGaeq4UdWgabaGaaGOmamaakaaabaGaamyBaiaadUga aSqabaaaaOGaeyO0H4Taeq4UdWMaeyypa0JaaGOmaiabgEna0kaaic dacaGGUaGaaGinaiabgEna0oaakaaabaGaaG4naiaaiIdacaaIWaGa aGimaiaaicdacqGHxdaTcaGGUaGaaGinaiaaisdacqGHxdaTcaaIZa GaaGimaiaaicdacaaIWaaaleqaaOGaeyypa0JaaGioaiaabUgacaqG obGaae4Caiaab+cacaqGTbaaaa@5CD2@

5.5 Solving differential equations for vibrating systems

Our goal in this course is to understand what the solutions to differential equations tell us about engineering problems we might need to solve. But if you have time on your hands, you might be interested in learning how to solve the differential equations. It’s fairly straightforward, if a little tedious algebraically. You will learn this material in future courses (applied math, and several more advanced engineering courses) whether you want to or not…

Review of complex numbers

It’s easiest to solve linear ODEs using complex variables. The following definitions and results are particularly useful:

Define i= −1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGPbGaeyypa0 ZaaOaaaeaacqGHsislcaaIXaaaleqaaaaa@392D@

Any complex number z can be split into imaginary and real parts as

z=a+ib MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG6bGaeyypa0 JaamyyaiabgUcaRiaadMgacaWGIbaaaa@3B18@

where a and b are two real numbers.

Define the complex conjugate as z ¯ =a−ib MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaceWG6bGbaebacq GH9aqpcaWGHbGaeyOeI0IaamyAaiaadkgaaaa@3B3B@

It follows that a=(z+ z ¯ )/2b=−i(z− z ¯ )/2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbGaeyypa0 JaaiikaiaadQhacqGHRaWkceWG6bGbaebacaGGPaGaai4laiaaikda caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caWGIbGaeyypa0JaeyOeI0IaamyA aiaacIcacaWG6bGaeyOeI0IabmOEayaaraGaaiykaiaac+cacaaIYa aaaa@@

The exponential of an imaginary number (Euler’s formula) is

e iθ =cosθ+isinθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGLbWaaWbaaS qabeaacaWGPbGaeqiUdehaaOGaeyypa0Jaci4yaiaac+gacaGGZbGa eqiUdeNaey4kaSIaamyAaiGacohacaGGPbGaaiOBaiabeI7aXbaa@@

You can prove this by taking the Taylor expansion of both sides of the formula

Euler’s formula enables us to write any complex number in polar form

a+ib=ρ e iθ ρ= a 2 + b 2 θ= tan −1 (b/a) a=ρcosθb=ρsinθ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadggacq GHRaWkcaWGPbGaamOyaiabg2da9iabeg8aYjaadwgadaahaaWcbeqa aiaadMgacqaH4oqCaaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeqyWdiNaeyypa0ZaaOaa aeaacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCa aaleqabaGaaGOmaaaaaeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabeI7aXjabg2da9i GacshacaGGHbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaakiaa cIcacaWGIbGaai4laiaadggacaGGPaaabaGaamyyaiabg2da9iabeg 8aYjGacogacaGGVbGaai4CaiabeI7aXjaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGIbGaey ypa0JaeqyWdiNaci4CaiaacMgacaGGUbGaeqiUdehaaaa@910E@

Euler’s formula also allows us to represent trig functions as complex exponentials

cosθ=( e iθ + e −iθ )/2sinθ=−i( e iθ − e −iθ )/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaciGGJbGaai4Bai aacohacqaH4oqCcqGH9aqpcaGGOaGaamyzamaaCaaaleqabaGaamyA aiabeI7aXbaakiabgUcaRiaadwgadaahaaWcbeqaaiabgkHiTiaadM gacqaH4oqCaaGccaGGPaGaai4laiaaikdacaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlGacohacaGGPbGaaiOBaiabeI7aXjabg2da 9iabgkHiTiaadMgacaGGOaGaamyzamaaCaaaleqabaGaamyAaiabeI 7aXbaakiabgkHiTiaadwgadaahaaWcbeqaaiabgkHiTiaadMgacqaH 4oqCaaGccaGGPaGaai4laiaaikdaaaa@705A@

Note that

d e iωt dt =iω e iωt d 2 e iωt d t 2 =− ω 2 e iωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWcaaqaaiaads gacaWGLbWaaWbaaSqabeaacaWGPbGaeqyYdCNaamiDaaaaaOqaaiaa dsgacaWG0baaaiabg2da9iaadMgacqaHjpWDcaWGLbWaaWbaaSqabe aacaWGPbGaeqyYdCNaamiDaaaakiaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGLbWa aWbaaSqabeaacaWGPbGaeqyYdCNaamiDaaaaaOqaaiaadsgacaWG0b WaaWbaaSqabeaacaaIYaaaaaaakiabg2da9iabgkHiTiabeM8a3naa CaaaleqabaGaaGOmaaaakiaadwgadaahaaWcbeqaaiaadMgacqaHjp WDcaWG0baaaaaa@6CDF@

Solution to the equation of motion for an undamped harmonic oscillator

Solve 1 ω n 2 d 2 x d t 2 +x=C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaamiEaiabg2da 9iaadoeaaaa@@  with initial conditions x= x 0 dx/dt= v 0 t=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 JaamiEamaaBaaaleaacaaIWaaabeaakiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caWGKbGaamiEaiaac+cacaWGKbGaam iDaiabg2da9iaadAhadaWgaaWcbaGaaGimaaqabaGccaaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaamiDaiabg2da9iaaicdaaaa@@  

Guess a solution of the form x=C+A e λt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeacaWGLbWaaWbaaSqabeaacqaH7oaBcaWG 0baaaaaa@3DAD@ where A and λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBaaa@372A@ are two complex numbers to be determined (this may seem a cheat, but actually there are only two ways to do an integral (1) guess a solution, differentiate it, and see if the answer is correct; and (2) rearrange the integral into another form with a known solution. We know an exponential is a good guess for x because when an exponential is differentiated it stays an exponential). Substitute this into our ODE

λ 2 ω n 2 A e λt +A e λt =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeq 4UdW2aaWbaaSqabeaacaaIYaaaaaGcbaGaeqyYdC3aa0baaSqaaiaa d6gaaeaacaaIYaaaaaaakiaadgeacaWGLbWaaWbaaSqabeaacqaH7o aBcaWG0baaaOGaey4kaSIaamyqaiaadwgadaahaaWcbeqaaiabeU7a SjaadshaaaGccqGH9aqpcaaIWaaaaa@486D@

We can satisfy this for any A by choosing λ 2 / ω n 2 =−1⇒λ=± −1 ω n =±i ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBdaahaa WcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaa caaIYaaaaOGaeyypa0JaeyOeI0IaaGymaiabgkDiElabeU7aSjabg2 da9iabgglaXoaakaaabaGaeyOeI0IaaGymaaWcbeaakiabeM8a3naa BaaaleaacaWGUbaabeaakiabg2da9iabgglaXkaadMgacqaHjpWDda WgaaWcbaGaamOBaaqabaaaaa@51C7@ . This gives us two families of solutions to the equations, one with x=C+Aexp(i ω n t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeaciGGLbGaaiiEaiaacchacaGGOaGaamyA aiabeM8a3naaBaaaleaacaWGUbaabeaakiaadshacaGGPaaaaa@42FA@ and another with x=Aexp(−i ω n t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 JaamyqaiGacwgacaGG4bGaaiiCaiaacIcacqGHsislcaWGPbGaeqyY dC3aaSbaaSqaaiaad6gaaeqaaOGaamiDaiaacMcaaaa@423D@ . The most general solution is the sum of these, with different coefficients

x=C+ A 1 e i ω n t + A 2 e −i ω n t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeadaWgaaWcbaGaaGymaaqabaGccaWGLbWa aWbaaSqabeaacaWGPbGaeqyYdC3aaSbaaWqaaiaad6gaaeqaaSGaam iDaaaakiabgUcaRiaadgeadaWgaaWcbaGaaGOmaaqabaGccaWGLbWa aWbaaSqabeaacqGHsislcaWGPbGaeqyYdC3aaSbaaWqaaiaad6gaae qaaSGaamiDaaaaaaa@4A57@

We need to find A 1 , A 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadgeadaWgaaWcbaGaaGOmaaqabaaa aa@398B@ : we can do this by substituting t=0 into x and using the given values of x at t=0

x(0)= A 1 + A 2 +C= x 0 ⇒ A 1 + A 2 = x 0 −C dx dt | t=0 =i ω n ( A 1 − A 2 )= v 0 ⇒ A 1 − A 2 =−i v 0 / ω n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadIhaca GGOaGaaGimaiaacMcacqGH9aqpcaWGbbWaaSbaaSqaaiaaigdaaeqa aOGaey4kaSIaamyqamaaBaaaleaacaaIYaaabeaakiabgUcaRiaado eacqGH9aqpcaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyO0H4Taamyq amaaBaaaleaacaaIXaaabeaakiabgUcaRiaadgeadaWgaaWcbaGaaG OmaaqabaGccqGH9aqpcaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyOe I0Iaam4qaaqaamaaeiaabaWaaSaaaeaacaWGKbGaamiEaaqaaiaads gacaWG0baaaaGaayjcSdWaaSbaaSqaaiaadshacqGH9aqpcaaIWaaa beaakiabg2da9iaadMgacqaHjpWDdaWgaaWcbaGaamOBaaqabaGcca GGOaGaamyqamaaBaaaleaacaaIXaaabeaakiabgkHiTiaadgeadaWg aaWcbaGaaGOmaaqabaGccaGGPaGaeyypa0JaamODamaaBaaaleaaca aIWaaabeaakiabgkDiElaadgeadaWgaaWcbaGaaGymaaqabaGccqGH sislcaWGbbWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaeyOeI0Iaam yAaiaadAhadaWgaaWcbaGaaGimaaqabaGccaGGVaGaeqyYdC3aaSba aSqaaiaad6gaaeqaaaaaaa@72E7@

Add and subtract these two equations to see that

A 1 = 1 2 ( ( x 0 −C)−i v 0 ω n ) A 2 = 1 2 ( ( x 0 −C)+i v 0 ω n ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaa daqadaqaaiaacIcacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0 Iaam4qaiaacMcacqGHsislcaWGPbWaaSaaaeaacaWG2bWaaSbaaSqa aiaaicdaaeqaaaGcbaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaaaO GaayjkaiaawMcaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyqamaaBaaaleaaca aIYaaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaeWa aeaacaGGOaGaamiEamaaBaaaleaacaaIWaaabeaakiabgkHiTiaado eacaGGPaGaey4kaSIaamyAamaalaaabaGaamODamaaBaaaleaacaaI WaaabeaaaOqaaiabeM8a3naaBaaaleaacaWGUbaabeaaaaaakiaawI cacaGLPaaaaaa@698D@

We can use Euler’s formula to re-write this as

A 1 =− i 2 X 0 e iϕ A 2 = i 2 X 0 e −iϕ X 0 = ( x 0 −C ) 2 + v 0 2 / ω n 2 sinϕ= x 0 −C ( x 0 −C ) 2 + v 0 2 / ω n 2 cosϕ= v 0 / ω n ( x 0 −C ) 2 + v 0 2 / ω n 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadgeada WgaaWcbaGaaGymaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaadMga aeaacaaIYaaaaiaadIfadaWgaaWcbaGaaGimaaqabaGccaWGLbWaaW baaSqabeaacaWGPbGaeqy1dygaaOGaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyqamaaBaaaleaaca aIYaaabeaakiabg2da9maalaaabaGaamyAaaqaaiaaikdaaaGaamiw amaaBaaaleaacaaIWaaabeaakiaadwgadaahaaWcbeqaaiabgkHiTi aadMgacqaHvpGzaaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8oabaGaamiwamaaBaaaleaaca aIWaaabeaakiabg2da9maakaaabaWaaeWaaeaacaWG4bWaaSbaaSqa aiaaicdaaeqaaOGaeyOeI0Iaam4qaaGaayjkaiaawMcaamaaCaaale qabaGaaGOmaaaakiabgUcaRiaadAhadaqhaaWcbaGaaGimaaqaaiaa ikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaa qabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua ci4CaiaacMgacaGGUbGaeqy1dyMaeyypa0ZaaSaaaeaacaWG4bWaaS baaSqaaiaaicdaaeqaaOGaeyOeI0Iaam4qaaqaamaakaaabaWaaeWa aeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0Iaam4qaaGaay jkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadAhadaqh aaWcbaGaaGimaaqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaai aad6gaaeaacaaIYaaaaaqabaaaaOGaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8Uaci4yaiaac+gacaGGZbGaeqy1dyMaeyypa0ZaaS aaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaOGaai4laiabeM8a3naa BaaaleaacaWGUbaabeaaaOqaamaakaaabaWaaeWaaeaacaWG4bWaaS baaSqaaiaaicdaaeqaaOGaeyOeI0Iaam4qaaGaayjkaiaawMcaamaa CaaaleqabaGaaGOmaaaakiabgUcaRiaadAhadaqhaaWcbaGaaGimaa qaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaI Yaaaaaqabaaaaaaaaa@BEAE@

(to see this just substitute X 0 ,ϕ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGybWaaSbaaS qaaiaaicdaaeqaaOGaaiilaiabew9aMbaa@39BB@ into the formulas and use Euler’s formula to show A 1 , A 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadgeadaWgaaWcbaGaaGOmaaqabaaa aa@398B@ are correct). Finally substitute A 1 , A 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadgeadaWgaaWcbaGaaGOmaaqabaaa aa@398B@ into the general solution for x to see that

x=C− i 2 X 0 e iϕ e i ω n t + i 2 X 0 e −iϕ e −i ω n t =C− i 2 X 0 ( e i( ω n t+ϕ) − e −i( ω n t+ϕ) ) =C+ X 0 sin( ω n t+ϕ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadIhacq GH9aqpcaWGdbGaeyOeI0YaaSaaaeaacaWGPbaabaGaaGOmaaaacaWG ybWaaSbaaSqaaiaaicdaaeqaaOGaamyzamaaCaaaleqabaGaamyAai abew9aMbaakiaadwgadaahaaWcbeqaaiaadMgacqaHjpWDdaWgaaad baGaamOBaaqabaWccaWG0baaaOGaey4kaSYaaSaaaeaacaWGPbaaba GaaGOmaaaacaWGybWaaSbaaSqaaiaaicdaaeqaaOGaamyzamaaCaaa leqabaGaeyOeI0IaamyAaiabew9aMbaakiaadwgadaahaaWcbeqaai abgkHiTiaadMgacqaHjpWDdaWgaaadbaGaamOBaaqabaWccaWG0baa aOGaeyypa0Jaam4qaiabgkHiTmaalaaabaGaamyAaaqaaiaaikdaaa GaamiwamaaBaaaleaacaaIWaaabeaakmaabmaabaGaamyzamaaCaaa leqabaGaamyAaiaacIcacqaHjpWDdaWgaaadbaGaamOBaaqabaWcca WG0bGaey4kaSIaeqy1dyMaaiykaaaakiabgkHiTiaadwgadaahaaWc beqaaiabgkHiTiaadMgacaGGOaGaeqyYdC3aaSbaaWqaaiaad6gaae qaaSGaamiDaiabgUcaRiabew9aMjaacMcaaaaakiaawIcacaGLPaaa aeaacqGH9aqpcaWGdbGaey4kaSIaamiwamaaBaaaleaacaaIWaaabe aakiGacohacaGGPbGaaiOBaiaacIcacqaHjpWDdaWgaaWcbaGaamOB aaqabaGccaWG0bGaey4kaSIaeqy1dyMaaiykaaaaaa@839F@

This agrees with the answer on the formula sheet.

Solution to the equation of motion for a free damped system

Solve 1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaaaa@4E43@ with initial conditions x= x 0 dx/dt= v 0 t=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 JaamiEamaaBaaaleaacaaIWaaabeaakiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caWGKbGaamiEaiaac+cacaWGKbGaam iDaiabg2da9iaadAhadaWgaaWcbaGaaGimaaqabaGccaaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaamiDaiabg2da9iaaicdaaaa@@

As before we guess a solution x=C+A e λt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeacaWGLbWaaWbaaSqabeaacqaH7oaBcaWG 0baaaaaa@3DAD@ where A and λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBaaa@372A@ are two complex numbers to be determined. Substituting into the equation:

( λ 2 ω n 2 + 2ςλ ω n +1 )A e λt =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaWaaS aaaeaacqaH7oaBdaahaaWcbeqaaiaaikdaaaaakeaacqaHjpWDdaqh aaWcbaGaamOBaaqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaIYa GaeqOWdyLaeq4UdWgabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaa kiabgUcaRiaaigdaaiaawIcacaGLPaaacaWGbbGaamyzamaaCaaale qabaGaeq4UdWMaamiDaaaakiabg2da9iaaicdaaaa@4E1A@

This gives a quadratic equation for λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBaaa@372A@ (it is called the ‘characteristic equation’ for the differential equation). It has solutions

λ=−ζ ω n ∓ ω n ζ 2 −1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcqGH9a qpcqGHsislcqaH2oGEcqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqWI tisBcqaHjpWDdaWgaaWcbaGaamOBaaqabaGcdaGcaaqaaiabeA7a6n aaCaaaleqabaGaaGOmaaaakiabgkHiTiaaigdaaSqabaaaaa@466C@

Depending on the value of ζ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEaaa@@ we find

· ζ>1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEcqGH+a GpcaaIXaaaaa@38F6@ (overdamped) – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ two real values of λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBaaa@372A@ λ=−ζ ω n ± ω d MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcqGH9a qpcqGHsislcqaH2oGEcqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH XcqScqaHjpWDdaWgaaWcbaGaamizaaqabaaaaa@42A0@

· ζ=1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEcqGH9a qpcaaIXaaaaa@38F4@ (critical damping): λ=− ω n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcqGH9a qpcqGHsislcqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaa@3C09@

· ζ<1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH2oGEcqGH8a apcaaIXaaaaa@38F2@ (underdamped) – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ two complex values of λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBaaa@372A@ λ=−ζ ω n ±i ω d MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH7oaBcqGH9a qpcqGHsislcqaH2oGEcqaHjpWDdaWgaaWcbaGaamOBaaqabaGccqGH XcqScaWGPbGaeqyYdC3aaSbaaSqaaiaadsgaaeqaaaaa@438E@

where we have defined ω d = ω n | ζ 2 −1 | MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHjpWDdaWgaa WcbaGaamizaaqabaGccqGH9aqpcqaHjpWDdaWgaaWcbaGaamOBaaqa baGcdaGcaaqaamaaemaabaGaeqOTdO3aaWbaaSqabeaacaaIYaaaaO GaeyOeI0IaaGymaaGaay5bSlaawIa7aaWcbeaaaaa@43F2@ To write the answers in terms of real valued functions we need to treat these cases separately.

Overdamped solution: We have that

x=C+ A 1 e (−ζ ω n + ω d )t + A 2 e (−ζ ω n − ω d )t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeadaWgaaWcbaGaaGymaaqabaGccaWGLbWa aWbaaSqabeaacaGGOaGaeyOeI0IaeqOTdONaeqyYdC3aaSbaaWqaai aad6gaaeqaaSGaey4kaSIaeqyYdC3aaSbaaWqaaiaadsgaaeqaaSGa aiykaiaadshaaaGccqGHRaWkcaWGbbWaaSbaaSqaaiaaikdaaeqaaO GaamyzamaaCaaaleqabaGaaiikaiabgkHiTiabeA7a6jabeM8a3naa BaaameaacaWGUbaabeaaliabgkHiTiabeM8a3naaBaaameaacaWGKb aabeaaliaacMcacaWG0baaaaaa@573F@

We can use the initial conditions to determine A 1 , A 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadgeadaWgaaWcbaGaaGOmaaqabaaa aa@398B@ :

x(0)=C+ A 1 + A 2 = x 0 dx dt | t=0 = A 1 ( ω d −ζ ω n )− A 1 ( ω d +ζ ω n )= v 0 ⇒ A 1 = v 0 +(ς ω n + ω d )( x 0 −C) 2 ω d A 2 =− v 0 +(ς ω n − ω d )( x 0 −C) 2 ω d MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadIhaca GGOaGaaGimaiaacMcacqGH9aqpcaWGdbGaey4kaSIaamyqamaaBaaa leaacaaIXaaabeaakiabgUcaRiaadgeadaWgaaWcbaGaaGOmaaqaba GccqGH9aqpcaWG4bWaaSbaaSqaaiaaicdaaeqaaaGcbaWaaqGaaeaa daWcaaqaaiaadsgacaWG4baabaGaamizaiaadshaaaaacaGLiWoada WgaaWcbaGaamiDaiabg2da9iaaicdaaeqaaOGaeyypa0Jaamyqamaa BaaaleaacaaIXaaabeaakiaacIcacqaHjpWDdaWgaaWcbaGaamizaa qabaGccqGHsislcqaH2oGEcqaHjpWDdaWgaaWcbaGaamOBaaqabaGc caGGPaGaeyOeI0IaamyqamaaBaaaleaacaaIXaaabeaakiaacIcacq aHjpWDdaWgaaWcbaGaamizaaqabaGccqGHRaWkcqaH2oGEcqaHjpWD daWgaaWcbaGaamOBaaqabaGccaGGPaGaeyypa0JaamODamaaBaaale aacaaIWaaabeaaaOqaaiabgkDiElaadgeadaWgaaWcbaGaaGymaaqa baGccqGH9aqpdaWcaaqaaiaadAhadaWgaaWcbaGaaGimaaqabaGccq GHRaWkcaGGOaGaeqOWdyLaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGa ey4kaSIaeqyYdC3aaSbaaSqaaiaadsgaaeqaaOGaaiykaiaacIcaca WG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0Iaam4qaiaacMcaaeaa caaIYaGaeqyYdC3aaSbaaSqaaiaadsgaaeqaaaaakiaaykW7caaMc8 UaaGPaVlaaykW7caWGbbWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0Ja eyOeI0YaaSaaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaS Iaaiikaiabek8awjabeM8a3naaBaaaleaacaWGUbaabeaakiabgkHi TiabeM8a3naaBaaaleaacaWGKbaabeaakiaacMcacaGGOaGaamiEam aaBaaaleaacaaIWaaabeaakiabgkHiTiaadoeacaGGPaaabaGaaGOm aiabeM8a3naaBaaaleaacaWGKbaabeaaaaaaaaa@A066@

Hence

x(t)=C+exp(−ς ω n t){ v 0 +(ς ω n + ω d )( x 0 −C) 2 ω d exp( ω d t)− v 0 +(ς ω n − ω d )( x 0 −C) 2 ω d exp(− ω d t) } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSIaciyzaiaacIhacaGG WbGaaiikaiabgkHiTiabek8awjabeM8a3naaBaaaleaacaWGUbaabe aakiaadshacaGGPaWaaiWaaeaadaWcaaqaaiaadAhadaWgaaWcbaGa aGimaaqabaGccqGHRaWkcaGGOaGaeqOWdyLaeqyYdC3aaSbaaSqaai aad6gaaeqaaOGaey4kaSIaeqyYdC3aaSbaaSqaaiaadsgaaeqaaOGa aiykaiaacIcacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0Iaam 4qaiaacMcaaeaacaaIYaGaeqyYdC3aaSbaaSqaaiaadsgaaeqaaaaa kiGacwgacaGG4bGaaiiCaiaacIcacqaHjpWDdaWgaaWcbaGaamizaa qabaGccaWG0bGaaiykaiabgkHiTmaalaaabaGaamODamaaBaaaleaa caaIWaaabeaakiabgUcaRiaacIcacqaHcpGvcqaHjpWDdaWgaaWcba GaamOBaaqabaGccqGHsislcqaHjpWDdaWgaaWcbaGaamizaaqabaGc caGGPaGaaiikaiaadIhadaWgaaWcbaGaaGimaaqabaGccqGHsislca WGdbGaaiykaaqaaiaaikdacqaHjpWDdaWgaaWcbaGaamizaaqabaaa aOGaciyzaiaacIhacaGGWbGaaiikaiabgkHiTiabeM8a3naaBaaale aacaWGKbaabeaakiaadshacaGGPaaacaGL7bGaayzFaaaaaa@85B9@

Critically damped solution: our guess for the critically damped solution gives only x=C+ A 1 e −ζ ω n t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeadaWgaaWcbaGaaGymaaqabaGccaWGLbWa aWbaaSqabeaacqGHsislcqaH2oGEcqaHjpWDdaWgaaadbaGaamOBaa qabaWccaWG0baaaaaa@428C@ which cannot satisfy the initial conditions on both x and dx/dt, so the solution is incomplete. We have to look around for another solution – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ it turns out that

x=C+ A 1 e − ω n t + A 2 t e − ω n t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeadaWgaaWcbaGaaGymaaqabaGccaWGLbWa aWbaaSqabeaacqGHsislcqaHjpWDdaWgaaadbaGaamOBaaqabaWcca WG0baaaOGaey4kaSIaamyqamaaBaaaleaacaaIYaaabeaakiaadsha caWGLbWaaWbaaSqabeaacqGHsislcqaHjpWDdaWgaaadbaGaamOBaa qabaWccaWG0baaaaaa@4A61@

will also satisfy the differential equation (this is a standard trick in situations where the characteristic equation has repeated roots). We can solve for A 1 , A 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadgeadaWgaaWcbaGaaGOmaaqabaaa aa@398B@ using the initial conditions:

x(0)=C+ A 1 = x 0 dx dt | t=0 =− ω n A 1 + A 2 = v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadIhaca GGOaGaaGimaiaacMcacqGH9aqpcaWGdbGaey4kaSIaamyqamaaBaaa leaacaaIXaaabeaakiabg2da9iaadIhadaWgaaWcbaGaaGimaaqaba aakeaadaabcaqaamaalaaabaGaamizaiaadIhaaeaacaWGKbGaamiD aaaaaiaawIa7amaaBaaaleaacaWG0bGaeyypa0JaaGimaaqabaGccq GH9aqpcqGHsislcqaHjpWDdaWgaaWcbaGaamOBaaqabaGccaWGbbWa aSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamyqamaaBaaaleaacaaIYa aabeaakiabg2da9iaadAhadaWgaaWcbaGaaGimaaqabaaaaaa@@

It follows that A 1 = x 0 −C A 2 = v 0 + ω n ( x 0 −C) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaeyypa0JaamiEamaaBaaaleaacaaIWaaabeaa kiabgkHiTiaadoeacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadg eadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaWG2bWaaSbaaSqaaiaa icdaaeqaaOGaey4kaSIaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaai ikaiaadIhadaWgaaWcbaGaaGimaaqabaGccqGHsislcaWGdbGaaiyk aaaa@@ so the solution is

x(t)=C+{ ( x 0 −C)+[ v 0 + ω n ( x 0 −C) ]t }exp(− ω n t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSYaaiWaaeaacaGGOaGa amiEamaaBaaaleaacaaIWaaabeaakiabgkHiTiaadoeacaGGPaGaey 4kaSYaamWaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIa eqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaaiikaiaadIhadaWgaaWcba GaaGimaaqabaGccqGHsislcaWGdbGaaiykaaGaay5waiaaw2faaiaa dshaaiaawUhacaGL9baaciGGLbGaaiiEaiaacchacaGGOaGaeyOeI0 IaeqyYdC3aaSbaaSqaaiaad6gaaeqaaOGaamiDaiaacMcaaaa@5AFF@

Underdamped solution: For this case

x=C+ A 1 e (−ζ ω n +i ω d )t + A 2 e (−ζ ω n −i ω d )t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadgeadaWgaaWcbaGaaGymaaqabaGccaWGLbWa aWbaaSqabeaacaGGOaGaeyOeI0IaeqOTdONaeqyYdC3aaSbaaWqaai aad6gaaeqaaSGaey4kaSIaamyAaiabeM8a3naaBaaameaacaWGKbaa beaaliaacMcacaWG0baaaOGaey4kaSIaamyqamaaBaaaleaacaaIYa aabeaakiaadwgadaahaaWcbeqaaiaacIcacqGHsislcqaH2oGEcqaH jpWDdaWgaaadbaGaamOBaaqabaWccqGHsislcaWGPbGaeqyYdC3aaS baaWqaaiaadsgaaeqaaSGaaiykaiaadshaaaaaaa@591B@

We can use the initial conditions to determine A 1 , A 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGbbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadgeadaWgaaWcbaGaaGOmaaqabaaa aa@398B@ (which are now complex):

x(0)=C+ A 1 + A 2 = x 0 dx dt | t=0 = A 1 (i ω d −ζ ω n )− A 1 (i ω d +ζ ω n )= v 0 ⇒ A 1 =−i v 0 +(ς ω n +i ω d )( x 0 −C) 2 ω d A 2 =i v 0 +(ς ω n −i ω d )( x 0 −C) 2 ω d MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadIhaca GGOaGaaGimaiaacMcacqGH9aqpcaWGdbGaey4kaSIaamyqamaaBaaa leaacaaIXaaabeaakiabgUcaRiaadgeadaWgaaWcbaGaaGOmaaqaba GccqGH9aqpcaWG4bWaaSbaaSqaaiaaicdaaeqaaaGcbaWaaqGaaeaa daWcaaqaaiaadsgacaWG4baabaGaamizaiaadshaaaaacaGLiWoada WgaaWcbaGaamiDaiabg2da9iaaicdaaeqaaOGaeyypa0Jaamyqamaa BaaaleaacaaIXaaabeaakiaacIcacaWGPbGaeqyYdC3aaSbaaSqaai aadsgaaeqaaOGaeyOeI0IaeqOTdONaeqyYdC3aaSbaaSqaaiaad6ga aeqaaOGaaiykaiabgkHiTiaadgeadaWgaaWcbaGaaGymaaqabaGcca GGOaGaamyAaiabeM8a3naaBaaaleaacaWGKbaabeaakiabgUcaRiab eA7a6jabeM8a3naaBaaaleaacaWGUbaabeaakiaacMcacqGH9aqpca WG2bWaaSbaaSqaaiaaicdaaeqaaaGcbaGaeyO0H4TaamyqamaaBaaa leaacaaIXaaabeaakiabg2da9iabgkHiTiaadMgadaWcaaqaaiaadA hadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaGGOaGaeqOWdyLaeqyY dC3aaSbaaSqaaiaad6gaaeqaaOGaey4kaSIaamyAaiabeM8a3naaBa aaleaacaWGKbaabeaakiaacMcacaGGOaGaamiEamaaBaaaleaacaaI WaaabeaakiabgkHiTiaadoeacaGGPaaabaGaaGOmaiabeM8a3naaBa aaleaacaWGKbaabeaaaaGccaaMc8UaaGPaVlaaykW7caaMc8Uaamyq amaaBaaaleaacaaIYaaabeaakiabg2da9iaadMgadaWcaaqaaiaadA hadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaGGOaGaeqOWdyLaeqyY dC3aaSbaaSqaaiaad6gaaeqaaOGaeyOeI0IaamyAaiabeM8a3naaBa aaleaacaWGKbaabeaakiaacMcacaGGOaGaamiEamaaBaaaleaacaaI WaaabeaakiabgkHiTiaadoeacaGGPaaabaGaaGOmaiabeM8a3naaBa aaleaacaWGKbaabeaaaaaaaaa@A5FA@

We can substitute this back into the solution and re-arrange the result

x(t)=C+exp(−ς ω n t){ ( x 0 −C) 1 2 ( e i ω d t + e −i ω d t )− v 0 +ς ω n ( x 0 −C) ω d i 2 ( e i ω d t − e −i ω d t ) } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSIaciyzaiaacIhacaGG WbGaaiikaiabgkHiTiabek8awjabeM8a3naaBaaaleaacaWGUbaabe aakiaadshacaGGPaWaaiWaaeaacaGGOaGaamiEamaaBaaaleaacaaI WaaabeaakiabgkHiTiaadoeacaGGPaWaaSaaaeaacaaIXaaabaGaaG OmaaaadaqadaqaaiaadwgadaahaaWcbeqaaiaadMgacqaHjpWDdaWg aaadbaGaamizaaqabaWccaWG0baaaOGaey4kaSIaamyzamaaCaaale qabaGaeyOeI0IaamyAaiabeM8a3naaBaaameaacaWGKbaabeaaliaa dshaaaaakiaawIcacaGLPaaacqGHsisldaWcaaqaaiaadAhadaWgaa WcbaGaaGimaaqabaGccqGHRaWkcqaHcpGvcqaHjpWDdaWgaaWcbaGa amOBaaqabaGccaGGOaGaamiEamaaBaaaleaacaaIWaaabeaakiabgk HiTiaadoeacaGGPaaabaGaeqyYdCNaaCjaVpaaBaaaleaacaWGKbaa beaaaaGcdaWcaaqaaiaadMgaaeaacaaIYaaaamaabmaabaGaamyzam aaCaaaleqabaGaamyAaiabeM8a3naaBaaameaacaWGKbaabeaaliaa dshaaaGccqGHsislcaWGLbWaaWbaaSqabeaacqGHsislcaWGPbGaeq yYdC3aaSbaaWqaaiaadsgaaeqaaSGaamiDaaaaaOGaayjkaiaawMca aaGaay5Eaiaaw2haaaaa@81C1@

Finally we recognize the combinations of complex exponentials as trig functions, giving

x(t)=C+exp(−ς ω n t){ ( x 0 −C)cos ω d t+ v 0 +ς ω n ( x 0 −C) ω d sin ω d t } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOa GaamiDaiaacMcacqGH9aqpcaWGdbGaey4kaSIaciyzaiaacIhacaGG WbGaaiikaiabgkHiTiabek8awjabeM8a3naaBaaaleaacaWGUbaabe aakiaadshacaGGPaWaaiWaaeaacaGGOaGaamiEamaaBaaaleaacaaI WaaabeaakiabgkHiTiaadoeacaGGPaGaci4yaiaac+gacaGGZbGaeq yYdC3aaSbaaSqaaiaadsgaaeqaaOGaamiDaiabgUcaRmaalaaabaGa amODamaaBaaaleaacaaIWaaabeaakiabgUcaRiabek8awjabeM8a3n aaBaaaleaacaWGUbaabeaakiaacIcacaWG4bWaaSbaaSqaaiaaicda aeqaaOGaeyOeI0Iaam4qaiaacMcaaeaacqaHjpWDcaWLa8+aaSbaaS qaaiaadsgaaeqaaaaakiGacohacaGGPbGaaiOBaiabeM8a3naaBaaa leaacaWGKbaabeaakiaadshaaiaawUhacaGL9baaaaa@6D57@

Solution to the equation of motion for a system subjected to harmonic external force

Solve 1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C+K F 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaiabgUcaRiaadUeacaWGgbWaaSbaaSqaaiaaicdaae qaaOGaci4CaiaacMgacaGGUbGaeqyYdCNaamiDaaaa@574E@ with initial conditions x= x 0 dx/dt= v 0 t=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 JaamiEamaaBaaaleaacaaIWaaabeaakiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caWGKbGaamiEaiaac+cacaWGKbGaam iDaiabg2da9iaadAhadaWgaaWcbaGaaGimaaqabaGccaaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaamiDaiabg2da9iaaicdaaaa@@

It is helpful to replace the trig function with its equivalent representation in terms of complex exponentials

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C−K F 0 i 2 ( e iωt − e −iωt ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaiabgkHiTiaadUeacaWGgbWaaSbaaSqaaiaaicdaae qaaOWaaSaaaeaacaWGPbaabaGaaGOmaaaadaqadaqaaiaadwgadaah aaWcbeqaaiaadMgacqaHjpWDcaWG0baaaOGaeyOeI0IaamyzamaaCa aaleqabaGaeyOeI0IaamyAaiabeM8a3jaadshaaaaakiaawIcacaGL Paaaaaa@@

We guess a solution of the form

x(t)= x p (t)+ x h (t) x p (t)=− i 2 ( B 1 e iωt − B 2 e −iωt ) x h =C+A e λt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaadIhaca GGOaGaamiDaiaacMcacqGH9aqpcaWG4bWaaSbaaSqaaiaadchaaeqa aOGaaiikaiaadshacaGGPaGaey4kaSIaamiEamaaBaaaleaacaWGOb aabeaakiaacIcacaWG0bGaaiykaaqaaiaadIhadaWgaaWcbaGaamiC aaqabaGccaGGOaGaamiDaiaacMcacqGH9aqpcqGHsisldaWcaaqaai aadMgaaeaacaaIYaaaamaabmaabaGaamOqamaaBaaaleaacaaIXaaa beaakiaadwgadaahaaWcbeqaaiaadMgacqaHjpWDcaWG0baaaOGaey OeI0IaamOqamaaBaaaleaacaaIYaaabeaakiaadwgadaahaaWcbeqa aiabgkHiTiaadMgacqaHjpWDcaWG0baaaaGccaGLOaGaayzkaaGaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaadIhadaWgaaWcbaGaamiAaaqabaGccqGH9aqpcaWGdb Gaey4kaSIaamyqaiaadwgadaahaaWcbeqaaiabeU7aSjaadshaaaaa aaa@750D@

where B 1 , B 2 ,A,λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGcbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadkeadaWgaaWcbaGaaGOmaaqabaGc caGGSaGaamyqaiaacYcacqaH7oaBaaa@3D71@ are complex numbers to be determined. Substituting into the ODE:

( λ 2 ω n 2 + 2ςλ ω n +1 )A e λt − i 2 ( ( 1− ω 2 ω n 2 + 2ςω ω n i ) B 1 e iωt −( 1− ω 2 ω n 2 − 2ςω ω n i ) B 2 e −iωt )=−K F 0 i 2 ( e iωt − e −iωt ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaWaaS aaaeaacqaH7oaBdaahaaWcbeqaaiaaikdaaaaakeaacqaHjpWDdaqh aaWcbaGaamOBaaqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaIYa GaeqOWdyLaeq4UdWgabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaa kiabgUcaRiaaigdaaiaawIcacaGLPaaacaWGbbGaamyzamaaCaaale qabaGaeq4UdWMaamiDaaaakiabgkHiTmaalaaabaGaamyAaaqaaiaa ikdaaaWaaeWaaeaadaqadaqaaiaaigdacqGHsisldaWcaaqaaiabeM 8a3naaCaaaleqabaGaaGOmaaaaaOqaaiabeM8a3naaDaaaleaacaWG UbaabaGaaGOmaaaaaaGccqGHRaWkdaWcaaqaaiaaikdacqaHcpGvcq aHjpWDaeaacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaOGaamyAaaGa ayjkaiaawMcaaiaadkeadaWgaaWcbaGaaGymaaqabaGccaWGLbWaaW baaSqabeaacaWGPbGaeqyYdCNaamiDaaaakiabgkHiTmaabmaabaGa aGymaiabgkHiTmaalaaabaGaeqyYdC3aaWbaaSqabeaacaaIYaaaaa GcbaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaaaakiabgkHi TmaalaaabaGaaGOmaiabek8awjabeM8a3bqaaiabeM8a3naaBaaale aacaWGUbaabeaaaaGccaWGPbaacaGLOaGaayzkaaGaamOqamaaBaaa leaacaaIYaaabeaakiaadwgadaahaaWcbeqaaiabgkHiTiaadMgacq aHjpWDcaWG0baaaaGccaGLOaGaayzkaaGaaGPaVlaaykW7caaMc8Ua eyypa0JaeyOeI0Iaam4saiaadAeadaWgaaWcbaGaaGimaaqabaGcda WcaaqaaiaadMgaaeaacaaIYaaaamaabmaabaGaamyzamaaCaaaleqa baGaamyAaiabeM8a3jaadshaaaGccqGHsislcaWGLbWaaWbaaSqabe aacqGHsislcaWGPbGaeqyYdCNaamiDaaaaaOGaayjkaiaawMcaaaaa @9CE0@

We can satisfy this by setting

B 1 ( 1− ω 2 ω n 2 + 2ςω ω n i )=K F 0 B 2 ( 1− ω 2 ω n 2 − 2ςω ω n i )=K F 0 ( λ 2 ω n 2 + 2ςλ ω n +1 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadkeadaWgaa WcbaGaaGymaaqabaGcdaqadaqaaiaaigdacqGHsisldaWcaaqaaiab eM8a3naaCaaaleqabaGaaGOmaaaaaOqaaiabeM8a3naaDaaaleaaca WGUbaabaGaaGOmaaaaaaGccqGHRaWkdaWcaaqaaiaaikdacqaHcpGv cqaHjpWDaeaacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaOGaamyAaa GaayjkaiaawMcaaiabg2da9iaadUeacaWGgbWaaSbaaSqaaiaaicda aeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaadkeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaaigdacq GHsisldaWcaaqaaiabeM8a3naaCaaaleqabaGaaGOmaaaaaOqaaiab eM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGccqGHsisldaWcaa qaaiaaikdacqaHcpGvcqaHjpWDaeaacqaHjpWDdaWgaaWcbaGaamOB aaqabaaaaOGaamyAaaGaayjkaiaawMcaaiabg2da9iaadUeacaWGgb WaaSbaaSqaaiaaicdaaeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7daqadaqaamaalaaabaGaeq4UdW2aaW baaSqabeaacaaIYaaaaaGcbaGaeqyYdC3aa0baaSqaaiaad6gaaeaa caaIYaaaaaaakiabgUcaRmaalaaabaGaaGOmaiabek8awjabeU7aSb qaaiabeM8a3naaBaaaleaacaWGUbaabeaaaaGccqGHRaWkcaaIXaaa caGLOaGaayzkaaGaeyypa0JaaGimaaaa@9E9C@

The first two equations show that

B 1 =K F 0 ( 1− ω 2 ω n 2 + 2ςω ω n i ) −1 =K F 0 M(ω/ ω n ,ζ) e iϕ B 2 =K F 0 ( 1− ω 2 ω n 2 − 2ςω ω n i ) −1 =K F 0 M(ω/ ω n ,ζ) e −iϕ M(ω/ ω n ,ζ)= 1 ( 1− ω 2 ω n 2 ) 2 + ( 2ςω ω n ) 2 ϕ= tan −1 −2ζω/ ω n (1− ω 2 / ω n 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaGPaVl aaykW7caaMc8UaamOqamaaBaaaleaacaaIXaaabeaakiabg2da9iaa dUeacaWGgbWaaSbaaSqaaiaaicdaaeqaaOWaaeWaaeaacaaIXaGaey OeI0YaaSaaaeaacqaHjpWDdaahaaWcbeqaaiaaikdaaaaakeaacqaH jpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaaaaOGaey4kaSYaaSaaae aacaaIYaGaeqOWdyLaeqyYdChabaGaeqyYdC3aaSbaaSqaaiaad6ga aeqaaaaakiaadMgaaiaawIcacaGLPaaadaahaaWcbeqaaiabgkHiTi aaigdaaaGccqGH9aqpcaWGlbGaamOramaaBaaaleaacaaIWaaabeaa kiaad2eacaGGOaGaeqyYdCNaai4laiabeM8a3naaBaaaleaacaWGUb aabeaakiaacYcacqaH2oGEcaGGPaGaamyzamaaCaaaleqabaGaamyA aiabew9aMbaaaOqaaiaaykW7caaMc8UaamOqamaaBaaaleaacaaIYa aabeaakiabg2da9iaadUeacaWGgbWaaSbaaSqaaiaaicdaaeqaaOWa aeWaaeaacaaIXaGaeyOeI0YaaSaaaeaacqaHjpWDdaahaaWcbeqaai aaikdaaaaakeaacqaHjpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaaa aOGaeyOeI0YaaSaaaeaacaaIYaGaeqOWdyLaeqyYdChabaGaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaaaakiaadMgaaiaawIcacaGLPaaadaah aaWcbeqaaiabgkHiTiaaigdaaaGccqGH9aqpcaWGlbGaamOramaaBa aaleaacaaIWaaabeaakiaad2eacaGGOaGaeqyYdCNaai4laiabeM8a 3naaBaaaleaacaWGUbaabeaakiaacYcacqaH2oGEcaGGPaGaamyzam aaCaaaleqabaGaeyOeI0IaamyAaiabew9aMbaaaOqaaiaad2eacaGG OaGaeqyYdCNaai4laiabeM8a3naaBaaaleaacaWGUbaabeaakiaacY cacqaH2oGEcaGGPaGaeyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaa daqadaqaaiaaigdacqGHsisldaWcaaqaaiabeM8a3naaCaaaleqaba GaaGOmaaaaaOqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaa aaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda qadaqaamaalaaabaGaaGOmaiabek8awjabeM8a3bqaaiabeM8a3naa BaaaleaacaWGUbaabeaaaaaakiaawIcacaGLPaaadaahaaWcbeqaai aaikdaaaaabeaaaaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7cqaHvpGzcqGH9aqpciGG0bGaaiyyai aac6gadaahaaWcbeqaaiabgkHiTiaaigdaaaGcdaWcaaqaaiabgkHi TiaaikdacqaH2oGEcqaHjpWDcaGGVaGaeqyYdC3aaSbaaSqaaiaad6 gaaeqaaaGcbaGaaiikaiaaigdacqGHsislcqaHjpWDdaahaaWcbeqa aiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYa aaaOGaaiykaaaaaaaa@DF39@

(we introduced M and ϕ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHvpGzaaa@373E@ to re-write B 1 , B 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGcbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadkeadaWgaaWcbaGaaGOmaaqabaaa aa@398D@ in polar form). Finally substitute back for B 1 , B 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGcbWaaSbaaS qaaiaaigdaaeqaaOGaaiilaiaadkeadaWgaaWcbaGaaGOmaaqabaaa aa@398D@ into the guess for x p (t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bWaaSbaaS qaaiaadchaaeqaaOGaaiikaiaadshacaGGPaaaaa@39F0@ and simplify the solution to see that

x p (t)=− i 2 K F 0 M(ω/ ω n ,ζ)( e i(ωt+ϕ) − e −i(ωt+ϕ) )=K F 0 M(ω/ ω n ,ζ)sin(ωt+ϕ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bWaaSbaaS qaaiaadchaaeqaaOGaaiikaiaadshacaGGPaGaeyypa0JaeyOeI0Ya aSaaaeaacaWGPbaabaGaaGOmaaaacaWGlbGaamOramaaBaaaleaaca aIWaaabeaakiaad2eacaGGOaGaeqyYdCNaai4laiabeM8a3naaBaaa leaacaWGUbaabeaakiaacYcacqaH2oGEcaGGPaWaaeWaaeaacaWGLb WaaWbaaSqabeaacaWGPbGaaiikaiabeM8a3jaadshacqGHRaWkcqaH vpGzcaGGPaaaaOGaeyOeI0IaamyzamaaCaaaleqabaGaeyOeI0Iaam yAaiaacIcacqaHjpWDcaWG0bGaey4kaSIaeqy1dyMaaiykaaaaaOGa ayjkaiaawMcaaiabg2da9iaadUeacaWGgbWaaSbaaSqaaiaaicdaae qaaOGaamytaiaacIcacqaHjpWDcaGGVaGaeqyYdC3aaSbaaSqaaiaa d6gaaeqaaOGaaiilaiabeA7a6jaacMcaciGGZbGaaiyAaiaac6gaca GGOaGaeqyYdCNaamiDaiabgUcaRiabew9aMjaacMcaaaa@@

Finally, we must determine x h (t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bWaaSbaaS qaaiaadIgaaeqaaOGaaiikaiaadshacaGGPaaaaa@39E8@ . By construction, our guess for x h (t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bWaaSbaaS qaaiaadIgaaeqaaOGaaiikaiaadshacaGGPaaaaa@39E8@ satisfies

1 ω n 2 d 2 x h d t 2 + 2ς ω n d x h dt + x h =C x h (0)= x 0 − x p (0)= x 0 − X 0 sinϕ d x h dt | t=0 = v 0 − d x p dt | t=0 = v 0 − X 0 ωcosϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaaSaaae aacaaIXaaabaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaaaa kmaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadIhadaWgaa WcbaGaamiAaaqabaaakeaacaWGKbGaamiDamaaCaaaleqabaGaaGOm aaaaaaGccqGHRaWkdaWcaaqaaiaaikdacqaHcpGvaeaacqaHjpWDda WgaaWcbaGaamOBaaqabaaaaOWaaSaaaeaacaWGKbGaamiEamaaBaaa leaacaWGObaabeaaaOqaaiaadsgacaWG0baaaiabgUcaRiaadIhada WgaaWcbaGaamiAaaqabaGccqGH9aqpcaWGdbaabaGaamiEamaaBaaa leaacaWGObaabeaakiaacIcacaaIWaGaaiykaiabg2da9iaadIhada WgaaWcbaGaaGimaaqabaGccqGHsislcaWG4bWaaSbaaSqaaiaadcha aeqaaOGaaiikaiaaicdacaGGPaGaeyypa0JaamiEamaaBaaaleaaca aIWaaabeaakiabgkHiTiaadIfadaWgaaWcbaGaaGimaaqabaGcciGG ZbGaaiyAaiaac6gacqaHvpGzaeaadaabcaqaamaalaaabaGaamizai aadIhadaWgaaWcbaGaamiAaaqabaaakeaacaWGKbGaamiDaaaaaiaa wIa7amaaBaaaleaacaWG0bGaeyypa0JaaGimaaqabaGccqGH9aqpca WG2bWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0YaaqGaaeaadaWcaaqa aiaadsgacaWG4bWaaSbaaSqaaiaadchaaeqaaaGcbaGaamizaiaads haaaaacaGLiWoadaWgaaWcbaGaamiDaiabg2da9iaaicdaaeqaaOGa eyypa0JaamODamaaBaaaleaacaaIWaaabeaakiabgkHiTiaadIfada WgaaWcbaGaaGimaaqabaGccqaHjpWDciGGJbGaai4BaiaacohacqaH vpGzaaaa@8B48@

This is identical to the differential equation for a damped free vibrating system (but with modified initial conditions), and we can just write down the solution from the preceding section.

Short-cut for calculating steady-state solutions for forced vibrating systems

For example, consider the base excited system

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C+K( 1+ 2ς ω n d dt ) Y 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaiabgUcaRiaadUeadaqadaqaaiaaigdacqGHRaWkda WcaaqaaiaaikdacqaHcpGvaeaacqaHjpWDdaWgaaWcbaGaamOBaaqa baaaaOWaaSaaaeaacaWGKbaabaGaamizaiaadshaaaaacaGLOaGaay zkaaGaamywamaaBaaaleaacaaIWaaabeaakiGacohacaGGPbGaaiOB aiabeM8a3jaadshaaaa@62C9@

We anticipate that the steady-state solution will have the form

x p (t)= X 0 sin(ωt+ϕ) X 0 =K Y 0 M(ω/ ω n ,ζ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bWaaSbaaS qaaiaadchaaeqaaOGaaiikaiaadshacaGGPaGaeyypa0Jaamiwamaa BaaaleaacaaIWaaabeaakiGacohacaGGPbGaaiOBaiaacIcacqaHjp WDcaWG0bGaey4kaSIaeqy1dyMaaiykaiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamiwamaaBaaaleaa caaIWaaabeaakiabg2da9iaadUeacaWGzbWaaSbaaSqaaiaaicdaae qaaOGaamytaiaacIcacqaHjpWDcaGGVaGaeqyYdC3aaSbaaSqaaiaa d6gaaeqaaOGaaiilaiabeA7a6jaacMcaaaa@701E@

so we only need to determine the magnification M MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGnbaaaa@@ and the phase ϕ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHvpGzaaa@373E@ . We can do this quickly by

(i)Replacing the harmonic function Y 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMfadaWgaa WcbaGaaGimaaqabaGcciGGZbGaaiyAaiaac6gacqaHjpWDcaWG0baa aa@3DA5@ by a complex exponential Y 0 e iωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGzbWaaSbaaS qaaiaaicdaaeqaaOGaamyzamaaCaaaleqabaGaamyAaiabeM8a3jaa dshaaaaaaa@3C0F@

(ii)Substituting x=C+K Y 0 M e iϕ e iωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG4bGaeyypa0 Jaam4qaiabgUcaRiaadUeacaWGzbWaaSbaaSqaaiaaicdaaeqaaOGa amytaiaadwgadaahaaWcbeqaaiaadMgacqaHvpGzaaGccaWGLbWaaW baaSqabeaacaWGPbGaeqyYdCNaamiDaaaaaaa@@ into the solution

This gives

K Y 0 M e iϕ ( 1− ω 2 ω n 2 +i 2ςω ω n ) e iωt =K( 1+i 2ςω ω n ) Y 0 e iωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadUeacaWGzb WaaSbaaSqaaiaaicdaaeqaaOGaamytaiaadwgadaahaaWcbeqaaiaa dMgacqaHvpGzaaGcdaqadaqaaiaaigdacqGHsisldaWcaaqaaiabeM 8a3naaCaaaleqabaGaaGOmaaaaaOqaaiabeM8a3naaDaaaleaacaWG UbaabaGaaGOmaaaaaaGccqGHRaWkcaWGPbWaaSaaaeaacaaIYaGaeq OWdyLaeqyYdChabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaaaOGa ayjkaiaawMcaaiaadwgadaahaaWcbeqaaiaadMgacqaHjpWDcaWG0b aaaOGaeyypa0Jaam4samaabmaabaGaaGymaiabgUcaRiaadMgadaWc aaqaaiaaikdacqaHcpGvcqaHjpWDaeaacqaHjpWDdaWgaaWcbaGaam OBaaqabaaaaaGccaGLOaGaayzkaaGaamywamaaBaaaleaacaaIWaaa beaakiaadwgadaahaaWcbeqaaiaadMgacqaHjpWDcaWG0baaaaaa@68C4@

Hence

M e iϕ = ( 1+i 2ςω ω n ) ( 1− ω 2 ω n 2 +i 2ςω ω n ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eacaWGLb WaaWbaaSqabeaacaWGPbGaeqy1dygaaOGaeyypa0ZaaSaaaeaadaqa daqaaiaaigdacqGHRaWkcaWGPbWaaSaaaeaacaaIYaGaeqOWdyLaeq yYdChabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaaaOGaayjkaiaa wMcaaaqaamaabmaabaGaaGymaiabgkHiTmaalaaabaGaeqyYdC3aaW baaSqabeaacaaIYaaaaaGcbaGaeqyYdC3aa0baaSqaaiaad6gaaeaa caaIYaaaaaaakiabgUcaRiaadMgadaWcaaqaaiaaikdacqaHcpGvcq aHjpWDaeaacqaHjpWDdaWgaaWcbaGaamOBaaqabaaaaaGccaGLOaGa ayzkaaaaaaaa@59F8@

Finally, write the complex numbers on the right hand side in polar form and read off M and ϕ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaHvpGzaaa@373E@

M= 1+ ( 2ςω ω n ) 2 ( 1− ω 2 ω n 2 ) 2 + ( 2ςω ω n ) 2 ϕ= tan −1 2ςω ω n − tan −1 2ςω ω n ( 1− ω 2 ω n 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eacqGH9a qpdaWcaaqaamaakaaabaGaaGymaiabgUcaRmaabmaabaWaaSaaaeaa caaIYaGaeqOWdyLaeqyYdChabaGaeqyYdC3aaSbaaSqaaiaad6gaae qaaaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaaGc baWaaOaaaeaadaqadaqaaiaaigdacqGHsisldaWcaaqaaiabeM8a3n aaCaaaleqabaGaaGOmaaaaaOqaaiabeM8a3naaDaaaleaacaWGUbaa baGaaGOmaaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaa GccqGHRaWkdaqadaqaamaalaaabaGaaGOmaiabek8awjabeM8a3bqa aiabeM8a3naaBaaaleaacaWGUbaabeaaaaaakiaawIcacaGLPaaada ahaaWcbeqaaiaaikdaaaaabeaaaaGccaaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlabew9aMjabg2da9iGacshacaGGHbGaaiOBamaaCaaale qabaGaeyOeI0IaaGymaaaakmaalaaabaGaaGOmaiabek8awjabeM8a 3bqaaiabeM8a3naaBaaaleaacaWGUbaabeaaaaGccqGHsislciGG0b Gaaiyyaiaac6gadaahaaWcbeqaaiabgkHiTiaaigdaaaGcdaWcaaqa amaalaaabaGaaGOmaiabek8awjabeM8a3bqaaiabeM8a3naaBaaale aacaWGUbaabeaaaaaakeaadaqadaqaaiaaigdacqGHsisldaWcaaqa aiabeM8a3naaCaaaleqabaGaaGOmaaaaaOqaaiabeM8a3naaDaaale aacaWGUbaabaGaaGOmaaaaaaaakiaawIcacaGLPaaaaaaaaa@85C5@

Similarly, to find the magnification and phase for the rotor-excited system, which has differential equation

1 ω n 2 d 2 x d t 2 + 2ς ω n dx dt +x=C− K ω n 2 d 2 y d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiabeM8a3naaDaaaleaacaWGUbaabaGaaGOmaaaaaaGcdaWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacaaI YaGaeqOWdyfabaGaeqyYdC3aaSbaaSqaaiaad6gaaeqaaaaakmaala aabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWG4bGa eyypa0Jaam4qaiabgkHiTmaalaaabaGaam4saaqaaiabeM8a3naaDa aaleaacaWGUbaabaGaaGOmaaaaaaGcdaWcaaqaaiaadsgadaahaaWc beqaaiaaikdaaaGccaWG5baabaGaamizaiaadshadaahaaWcbeqaai aaikdaaaaaaaaa@@

we make the substitutions (i) and (ii) above and simplify the result to see that:

M e iϕ = ω 2 / ω n 2 ( 1− ω 2 ω n 2 +i 2ςω ω n ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eacaWGLb WaaWbaaSqabeaacaWGPbGaeqy1dygaaOGaeyypa0ZaaSaaaeaacqaH jpWDdaahaaWcbeqaaiaaikdaaaGccaGGVaGaeqyYdC3aa0baaSqaai aad6gaaeaacaaIYaaaaaGcbaWaaeWaaeaacaaIXaGaeyOeI0YaaSaa aeaacqaHjpWDdaahaaWcbeqaaiaaikdaaaaakeaacqaHjpWDdaqhaa WcbaGaamOBaaqaaiaaikdaaaaaaOGaey4kaSIaamyAamaalaaabaGa aGOmaiabek8awjabeM8a3bqaaiabeM8a3naaBaaaleaacaWGUbaabe aaaaaakiaawIcacaGLPaaaaaaaaa@55D6@

Re-write the right hand side in polar form

M= ω 2 / ω n 2 ( 1− ω 2 ω n 2 ) 2 + ( 2ςω ω n ) 2 ϕ= tan −1 −2ςω/ ω n ( 1− ω 2 ω n 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eacqGH9a qpdaWcaaqaaiabeM8a3naaCaaaleqabaGaaGOmaaaakiaac+cacqaH jpWDdaqhaaWcbaGaamOBaaqaaiaaikdaaaaakeaadaGcaaqaamaabm aabaGaaGymaiabgkHiTmaalaaabaGaeqyYdC3aaWbaaSqabeaacaaI YaaaaaGcbaGaeqyYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaaaaaO GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaa baWaaSaaaeaacaaIYaGaeqOWdyLaeqyYdChabaGaeqyYdC3aaSbaaS qaaiaad6gaaeqaaaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOm aaaaaeqaaaaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeqy1dy Maeyypa0JaciiDaiaacggacaGGUbWaaWbaaSqabeaacqGHsislcaaI XaaaaOWaaSaaaeaacqGHsislcaaIYaGaeqOWdyLaeqyYdCNaai4lai abeM8a3naaBaaaleaacaWGUbaabeaaaOqaamaabmaabaGaaGymaiab gkHiTmaalaaabaGaeqyYdC3aaWbaaSqabeaacaaIYaaaaaGcbaGaeq yYdC3aa0baaSqaaiaad6gaaeaacaaIYaaaaaaaaOGaayjkaiaawMca aaaaaaa@764D@

You will learn even faster tricks for solving differential equations in circuits next semester, and perhaps in more advanced level linear systems and control theory courses. In fact, the pros know tricks that avoid writing down the differential equation altogether – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ they can just go straight to the solution! If you want to develop these superpowers, stick with engineering, and keep writing those generous tuition checks!

5.6 Introduction to vibration of systems with many degrees of freedom

The simple 1DOF systems analyzed in the preceding section are very helpful to develop a feel for the general characteristics of vibrating systems. They are too simple to approximate most real systems, however. Real systems have more than just one degree of freedom. Real systems are also very rarely linear. You may be feeling cheated – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ are the simple idealizations that you get to see in intro courses really any use? It turns out that they are, but you can only really be convinced of this if you know how to analyze more realistic problems, and see that they often behave just like the simple idealizations.

The motion of systems with many degrees of freedom, or nonlinear systems, cannot usually be described using simple formulas. Even when they can, the formulas are so long and complicated that you need a computer to evaluate them. For this reason, introductory courses typically avoid these topics. However, if you are willing to use a computer, analyzing the motion of these complex systems is actually quite straightforward – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ in fact, often easier than using the nasty formulas we derived for 1DOF systems.

This section of the notes is intended mostly for advanced students, who may be insulted by simplified models. If you are feeling insulted, read on…

5.6.1 Equations of motion for undamped linear systems with many degrees of freedom.

We always express the equations of motion for a system with many degrees of freedom in a standard form. The two degree of freedom system shown in the picture can be used as an example. We won’t go through the calculation in detail here (you should be able to derive it for yourself – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ draw a FBD, use Newton’s law and all that tedious stuff), but here is the final answer:

m 1 d 2 x 1 d t 2 +( k 1 + k 2 ) x 1 − k 2 x 2 =0 m 2 d 2 x 2 d t 2 − k 2 x 1 +( k 2 + k 3 ) x 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyBam aaBaaaleaacaaIXaaabeaakmaalaaabaGaamizamaaCaaaleqabaGa aGOmaaaakiaadIhadaWgaaWcbaGaaGymaaqabaaakeaacaWGKbGaam iDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaqadaqaaiaadUga daWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGRbWaaSbaaSqaaiaaik daaeqaaaGccaGLOaGaayzkaaGaamiEamaaBaaaleaacaaIXaaabeaa kiabgkHiTiaadUgadaWgaaWcbaGaaGOmaaqabaGccaWG4bWaaSbaaS qaaiaaikdaaeqaaOGaeyypa0JaaGimaaqaaiaad2gadaWgaaWcbaGa aGOmaaqabaGcdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGcca WG4bWaaSbaaSqaaiaaikdaaeqaaaGcbaGaamizaiaadshadaahaaWc beqaaiaaikdaaaaaaOGaeyOeI0Iaam4AamaaBaaaleaacaaIYaaabe aakiaadIhadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaGGOaGaam4A amaaBaaaleaacaaIYaaabeaakiabgUcaRiaadUgadaWgaaWcbaGaaG 4maaqabaGccaGGPaGaamiEamaaBaaaleaacaaIYaaabeaakiabg2da 9iaaicdaaaaa@@

To solve vibration problems, we always write the equations of motion in matrix form. For an undamped system, the matrix equation of motion always looks like this

M d 2 x d t 2 +Kx=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah2eadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWH4baabaGaamizaiaa dshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaaC4saiaahIhacq GH9aqpcaWHWaaaaa@@

where x is a vector of the variables describing the motion,M is called the ‘mass matrix’ and K is called the ‘Stiffness matrix’ for the system. For the two spring-mass example, the equation of motion can be written in matrix form as

[ m 1 0 0 m 2 ] d 2 d t 2 [ x 1 x 2 ]+[ k 1 + k 2 − k 2 − k 2 k 2 + k 3 ][ x 1 x 2 ]=[ 0 0 ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaadmaabaqbae qabiGaaaqaaiaad2gadaWgaaWcbaGaaGymaaqabaaakeaacaaIWaaa baGaaGimaaqaaiaad2gadaWgaaWcbaGaaGOmaaqabaaaaaGccaGLBb GaayzxaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaaGcbaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOWaamWaaeaafaqabe GabaaabaGaamiEamaaBaaaleaacaaIXaaabeaaaOqaaiaadIhadaWg aaWcbaGaaGOmaaqabaaaaaGccaGLBbGaayzxaaGaey4kaSYaamWaae aafaqabeGacaaabaGaam4AamaaBaaaleaacaaIXaaabeaakiabgUca RiaadUgadaWgaaWcbaGaaGOmaaqabaaakeaacqGHsislcaWGRbWaaS baaSqaaiaaikdaaeqaaaGcbaGaeyOeI0Iaam4AamaaBaaaleaacaaI YaaabeaaaOqaaiaadUgadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkca WGRbWaaSbaaSqaaiaaiodaaeqaaaaaaOGaay5waiaaw2faamaadmaa baqbaeqabiqaaaqaaiaadIhadaWgaaWcbaGaaGymaaqabaaakeaaca WG4bWaaSbaaSqaaiaaikdaaeqaaaaaaOGaay5waiaaw2faaiabg2da 9maadmaabaqbaeqabiqaaaqaaiaaicdaaeaacaaIWaaaaaGaay5wai aaw2faaaaa@@

For a system with two masses (or more generally, two degrees of freedom), M and K are 2x2 matrices. For a system with n degrees of freedom, they are nxn matrices.

The spring-mass system is linear. A nonlinear system has more complicated equations of motion, but these can always be arranged into the standard matrix form by assuming that the displacement of the system is small, and linearizing the equation of motion. For example, the full nonlinear equations of motion for the double pendulum shown in the figure are

( m 1 + m 2 ) L 1 θ ¨ 1 + m 2 L 2 θ ˙ 2 2 sin( θ 1 − θ 2 )+ m 2 L 2 θ ¨ 2 cos( θ 1 − θ 2 )+( m 1 + m 2 )gsin θ 1 =0 m 2 L 2 θ ¨ 2 + m 2 L 1 θ ¨ 1 cos( θ 1 − θ 2 )− m 2 L 1 θ ˙ 1 2 sin( θ 1 − θ 2 )+ m 2 gsin θ 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaiikai aad2gadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqa aiaaikdaaeqaaOGaaiykaiaadYeadaWgaaWcbaGaaGymaaqabaGccu aH4oqCgaWaamaaBaaaleaacaaIXaaabeaakiabgUcaRiaad2gadaWg aaWcbaGaaGOmaaqabaGccaWGmbWaaSbaaSqaaiaaikdaaeqaaOGafq iUdeNbaiaadaqhaaWcbaGaaGOmaaqaaiaaikdaaaGcciGGZbGaaiyA aiaac6gacaGGOaGaeqiUde3aaSbaaSqaaiaaigdaaeqaaOGaeyOeI0 IaeqiUde3aaSbaaSqaaiaaikdaaeqaaOGaaiykaiabgUcaRiaad2ga daWgaaWcbaGaaGOmaaqabaGccaWGmbWaaSbaaSqaaiaaikdaaeqaaO GafqiUdeNbamaadaWgaaWcbaGaaGOmaaqabaGcciGGJbGaai4Baiaa cohacaGGOaGaeqiUde3aaSbaaSqaaiaaigdaaeqaaOGaeyOeI0Iaeq iUde3aaSbaaSqaaiaaikdaaeqaaOGaaiykaiabgUcaRiaacIcacaWG TbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamyBamaaBaaaleaaca aIYaaabeaakiaacMcacaWGNbGaci4CaiaacMgacaGGUbGaeqiUde3a aSbaaSqaaiaaigdaaeqaaOGaeyypa0JaaGimaaqaaiaad2gadaWgaa WcbaGaaGOmaaqabaGccaWGmbWaaSbaaSqaaiaaikdaaeqaaOGafqiU deNbamaadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaWGTbWaaSbaaS qaaiaaikdaaeqaaOGaamitamaaBaaaleaacaaIXaaabeaakiqbeI7a XzaadaWaaSbaaSqaaiaaigdaaeqaaOGaci4yaiaac+gacaGGZbGaai ikaiabeI7aXnaaBaaaleaacaaIXaaabeaakiabgkHiTiabeI7aXnaa BaaaleaacaaIYaaabeaakiaacMcacqGHsislcaWGTbWaaSbaaSqaai aaikdaaeqaaOGaamitamaaBaaaleaacaaIXaaabeaakiqbeI7aXzaa caWaa0baaSqaaiaaigdaaeaacaaIYaaaaOGaci4CaiaacMgacaGGUb GaaiikaiabeI7aXnaaBaaaleaacaaIXaaabeaakiabgkHiTiabeI7a XnaaBaaaleaacaaIYaaabeaakiaacMcacqGHRaWkcaWGTbWaaSbaaS qaaiaaikdaaeqaaOGaam4zaiGacohacaGGPbGaaiOBaiabeI7aXnaa BaaaleaacaaIYaaabeaakiabg2da9iaaicdaaaaa@A925@

Here, a single dot over a variable represents a time derivative, and a double dot represents a second time derivative (i.e. acceleration). These equations look horrible (and indeed they are – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ the motion of a double pendulum can even be chaotic), but if we assume that if θ 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIXaaabeaaaaa@38D7@ , θ 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXnaaBa aaleaacaaIYaaabeaaaaa@38D8@ , and their time derivatives are all small, so that terms involving squares, or products, of these variables can all be neglected, that and recall that cos(x)≈1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacogacaGGVb Gaai4CaiaacIcacaWG4bGaaiykaiabgIKi7kaaigdaaaa@3DCF@ and sin(x)≈x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacohacaGGPb GaaiOBaiaacIcacaWG4bGaaiykaiabgIKi7kaadIhaaaa@3E16@ for small x, the equations simplify to

( m 1 + m 2 ) L 1 θ ¨ 1 + m 2 L 2 θ ¨ 2 +( m 1 + m 2 )g θ 1 =0 m 2 L 2 θ ¨ 2 + m 2 L 1 θ ¨ 1 + m 2 g θ 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaiikai aad2gadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqa aiaaikdaaeqaaOGaaiykaiaadYeadaWgaaWcbaGaaGymaaqabaGccu aH4oqCgaWaamaaBaaaleaacaaIXaaabeaakiabgUcaRiaad2gadaWg aaWcbaGaaGOmaaqabaGccaWGmbWaaSbaaSqaaiaaikdaaeqaaOGafq iUdeNbamaadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaGGOaGaamyB amaaBaaaleaacaaIXaaabeaakiabgUcaRiaad2gadaWgaaWcbaGaaG OmaaqabaGccaGGPaGaam4zaiabeI7aXnaaBaaaleaacaaIXaaabeaa kiabg2da9iaaicdaaeaacaWGTbWaaSbaaSqaaiaaikdaaeqaaOGaam itamaaBaaaleaacaaIYaaabeaakiqbeI7aXzaadaWaaSbaaSqaaiaa ikdaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIYaaabeaakiaadY eadaWgaaWcbaGaaGymaaqabaGccuaH4oqCgaWaamaaBaaaleaacaaI XaaabeaakiabgUcaRiaad2gadaWgaaWcbaGaaGOmaaqabaGccaWGNb GaeqiUde3aaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaaGimaaaaaa@69D4@

Or, in matrix form

[ ( m 1 + m 2 ) L 1 m 2 L 2 m 2 L 1 m 2 L 2 ] d 2 d t 2 [ θ 1 θ 2 ]+[ ( m 1 + m 2 )g 0 0 m 2 g ][ θ 1 θ 2 ]=[ 0 0 ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaadmaabaqbae qabiGaaaqaamaabmaabaGaamyBamaaBaaaleaacaaIXaaabeaakiab gUcaRiaad2gadaWgaaWcbaGaaGOmaaqabaaakiaawIcacaGLPaaaca WGmbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamyBamaaBaaaleaacaaI YaaabeaakiaadYeadaWgaaWcbaGaaGOmaaqabaaakeaacaWGTbWaaS baaSqaaiaaikdaaeqaaOGaamitamaaBaaaleaacaaIXaaabeaaaOqa aiaad2gadaWgaaWcbaGaaGOmaaqabaGccaWGmbWaaSbaaSqaaiaaik daaeqaaaaaaOGaay5waiaaw2faamaalaaabaGaamizamaaCaaaleqa baGaaGOmaaaaaOqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaa aakmaadmaabaqbaeqabiqaaaqaaiabeI7aXnaaBaaaleaacaaIXaaa beaaaOqaaiabeI7aXnaaBaaaleaacaaIYaaabeaaaaaakiaawUfaca GLDbaacqGHRaWkdaWadaqaauaabeqaciaaaeaacaGGOaGaamyBamaa BaaaleaacaaIXaaabeaakiabgUcaRiaad2gadaWgaaWcbaGaaGOmaa qabaGccaGGPaGaam4zaaqaaiaaicdaaeaacaaIWaaabaGaamyBamaa BaaaleaacaaIYaaabeaakiaadEgaaaaacaGLBbGaayzxaaWaamWaae aafaqabeGabaaabaGaeqiUde3aaSbaaSqaaiaaigdaaeqaaaGcbaGa eqiUde3aaSbaaSqaaiaaikdaaeqaaaaaaOGaay5waiaaw2faaiabg2 da9maadmaabaqbaeqabiqaaaqaaiaaicdaaeaacaaIWaaaaaGaay5w aiaaw2faaaaa@@

This is again in the standard form.

Throughout the rest of this section, we will focus on exploring the behavior of systems of springs and masses. This is not because spring/mass systems are of any particular interest, but because they are easy to visualize, and, more importantly the equations of motion for a spring-mass system are identical to those of any linear system. This could include a realistic mechanical system, an electrical system, or anything that catches your fancy. (Then again, your fancy may tend more towards nonlinear systems, but if so, you should keep that to yourself).

5.6.2 Natural frequencies and mode shapes for undamped linear systems with many degrees of freedom.

First, let’s review the definition of natural frequencies and mode shapes. Recall that we can set a system vibrating by displacing it slightly from its static equilibrium position, and then releasing it. In general, the resulting motion will not be harmonic. However, there are certain special initial displacements that will cause harmonic vibrations. These special initial deflections are called mode shapes, and the corresponding frequencies of vibration are called natural frequencies.

The natural frequencies of a vibrating system are its most important property. It is helpful to have a simple way to calculate them.

Fortunately, calculating natural frequencies turns out to be quite easy (at least on a computer). Recall that the general form of the equation of motion for a vibrating system is

M d 2 x d t 2 +Kx=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah2eadaWcaa qaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWH4baabaGaamizaiaa dshadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaaC4saiaahIhacq GH9aqpcaWHWaaaaa@@

where x is a time dependent vector that describes the motion, and M and K are mass and stiffness matrices. Since we are interested in finding harmonic solutions for x, we can simply assume that the solution has the form Xsinωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfaciGGZb GaaiyAaiaac6gacqaHjpWDcaWG0baaaa@3CB9@ , and substitute into the equation of motion

−MX ω 2 sinωt+KXsinωt=0⇒KX= ω 2 MX MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaah2 eacaWHybGaeqyYdC3aaWbaaSqabeaacaaIYaaaaOGaci4CaiaacMga caGGUbGaeqyYdCNaamiDaiabgUcaRiaahUeacaWHybGaci4CaiaacM gacaGGUbGaeqyYdCNaamiDaiabg2da9iaahcdacaaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaey O0H4TaaC4saiaahIfacqGH9aqpcqaHjpWDdaahaaWcbeqaaiaaikda aaGccaWHnbGaaCiwaaaa@642D@

The vectors u and scalars λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ that satisfy a matrix equation of the form Ku=λMu MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahUeacaWH1b Gaeyypa0Jaeq4UdWMaaCytaiaahwhaaaa@3C9A@ are called ‘generalized eigenvectors’ and ‘generalized eigenvalues’ of the equation. It is impossible to find exact formulas for λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ and u for a large matrix (formulas exist for up to 5x5 matrices, but they are so messy they are useless), but MATLAB has built-in functions that will compute generalized eigenvectors and eigenvalues given numerical values for M and K.

The special values of λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ satisfying KX=λMX MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahUeacaWHyb Gaeyypa0Jaeq4UdWMaaCytaiaahIfaaaa@3C60@ are related to the natural frequencies by ω i = λ i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGPbaabeaakiabg2da9maakaaabaGaeq4UdW2aaSbaaSqa aiaadMgaaeqaaaqabaaaaa@3D0F@

The special vectors X are the ‘Mode shapes’ of the system. These are the special initial displacements that will cause the mass to vibrate harmonically.

If you only want to know the natural frequencies (common) you can use the MATLAB command

d = eig(K,M)

This returns a vector d, containing all the values of λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ satisfying Ku=λMu MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahUeacaWH1b Gaeyypa0Jaeq4UdWMaaCytaiaahwhaaaa@3C9A@ (for an nxn matrix, there are usually n different values). The natural frequencies follow as ω i = λ i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGPbaabeaakiabg2da9maakaaabaGaeq4UdW2aaSbaaSqa aiaadMgaaeqaaaqabaaaaa@3D0F@ .

If you want to find both the eigenvalues and eigenvectors, you must use

[V,D] = eig(K,M)

This returns two matrices, V and D. Each column of the matrix V corresponds to a vector u that satisfies the equation, and the diagonal elements of D contain the corresponding value of λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ . To extract the ith frequency and mode shape, use

omega = sqrt(D(i,i))

X = V(:,i)

For example, here is a MATLAB function that uses this function to automatically compute the natural frequencies of the spring-mass system shown in the figure.

function [freqs,modes] = compute_frequencies(k1,k2,k3,m1,m2)

M = [m1,0;0,m2];

K = [k1+k2,-k2;-k2,k2+k3];

[V,D] = eig(K,M);

for i = 1:2

freqs(i) = sqrt(D(i,i));

end

modes = V;

end

You could try running this with

>> [freqs,modes] = compute_frequencies(2,1,1,1,1)

This gives the natural frequencies as ω 1 =1, ω 2 =2.236 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaaIXaaabeaakiabg2da9iaaigdacaGGSaGaeqyYdC3aaSba aSqaaiaaikdaaeqaaOGaeyypa0JaaGOmaiaac6cacaaIYaGaaG4mai aaiAdaaaa@42D5@ , and the mode shapes as X 1 =(−0.707,−0,707) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfadaWgaa WcbaGaaGymaaqabaGccqGH9aqpcaGGOaGaeyOeI0IaaGimaiaac6ca caaI3aGaaGimaiaaiEdacaGGSaGaeyOeI0IaaGimaiaacYcacaaI3a GaaGimaiaaiEdacaGGPaaaaa@@ (i.e. both masses displace in the same direction) and X 2 =(−0.707,0.707) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpcaGGOaGaeyOeI0IaaGimaiaac6ca caaI3aGaaGimaiaaiEdacaGGSaGaaGimaiaac6cacaaI3aGaaGimai aaiEdacaGGPaaaaa@@ (the two masses displace in opposite directions.

If you read textbooks on vibrations, you will find that they may give different formulas for the natural frequencies and vibration modes. (If you read a lot of textbooks on vibrations there is probably something seriously wrong with your social life). This is partly because solving Ku=λMu MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahUeacaWH1b Gaeyypa0Jaeq4UdWMaaCytaiaahwhaaaa@3C9A@ for λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ and u is rather complicated (especially if you have to do the calculation by hand), and partly because this formula hides some subtle mathematical features of the equations of motion for vibrating systems. For example, the solutions to Ku=λMu MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahUeacaWH1b Gaeyypa0Jaeq4UdWMaaCytaiaahwhaaaa@3C9A@ are generally complex ( λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ and u have real and imaginary parts), so it is not obvious that our guess Xsinωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfaciGGZb GaaiyAaiaac6gacqaHjpWDcaWG0baaaa@3CB9@ actually satisfies the equation of motion. It turns out, however, that the equations of motion for a vibrating system can always be arranged so that M and K are symmetric. In this case λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ and u are real, and λ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeU7aSbaa@37EE@ is always positive or zero. The old fashioned formulas for natural frequencies and vibration modes show this more clearly. But our approach gives the same answer, and can also be generalized rather easily to solve damped systems (see Section 5.5.5), whereas the traditional textbook methods cannot.

5.6.3 Free vibration of undamped linear systems with many degrees of freedom.

As an example, consider a system with n identical masses with mass m, connected by springs with stiffness k, as shown in the picture. Suppose that at time t=0 the masses are displaced from their static equilibrium position by distances u 1 , u 2 ... u n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwhadaWgaa WcbaGaaGymaaqabaGccaGGSaGaamyDamaaBaaaleaacaaIYaaabeaa kiaac6cacaGGUaGaaiOlaiaadwhadaWgaaWcbaGaamOBaaqabaaaaa@3EF0@ , and have initial speeds v 1 , v 2 .... v n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhadaWgaa WcbaGaaGymaaqabaGccaGGSaGaamODamaaBaaaleaacaaIYaaabeaa kiaac6cacaGGUaGaaiOlaiaac6cacaWG2bWaaSbaaSqaaiaad6gaae qaaaaa@3FA5@ . We would like to calculate the motion of each mass x 1 (t), x 2 (t)... x n (t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGymaaqabaGccaGGOaGaamiDaiaacMcacaGGSaGaamiEamaa BaaaleaacaaIYaaabeaakiaacIcacaWG0bGaaiykaiaac6cacaGGUa GaaiOlaiaadIhadaWgaaWcbaGaamOBaaqabaGccaGGOaGaamiDaiaa cMcaaaa@45F9@ as a function of time.

It is convenient to represent the initial displacement and velocity as n dimensional vectors u and v, as

u=[ u 1 , u 2 ... u n ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwhacqGH9a qpcaGGBbGaamyDamaaBaaaleaacaaIXaaabeaakiaacYcacaWG1bWa aSbaaSqaaiaaikdaaeqaaOGaaiOlaiaac6cacaGGUaGaamyDamaaBa aaleaacaWGUbaabeaakiaac2faaaa@42BE@ , and v=[ v 1 , v 2 ... v n ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAhacqGH9a qpcaGGBbGaamODamaaBaaaleaacaaIXaaabeaakiaacYcacaWG2bWa aSbaaSqaaiaaikdaaeqaaOGaaiOlaiaac6cacaGGUaGaamODamaaBa aaleaacaWGUbaabeaakiaac2faaaa@42C2@ . In addition, we must calculate the natural frequencies ω i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBa aaleaacaWGPbaabeaaaaa@@ and mode shapes X i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfadaWgaa WcbaGaamyAaaqabaaaaa@@ , i=1..n for the system. The motion can then be calculated using the following formula

x(t)= ∑ i=1 n A i X i cos ω i t+ B i X i sin ω i t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIhacaGGOa GaamiDaiaacMcacqGH9aqpdaaeWbqaaiaadgeadaWgaaWcbaGaamyA aaqabaGccaWHybWaaSbaaSqaaiaadMgaaeqaaOGaci4yaiaac+gaca GGZbGaeqyYdC3aaSbaaSqaaiaadMgaaeqaaOGaamiDaiabgUcaRiaa dkeadaWgaaWcbaGaamyAaaqabaGccaWHybWaaSbaaSqaaiaadMgaae qaaOGaci4CaiaacMgacaGGUbGaeqyYdC3aaSbaaSqaaiaadMgaaeqa aOGaamiDaaWcbaGaamyAaiabg2da9iaaigdaaeaacaWGUbaaniabgg HiLdaaaa@56B7@

where

A i = u⋅ X i X i ⋅ X i B i = v⋅ X i ω i X i ⋅ X i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadgeadaWgaa WcbaGaamyAaaqabaGccqGH9aqpdaWcaaqaaiaahwhacqGHflY1caWH ybWaaSbaaSqaaiaadMgaaeqaaaGcbaGaaCiwamaaBaaaleaacaWGPb aabeaakiabgwSixlaahIfadaWgaaWcbaGaamyAaaqabaaaaOGaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaadkeadaWgaaWcbaGaamyAaaqa baGccqGH9aqpdaWcaaqaaiaahAhacqGHflY1caWHybWaaSbaaSqaai aadMgaaeqaaaGcbaGaeqyYdC3aaSbaaSqaaiaadMgaaeqaaOGaaCiw amaaBaaaleaacaWGPbaabeaakiabgwSixlaahIfadaWgaaWcbaGaam yAaaqabaaaaaaa@6A74@

Here, the dot represents an n dimensional dot product (to evaluate it in matlab, just use the dot() command).

This expression tells us that the general vibration of the system consists of a sum of all the vibration modes, (which all vibrate at their own discrete frequencies). You can control how big the contribution is from each mode by starting the system with different initial conditions. The mode shapes X i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfadaWgaa WcbaGaamyAaaqabaaaaa@@ have the curious property that the dot product of two different mode shapes is always zero ( X 1 ⋅ X 2 =0 X 1 ⋅ X 3 =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahIfadaWgaa WcbaGaaGymaaqabaGccqGHflY1caWHybWaaSbaaSqaaiaaikdaaeqa aOGaeyypa0JaaGimaiaaykW7caaMc8UaaGPaVlaahIfadaWgaaWcba GaaGymaaqabaGccqGHflY1caWHybWaaSbaaSqaaiaaiodaaeqaaOGa eyypa0JaaGimaaaa@4A3A@ , etc) – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ so you can see that if the initial displacements u happen to be the same as a mode shape, the vibration will be harmonic.

The figure on the right animates the motion of a system with 6 masses, which is set in motion by displacing the leftmost mass and releasing it. The graph shows the displacement of the leftmost mass as a function of time. You can download the MATLAB code for this computation here, and see how the formulas listed in this section are used to compute the motion. The program will predict the motion of a system with an arbitrary number of masses, and since you can easily edit the code to type in a different mass and stiffness matrix, it effectively solves any transient vibration problem.

5.6.4 Forced vibration of lightly damped linear systems with many degrees of freedom.

It is quite simple to find a formula for the motion of an undamped system subjected to time varying forces. The predictions are a bit unsatisfactory, however, because their vibration of an undamped system always depends on the initial conditions. In a real system, damping makes the steady-state response independent of the initial conditions. However, we can get an approximate solution for lightly damped systems by finding the solution for an undamped system, and then neglecting the part of the solution that depends on initial conditions.

As an example, we will consider the system with two springs and masses shown in the picture. Each mass is subjected to a harmonic force, which vibrates with some frequency ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@@ (the forces acting on the different masses all vibrate at the same frequency). The equations of motion are

m 1 d 2 x 1 d t 2 +( k 1 + k 2 ) x 1 − k 2 x 2 = F 1 cosωt m 2 d 2 x 2 d t 2 − k 2 x 1 +( k 2 + k 3 ) x 2 = F 2 cosωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyBam aaBaaaleaacaaIXaaabeaakmaalaaabaGaamizamaaCaaaleqabaGa aGOmaaaakiaadIhadaWgaaWcbaGaaGymaaqabaaakeaacaWGKbGaam iDamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaqadaqaaiaadUga daWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGRbWaaSbaaSqaaiaaik daaeqaaaGccaGLOaGaayzkaaGaamiEamaaBaaaleaacaaIXaaabeaa kiabgkHiTiaadUgadaWgaaWcbaGaaGOmaaqabaGccaWG4bWaaSbaaS qaaiaaikdaaeqaaOGaeyypa0JaamOramaaBaaaleaacaaIXaaabeaa kiGacogacaGGVbGaai4CaiabeM8a3jaadshaaeaacaWGTbWaaSbaaS qaaiaaikdaaeqaaOWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamiEamaaBaaaleaacaaIYaaabeaaaOqaaiaadsgacaWG0bWaaW baaSqabeaacaaIYaaaaaaakiabgkHiTiaadUgadaWgaaWcbaGaaGOm aaqabaGccaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaaiikai aadUgadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaWGRbWaaSbaaSqa aiaaiodaaeqaaOGaaiykaiaadIhadaWgaaWcbaGaaGOmaaqabaGccq GH9aqpcaWGgbWaaSbaaSqaaiaaikdaaeqaaOGaci4yaiaac+gacaGG ZbGaeqyYdCNaamiDaaaaaa@@

We can write these in matrix form as

[ m 1 0 0 m 2 ] d 2 d t 2 [ x 1 x 2 ]+[ k 1 + k 2 − k 2 − k 2 k 2 + k 3 ][ x 1 x 2 ]=[ F 1 F 2 ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaadmaabaqbae qabiGaaaqaaiaad2gadaWgaaWcbaGaaGymaaqabaaakeaacaaIWaaa baGaaGimaaqaaiaad2gadaWgaaWcbaGaaGOmaaqabaaaaaGccaGLBb GaayzxaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaaGcbaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOWaamWaaeaafaqabe GabaaabaGaamiEamaaBaaaleaacaaIXaaabeaaaOqaaiaadIhadaWg aaWcbaGaaGOmaaqabaaaaaGccaGLBbGaayzxaaGaey4kaSYaamWaae aafaqabeGacaaabaGaam4AamaaBaaaleaacaaIXaaabeaakiabgUca RiaadUgadaWgaaWcbaGaaGOmaaqabaaakeaacqGHsislcaWGRbWaaS baaSqaaiaaikdaaeqaaaGcbaGaeyOeI0Iaam4AamaaBaaaleaacaaI YaaabeaaaOqaaiaadUgadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkca WGRbWaaSbaaSqaaiaaiodaaeqaaaaaaOGaay5waiaaw2faamaadmaa baqbaeqabiqaaaqaaiaadIhadaWgaaWcbaGaaGymaaqabaaakeaaca WG4bWaaSbaaSqaaiaaikdaaeqaaaaaaOGaay5waiaaw2faaiabg2da 9maadmaabaqbaeqabiqaaaqaaiaadAeadaWgaaWcbaGaaGymaaqaba aakeaacaWGgbWaaSbaaSqaaiaaikdaaeqaaaaaaOGaay5waiaaw2fa aaaa@663C@

or, more generally,

To find the steady-state solution, we simply assume that the masses will all vibrate harmonically at the same frequency as the forces. This means that , , where are the (unknown) amplitudes of vibration of the two masses. In vector form we could write , where . Substituting this into the equation of motion gives

This is a system of linear equations for X. They can easily be solved using MATLAB. As an example, here is a simple MATLAB function that will calculate the vibration amplitude for a linear system with many degrees of freedom, given the stiffness and mass matrices, and the vector of forces f.

function X = forced_vibration(K,M,f,omega)

% Function to calculate steady state amplitude of

% a forced linear system.

% K is nxn the stiffness matrix

% M is the nxn mass matrix

% f is the n dimensional force vector

% omega is the forcing frequency, in radians/sec.

% The function computes a vector X, giving the amplitude of

% each degree of freedom

%

X = (K-M*omega^2)\f;

end

The function is only one line long!

As an example, the graph below shows the predicted steady-state vibration amplitude for the spring-mass system, for the special case where the masses are all equal , and the springs all have the same stiffness . The first mass is subjected to a harmonic force , and no force acts on the second mass. Note that the graph shows the magnitude of the vibration amplitude – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ the formula predicts that for some frequencies some masses have negative vibration amplitudes, but the negative sign has been ignored, as the negative sign just means that the mass vibrates out of phase with the force.

Several features of the result are worth noting:

If the forcing frequency is close to any one of the natural frequencies of the system, huge vibration amplitudes occur. This phenomenon is known as resonance. You can check the natural frequencies of the system using the little matlab code in section 5.5.2 – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ they turn out to be and . At these frequencies the vibration amplitude is theoretically infinite.

The figure predicts an intriguing new phenomenon – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ at a magic frequency, the amplitude of vibration of mass 1 (that’s the mass that the force acts on) drops to zero. This is called ‘Anti-resonance,’ and it has an important engineering application. Suppose that we have designed a system with a serious vibration problem (like the London Millenium bridge). Usually, this occurs because some kind of unexpected force is exciting one of the vibration modes in the system. We can idealize this behavior as a mass-spring system subjected to a force, as shown in the figure. So how do we stop the system from vibrating? Our solution for a 2DOF system shows that a system with two masses will have an anti-resonance. So we simply turn our 1DOF system into a 2DOF system by adding another spring and a mass, and tune the stiffness and mass of the new elements so that the anti-resonance occurs at the appropriate frequency. Of course, adding a mass will create a new vibration mode, but we can make sure that the new natural frequency is not at a bad frequency. We can also add a dashpot in parallel with the spring, if we want – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ this has the effect of making the anti-resonance phenomenon somewhat less effective (the vibration amplitude will be small, but finite, at the ‘magic’ frequency), but the new vibration modes will also have lower amplitudes at resonance. The added spring – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ mass system is called a ‘tuned vibration absorber.’ This approach was used to solve the Millenium Bridge vibration problem.

5.6.5 The effects of damping

In most design calculations, we don’t worry about accounting for the effects of damping very accurately. This is partly because it’s very difficult to find formulas that model damping realistically, and even more difficult to find values for the damping parameters. Also, the mathematics required to solve damped problems is a bit messy. Old textbooks don’t cover it, because for practical purposes it is only possible to do the calculations using a computer. It is not hard to account for the effects of damping, however, and it is helpful to have a sense of what its effect will be in a real system. We’ll go through this rather briefly in this section.

Equations of motion: The figure shows a damped spring-mass system. The equations of motion for the system can easily be shown to be

To solve these equations, we have to reduce them to a system that MATLAB can handle, by re-writing them as first order equations. We follow the standard procedure to do this – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ define and as new variables, and then write the equations in matrix form as

(This result might not be obvious to you – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ if so, multiply out the vector-matrix products to see that the equations are all correct). This is a matrix equation of the form

where y is a vector containing the unknown velocities and positions of the mass.

Free vibration response: Suppose that at time t=0 the system has initial positions and velocities , and we wish to calculate the subsequent motion of the system. To do this, we must solve the equation of motion. We start by guessing that the solution has the form (the negative sign is introduced because we expect solutions to decay with time). Here, is a constant vector, to be determined. Substituting this into the equation of motion gives

This is another generalized eigenvalue problem, and can easily be solved with MATLAB. The solution is much more complicated for a damped system, however, because the possible values of and that satisfy the equation are in general complex – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ that is to say, each can be expressed as , where and are positive real numbers, and . This makes more sense if we recall Euler’s formula

(if you haven’t seen Euler’s formula, try doing a Taylor expansion of both sides of the equation – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ you will find they are magically equal. If you don’t know how to do a Taylor expansion, you probably stopped reading this ages ago, but if you are still hanging in there, just trust me…). So, the solution is predicting that the response may be oscillatory, as we would expect. Once all the possible vectors and have been calculated, the response of the system can be calculated as follows:

1.Construct a matrix H , in which each column is one of the possible values of (MATLAB constructs this matrix automatically)

2.Construct a diagonal matrix (t), which has the form

where each is one of the solutions to the generalized eigenvalue equation.

3.Calculate a vector a (this represents the amplitudes of the various modes in the vibration response) that satisfies

4.The vibration response then follows as

All the matrices and vectors in these formulas are complex valued – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ but all the imaginary parts magically disappear in the final answer.

HEALTH WARNING: The formulas listed here only work if all the generalized eigenvalues satisfying are different. For some very special choices of damping, some eigenvalues may be repeated. In this case the formula won’t work. A quick and dirty fix for this is just to change the damping very slightly, and the problem disappears. Your applied math courses will hopefully show you a better fix, but we won’t worry about that here.

This all sounds a bit involved, but it actually only takes a few lines of MATLAB code to calculate the motion of any damped system. As an example, a MATLAB code that animates the motion of a damped spring-mass system shown in the figure (but with an arbitrary number of masses) can be downloaded here. You can use the code to explore the behavior of the system. In addition, you can modify the code to solve any linear free vibration problem by modifying the matrices M and D.

Here are some animations that illustrate the behavior of the system. The animations below show vibrations of the system with initial displacements corresponding to the three mode shapes of the undamped system (calculated using the procedure in Section 5.5.2). The results are shown for k=m=1 . In each case, the graph plots the motion of the three masses – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ if a color doesn’t show up, it means one of the other masses has the exact same displacement.

Mode 1 Mode 2 Mode 3

Notice that

1.For each mode, the displacement history of any mass looks very similar to the behavior of a damped, 1DOF system.

2.The amplitude of the high frequency modes die out much faster than the low frequency mode.

This explains why it is so helpful to understand the behavior of a 1DOF system. If a more complicated system is set in motion, its response initially involves contributions from all its vibration modes. Soon, however, the high frequency modes die out, and the dominant behavior is just caused by the lowest frequency mode. The animation to the right demonstrates this very nicely – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ here, the system was started by displacing only the first mass. The initial response is not harmonic, but after a short time the high frequency modes stop contributing, and the system behaves just like a 1DOF approximation. For design purposes, idealizing the system as a 1DOF damped spring-mass system is usually sufficient.

Notice also that light damping has very little effect on the natural frequencies and mode shapes – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ so the simple undamped approximation is a good way to calculate these.

Of course, if the system is very heavily damped, then its behavior changes completely – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ the system no longer vibrates, and instead just moves gradually towards its equilibrium position. You can simulate this behavior for yourself using the matlab code – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ try running it with or higher. Systems of this kind are not of much practical interest.

Steady-state forced vibration response. Finally, we take a look at the effects of damping on the response of a spring-mass system to harmonic forces. The equations of motion for a damped, forced system are

This is an equation of the form

where we have used Euler’s famous formula again. We can find a solution to

by guessing that , and substituting into the matrix equation

This equation can be solved for . Similarly, we can solve

by guessing that , which gives an equation for of the form . You actually don’t need to solve this equation – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ you can simply calculate by just changing the sign of all the imaginary parts of . The full solution follows as

This is the steady-state vibration response. Just as for the 1DOF system, the general solution also has a transient part, which depends on initial conditions. We know that the transient solution will die away, so we ignore it.

The solution for y(t) looks peculiar, because of the complex numbers. If we just want to plot the solution as a function of time, we don’t have to worry about the complex numbers, because they magically disappear in the final answer. In fact, if we use MATLAB to do the computations, we never even notice that the intermediate formulas involve complex numbers. If we do plot the solution, it is obvious that each mass vibrates harmonically, at the same frequency as the force (this is obvious from the formula too). It’s not worth plotting the function – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ we are really only interested in the amplitude of vibration of each mass. This can be calculated as follows

1.Let , denote the components of and

2.The vibration of the jth mass then has the form

where

are the amplitude and phase of the harmonic vibration of the mass.

If you know a lot about complex numbers you could try to derive these formulas for yourself. If not, just trust me – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ your math classes should cover this kind of thing. MATLAB can handle all these computations effortlessly. As an example, here is a simple MATLAB script that will calculate the steady-state amplitude of vibration and phase of each degree of freedom of a forced n degree of freedom system, given the force vector f, and the matrices M and D that describe the system.

function [amp,phase] = damped_forced_vibration(D,M,f,omega)

% Function to calculate steady state amplitude of

% a forced linear system.

% D is 2nx2n the stiffness/damping matrix

% M is the 2nx2n mass matrix

% f is the 2n dimensional force vector

% omega is the forcing frequency, in radians/sec.

% The function computes a vector ‘amp’, giving the amplitude of

% each degree of freedom, and a second vector ‘phase’,

% which gives the phase of each degree of freedom

%

Y0 = (D+M*i*omega)\f; % The i here is sqrt(-1)

% We dont need to calculate Y0bar - we can just change the sign of

% the imaginary part of Y0 using the 'conj' command

for j =1:length(f)/2

amp(j) = sqrt(Y0(j)*conj(Y0(j)));

phase(j) = log(conj(Y0(j))/Y0(j))/(2*i);

end

end

Again, the script is very simple.

Here is a graph showing the predicted vibration amplitude of each mass in the system shown. Note that only mass 1 is subjected to a force.

The important conclusions to be drawn from these results are:

1.We observe two resonances, at frequencies very close to the undamped natural frequencies of the system.

2.For light damping, the undamped model predicts the vibration amplitude quite accurately, except very close to the resonance itself (where the undamped model has an infinite vibration amplitude)

3.In a damped system, the amplitude of the lowest frequency resonance is generally much greater than higher frequency modes. For this reason, it is often sufficient to consider only the lowest frequency mode in design calculations. This means we can idealize the system as just a single DOF system, and think of it as a simple spring-mass system as described in the early part of this chapter. The relative vibration amplitudes of the various resonances do depend to some extent on the nature of the force – MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@@ it is possible to choose a set of forces that will excite only a high frequency mode, in which case the amplitude of this special excited mode will exceed all the others. But for most forcing, the lowest frequency one is the one that matters.

Passive isolation is the simplest and most widely used technique for minimizing vibrations. It involves placing elastic or viscoelastic materials, such as rubber, springs, or dampers, between the source of vibration and the receiver. These materials act as buffers that absorb and dissipate some of the vibrational energy, thus reducing its impact on the receiver. Passive isolation is suitable for low to moderate frequency vibrations, but it has some limitations, such as temperature sensitivity, nonlinear behavior, and degradation over time.

One technique of passive isolation is increasing unsprung mass. When there is more mass to move prior to suspension/powertrain mounts and bushings, less energy is transmitted into the vehicle. This might lead to using cast iron knuckles and brake calipers instead of higher cost cast aluminum.

Engine mounts- The engine mounts play a major role in damping the vibrations generated by the engine. Suspensions- All suspension systems are constantly working on absorbing the road surface impacts and keeping the car planted. Tyres- just like suspension tyres absorb the majority of impact due to road undulations. Dead weights- dead weights are commonly used on crankshafts of 3-cylinder cars to reduce down engine vibrations. Design optimization- Aerodynamic shape, lower center of gravity, and equal weight distribution, are some factors an engineer goes through while designing a car to reduce vibration. Vehicle chassis- A well designed, tuned chassis reduces great amount of vibrations. Sounddamping- foam fabric rubber shits absorb vibration

Passive isolation techniques and materials play a crucial role in creating quieter and more comfortable environments, improving the efficiency and longevity of machinery, and enhancing the overall quality of products and structures. However more research is needed to explore materials such as viscoelastics to enable applications in insulation and other isolation purposes.

Active isolation is a more advanced and sophisticated technique that uses sensors, actuators, and controllers to measure and counteract vibrations. The sensors detect the vibration signals and send them to the controller, which calculates the appropriate response and commands the actuators to generate an opposite force or displacement. This way, the active isolation system cancels out the vibration at the source or at the receiver, depending on the configuration. Active isolation is effective for high frequency vibrations, but it requires more complex and expensive components, as well as more power and maintenance.

I believe another simple way to an active isolation is to implement a vibration absorber, similar to those found on airplane wings. They prevent the wings, for example, to vibrate on the same high frequencies and brake. You don’t need to add a damping, because it absorbs the high frequencies and it won’t vibrate in the natural frequency.

Structural damping is another technique that aims to reduce vibrations by modifying the properties of the materials or the geometry of the components. Structural damping can be achieved by adding damping layers or coatings, such as viscoelastic polymers, metals, or ceramics, to the surfaces or interfaces of the vibrating parts. These layers or coatings increase the internal friction and dissipation of the vibrational energy, thus reducing the amplitude and frequency of the vibration. Structural damping can also be achieved by changing the shape or size of the components, such as adding ribs, holes, or slots, to alter their stiffness and natural frequencies.

Most panel radiation is a forced response. Avoid simple panel designs which some designers gravitate toward. Use complex non-symmetric shapes and as mentioned previously ribs. Pay close attention to the cavity mode of the passenger compartment. This will require usually very little energy to excite. Again, insure panels are designed to avoid all simple modes and simple symmetry. Mount masses and/or components off center on panels to decrease the ease at which those panels can be excited. This is all done to avoid panel participation in the cavity mode.

after reading the content of structural damping, I also get to know how the parts(belongs to suspension & its mounts) have to be in vehicle. Im not familiar with structural damping any other points in above mentioned, Along with my driving experience and big fan of suspension engineering, To reduce the vibration transfer from various components within the vehicle have to do design optimization, using different type Of noise isolation packages installation, Sort out the right Spring rate, Damper(shock absorber) what are the modification point required to get optimized damper motion(compression & retard motion). This are the areas have put the effort to restrict the vibrations transferring, choosing right elastomer also.

All the ideas are great for the structural damping. I worked on a project that I added ribs and changed the natural frequency of the chassis, because increased the damping constant. We can improve the structure to have more damping, but still need to be careful about the new natural frequency.

Acoustic damping is a technique that focuses on reducing the noise generated by vibrations, rather than the vibrations themselves. Acoustic damping involves using sound-absorbing or sound-insulating materials, such as foams, fabrics, or composites, to cover or fill the cavities or spaces where noise propagates. These materials reduce the reflection and transmission of sound waves, thus lowering the sound pressure level and improving the acoustic comfort of the vehicle. Acoustic damping can also involve using active noise control systems, which use microphones, speakers, and controllers to generate and emit sound waves that are out of phase with the noise, thus canceling it out.

My anecdotal experience (25+ programs) is first, that all panel radiation is the result of a forced response and is attenuated by mass damping. The second anecdotal discovery is you must develop your damping material optimization with a fully trimmed vehicle. I found the best way to add the least amount of damping material was to release a fully covered pattern. Then in the hardware phase on-road have a test vehicle built with no damping material. Systematically add panel damping based on on-road noise results/targets. You will find carpet, other sound barriers and normal vehicle hardware (like seats) add some parasitic panel damping. They all change the radiation efficiency of the panels they are mounted on. Results: lower mass and cost.

Design optimization is a technique that involves using mathematical models and computational tools to analyze and improve the vibration characteristics of the automotive system. Design optimization can help engineers to identify and eliminate the sources and modes of vibration, as well as to optimize the parameters and variables that affect the vibration behavior, such as mass, stiffness, damping, geometry, and location. Design optimization can also help engineers to evaluate and compare different techniques and solutions for isolating and damping vibrations, as well as to test and validate their performance and feasibility.

Maintenance and monitoring are essential techniques for ensuring the long-term effectiveness and reliability of the vibration isolation and damping systems. Maintenance involves inspecting and replacing the components and materials that are subject to wear and tear, such as rubber mounts, springs, dampers, and damping layers. Monitoring involves using sensors and instruments to measure and record the vibration levels and frequencies, as well as to detect and diagnose any faults or anomalies. Maintenance and monitoring can help engineers to prevent and correct any problems that may compromise the vibration quality and safety of the vehicle.

Vibration can also cause cracks on the structure. There are sensors and also techniques, such as TPA method (transfer path analysis) to identify possible cracks. Once the crack is find, it’s possible to weld the surface to prevent a bigger one. The study of TPA for cracks is a great way to prevent more damage.

In my experience, there isn't a definitive answer; it's a intricate interplay of factors such as design, materials, performance requirements, and local regulations. Early design considerations can prevent unforeseen costs, but not everything can be anticipated, requiring some passive dampening. The key is understanding the needs and navigating the challenge to identify the optimal market solution. If you're interested, Henkel offers engineering support from early design to later-stage solutions.

Can your customers hear your improvements? My experience is when it comes to NVH treatments, there was never enough (per NVH engineers). Even though, there were no real customer concerns. Each vehicle has a background (ambient) noise level. If your additional treatments do not specifically drop the ambient noise in the vehicle, no one will hear your additional treatment (added cost for nothing). Leaving the treatments in is not additive. The more you put in will not necessarily make the vehicle quieter, unless that ambient threshold is breached. Given the optimization of this is not achievable in software, adequate time and resources need to be dedicated in the hardware phase to achieve the quietest, lowest mass and cost vehicle.

In my thinking, First, the engine should be designed to produce less vibrations. Second, the aerodynamic body structure can provide some control. Third, the use of well formulated polymer compound for dampners. These above factors can help in dampening the vibrations and saving the cost incurred in use of dampeners. Thanks. Prabhjot singh.

Since one can rarely eliminate the source and relying on additive solutions have limits in the reduction. I prefer to look at the transfer path as a whole before starting to add to the system. Never forget the transfer path in many cases consists of elements that make up filter systems already. Many times small changes can greatly reduce the output. Sometimes even at a cost or weight savings.

For more information, please visit adss hardware.