Chapter 5. Vibrations

Transcription

1 Chapter 5 Vibrations 5.1 Overview of Vibrations Exaples of practical vibration probles Vibration is a continuous cyclic otion of a structure or a coponent. Generally, engineers try to avoid vibrations, because vibrations have a nuber of unpleasant effects: Cyclic otion iplies cyclic forces. Cyclic forces are very daaging to aterials. Even odest levels of vibration can cause extree discofort; Vibrations generally lead to a loss of precision in controlling achinery. Exaples where vibration suppression is an issue include: Structural vibrations. Most buildings are ounted on top of special rubber pads, which are intended to isolate the building fro ground vibrations. The figure on the right shows vibration isolators being installed under the floor of a building during construction (fro ) No vibrations course is coplete without a ention of the Tacoa Narrows suspension bridge. This bridge, constructed in the 1940s, was at the tie the longest suspension bridge in the world. Because it was a new design, it suffered fro 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 To the credit of the designers, the bridge survived for an aazingly long tie before it finally failed. It is thought that the vibrations were a for of selfexcited vibration known as `flutter, or galloping A siilar for of vibration is known to occur

2 in aircraft wings. Interestingly, odern cable stayed bridges that also suffer fro a new vibration proble: the cables are very lightly daped and can vibrate badly in high winds (this is a resonance proble, not flutter). You can find a detailed article on the subject at Soe bridge designs go as far as to incorporate active vibration suppression systes in their cables. Vehicle suspension systes are failiar to everyone, but continue to evolve as engineers work to iprove vehicle handling and ride (the figure above is fro A radical new approach to suspension design eerged in 003 when a research group led by Malcol Sith at Cabridge University invented a new echanical suspension eleent 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 displaceent 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 The device was adopted in secret by the McLaren Forula 1 racing tea in 005 (they called it the J daper, and a scandal erupted in Forula 1 racing when the Renault tea anaged 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 becoe a standard eleent in forula 1 racing. It is only a atter of tie before it appears on vehicles available to the rest of us. Precision Machinery: The picture on the right shows one exaple of a precision instruent. It is essential to isolate electron icroscopes fro vibrations. A typical transission electron icroscope is designed to resolve features of aterials down to atoic length scales. If the specien vibrates by ore than a few atoic spacings, it will be ipossible to see! This is one reason that electron icroscopes are always located in the baseent the baseent of a building vibrates uch less than the upper floors. Professor K.-S. Ki at Brown recently invented and patented a new vibration isolation syste to support his atoic force icroscope on the 7 th floor of the Barus- Holley building you can find the patent at United States Patent, Patent Nuber 7,543,791. Here is another precision instruent that is very sensitive to vibrations.

3 3 The picture shows features of a typical hard disk drive. It is particularly iportant to prevent vibrations in the disk stack assebly and in the disk head positioner, since any relative otion between these two coponents will ake it ipossible to read data. The spinning disk stack assebly has soe 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. Exaples of beneficial applications of vibrations include ultrasonic probes, both for edical application and for nondestructive testing. The picture shows a edical application of ultrasound: it is an iage of soeone s colon. This type of instruent can resolve features down to a fraction of a illieter, and is infinitely preferable to exploratory surgery. Ultrasound is also used to detect cracks in aircraft and structures. Musical instruents and loudspeakers are a second exaple of systes which put vibrations to good use. Finally, ost echanical clocks use vibrations to easure tie Vibration Measureent When faced with a vibration proble, engineers generally start by aking soe easureents to try to isolate the cause of the proble. There are two coon ways to easure vibrations: 1. An acceleroeter is a sall electro-echanical device that gives an electrical signal proportional to its acceleration. The picture shows a typical 3 axis acceleroeter.. A displaceent transducer is siilar to an acceleroeter, but gives an electrical signal proportional to its displaceent. Displaceent transducers are generally preferable if you need to easure low frequency vibrations; acceleroeters behave better at high frequencies. The ost coon procedure is to ount three acceleroeters at a point on the vibrating structure, so as to easure accelerations in three utually perpendicular directions. The velocity and displaceent are then deduced by integrating the accelerations.

4 Features of a Typical Vibration Response The picture below shows a typical signal that you ight record using an acceleroeter or displaceent transducer. Iportant features of the response are The signal is often (although not always) periodic: that is to say, it repeats itself at fixed intervals of tie. Vibrations that do not repeat theselves in this way are said to be rando. All the systes we consider in this course will exhibit periodic vibrations. Displaceent or Acceleration y(t) Period, T Peak to Peak Aplitude A tie The PERIOD of the signal, T, is the tie required for one coplete cycle of oscillation, as shown in the picture. The FREQUENCY of the signal, f, is the nuber of cycles of oscillation per second. Cycles per second is often given the nae Hertz: thus, a signal which repeats 100 ties per second is said to oscillate at 100 Hertz. The ANGULAR FREQUENCY of the signal,, is defined as f. We specify angular frequency in radians per second. Thus, a signal that oscillates at 100 Hz has angular frequency 00 radians per second. Period, frequency and angular frequency are related by f 1 f T T The PEAK-TO-PEAK AMPLITUDE of the signal, A, is the difference between its axiu value and its iniu value, as shown in the picture The AMPLITUDE of the signal is generally taken to ean half its peak to peak aplitude. Engineers soeties use aplitude as an abbreviation for peak to peak aplitude, however, so be careful. The ROOT MEAN SQUARE AMPLITUDE or RMS aplitude is defined as T 1/ 1 y( t) T 0

5 Haronic Oscillations Haronic oscillations are a particularly siple for of vibration response. Consider the spring-ass syste shown below (you will only see the spring-ass syste if your browser supports Java). If the spring is perturbed fro its static equilibriu position, it vibrates (press `start to watch the vibration). We will analyze the otion of the spring ass syste soon. We will find that the displaceent of the ass fro its static equilibriu position, x, has the for x( t) X sin( t ) 0 Here, X0 is the aplitude of the displaceent, is the frequency of oscillations in radians per second, and (in radians) is known as the `phase of the vibration. Vibrations of this for are said to be Haronic. Typical values for aplitude and frequency are listed in the table below Frequency /Hz Aplitude/ Atoic Vibration Threshold of huan perception Machinery and building vibes Swaying of tall buildings We can also express the displaceent in ters of its period of oscillation T x( t) X0 sin t T The velocity v and acceleration a of the ass follow as v( t) V0 sin t V0 X0 cos t a( t) A0 sin t A0 V0 X0 Here, V0 is the aplitude of the velocity, and A0 is the aplitude of the acceleration. Note the siple relationships between acceleration, velocity and displaceent aplitudes. Experient with the Java applet shown until you feel cofortable with the concepts of aplitude, frequency, period and phase of a signal. Surprisingly, any coplex engineering systes behave just like the spring ass syste we are looking at here. To describe the behavior of the syste, then, we need to know three things (in order of iportance): (1) The frequency (or period) of the vibrations () The aplitude of the vibrations (3) Occasionally, we ight be interested in the phase, but this is rare. So, our next proble is to find a way to calculate these three quantities for engineering systes. We will do this in stages. First, we will analyze a nuber of freely vibrating, conservative systes. Second, we will exaine free vibrations in a dissipative syste, to show the influence of energy losses in a echanical syste. Finally, we will discuss the behavior of echanical systes when they are subjected to oscillating forces.

6 6 5. Free vibration of conservative, single degree of freedo, linear systes. First, we will explain what is eant by the title of this section. Recall that a syste is conservative if energy is conserved, i.e. potential energy + kinetic energy = constant during otion. Free vibration eans that no tie varying external forces act on the syste. A syste has one degree of freedo if its otion can be copletely described by a single scalar variable. We ll discuss this in a bit ore detail later. A syste is said to be linear if its equation of otion is linear. We will see what this eans shortly. It turns out that all 1DOF, linear conservative systes behave in exactly the sae way. By analyzing the otion of one representative syste, we can learn about all others. s We will follow standard procedure, and use a spring-ass syste as k, d our representative exaple. Proble: The figure shows a spring ass syste. The spring has stiffness k and unstretched length L 0. The ass is released with velocity v 0 fro position s 0 at tie t 0. Find s( t ). There is a standard approach to solving probles like this (i) Get a differential equation for s using F=a (or other ethods to be discussed) (ii) Solve the differential equation. The picture shows a free body diagra for the ass. Newton s law of otion states that d s F a Fsi ( N g) j i Fs k( s L ) The spring force is related to the length of the spring by 0. The i coponent of the equation of otion and this equation then shows that This is our equation of otion for s. d s ks kl 0 Now, we need to solve this equation. We could, of course, use Matheatica to do this in fact here is the Matheatica solution. F s g N

7 7 This is the correct solution but Matheatica gives the result in a rather ore coplicated for than necessary. For nearly all siple vibration probles, it is actually sipler just to write down the solution. We will discuss the general procedure you should follow to do this in the next section How to solve equations of otion for vibration probles Note that all vibrations probles have siilar equations of otion. Consequently, we can just solve the equation once, record the solution, and use it to solve any vibration proble we ight be interested in. The procedure to solve any vibration proble is: 1. Derive the equation of otion, using Newton s laws (or soeties you can use energy ethods, as discussed in Section 5.3). Do soe algebra to arrange the equation of otion into a standard for 3. Look up the solution to this standard for in a table of solutions to vibration probles. We have provided a table of standard solutions as a separate docuent that you can download and print for future reference. We will illustrate the procedure using any exaples. 5.. Solution to the equation of otion for an undaped spring-ass syste We would like to solve d s ks kl 0 ds with initial conditions v0 fro position s0 at tie t 0. s k, L 0

8 8 We therefore consult our list of solutions to differential equations, and observe that it gives the solution to the following equation 1 d x x 0 n This is very siilar to our equation, but not identical. To see how to get our equation into this for, note that (i) the standard equation has no coefficient in front of the x; and (ii) its right hand side is zero. We can get our equation to look like this if we divide both sides by k, and subtract d fro both sides of the equation. This gives Finally, we see that if we define then our equation is equivalent to the standard one. d s s L 0 0 k 1 x s L 0 s x L0 k n HEALTH WARNING: it is iportant to note that this substitution only works if L0 derivative is zero. is constant, so its tie The solution for x is x X0 sin nt 1 x0 X0 x0 v0 / tan n n v0 Here, x0 and v 0 are the initial value of x and dx / its tie derivative, which ust be coputed fro the initial values of s and its tie derivative d ds x0 s0 L0 v0 ( s L0 ) v0 When we present the solution, we have a choice of writing down the solution for x, and giving forulas for the various ters in the solution (this is what is usually done): x X sin t 0 k 1 ( s0 L0 ) / tan n n X s L v n v0 Alternatively, we can express all the variables in the standard solution in ters of s k 1 ( s0 L0 ) s L0 s0 L0 v0 / sin tan n n t v 0 But this solution looks very essy (ore like the Matheatica solution). Observe that: The ass oscillates haronically, as discussed in the preceding section; The angular frequency of oscillation, n, is a characteristic property of the syste, and is independent of the initial position or velocity of the ass. This is a very iportant observation, and n

9 9 we will expand upon it below. The characteristic frequency is known as the natural frequency of the syste. Increasing the stiffness of the spring increases the natural frequency of the syste; Increasing the ass reduces the natural frequency of the syste Natural Frequencies and Mode Shapes. We saw that the spring ass syste 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 echanical systes. All stable, unforced, echanical systes vibrate haronically at certain discrete frequencies, known as natural frequencies of the syste. For the spring ass syste, we found only one natural frequency. More coplex systes have several natural frequencies. For exaple, the syste of two asses shown below has two natural frequencies, given by k 3k 1, k x 1 x k k A syste with three asses would have three natural frequencies, and so on. In general, a syste with ore than one natural frequency will not vibrate haronically. For exaple, suppose we start the two ass syste vibrating, with initial conditions o dx1 x1 x1 0 t 0 o dx x x 0 The response ay be shown (see sect 5.5 if you want to know how) to be x A sin t A sin t with sin x A1 sin 1t 1 A t 1 o o 1 o o A1 x1 x A x1 x 1 In general, the vibration response will look coplicated, and is not haronic. The aniation above shows a typical exaple (if you are using the pdf version of these notes the aniation will not work - you can download the atlab code that creates this aniation here and run it for yourself)

10 10 However, if we choose the special initial conditions: o o x1 X0 x X0 then the response is siply x X sin t x X0 sin 1t 1 i.e., both asses vibrate haronically, at the first natural frequency, as shown in the aniation to the right. (To repeat this in the MATLAB code, edit the file to set A1=0.3 and A=0) Siilarly, if we choose then o o x1 X0 x X0 x1 X0 sin t x X0 sin t i.e., the syste vibrates haronically, at the second natural frequency. (To repeat this in the MATLAB code, edit the file to set A=0.3 and A1 = 0) The special initial displaceents of a syste that cause it to vibrate haronically are called `ode shapes for the syste. If a syste has several natural frequencies, there is a corresponding ode of vibration for each natural frequency. The natural frequencies are arguably the single ost iportant property of any echanical syste. This is because, as we shall see, the natural frequencies coincide (alost) with the syste s resonant frequencies. That is to say, if you apply a tie varying force to the syste, and choose the frequency of the force to be equal to one of the natural frequencies, you will observe very large aplitude vibrations. When designing a structure or coponent, you generally want to control its natural vibration frequencies very carefully. For exaple, if you wish to stop a syste fro vibrating, you need to ake sure that all its natural frequencies are uch greater than the expected frequency of any forces that are likely to act on the structure. If you are designing a vibration isolation platfor, you generally want to ake its natural frequency uch 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 coponents. Once a prototype has been built, it is usual to easure the natural frequencies and ode shapes for a syste. This is done by attaching a nuber of acceleroeters to the syste, and then hitting it with a haer (this is usually a regular rubber tipped haer, which ight be instruented to easure the ipulse exerted by the haer during the ipact). By trial and error, one can find a spot to hit the device so as to excite each ode of vibration independent of any other. You can tell when you have found such a spot, because the whole syste vibrates haronically. The natural frequency and ode shape of each vibration ode is then deterined fro the acceleroeter readings. Ipulse haer tests can even be used on big structures like bridges or buildings but you need a big haer. In a recent test on a new cable stayed bridge in France, the bridge was excited by first attaching a

11 11 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 Calculating the nuber of degrees of freedo (and natural frequencies) of a syste When you analyze the behavior a syste, it is helpful to know ahead of tie how any vibration frequencies you will need to calculate. There are various ways to do this. Here are soe rules that you can apply: The nuber of degrees of freedo is equal to the nuber of independent coordinates required to describe the otion. This is only helpful if you can see by inspection how to describe your syste. For the spring-ass syste in the preceding section, we know that the ass can only ove in one direction, and so specifying the length of the spring s will copletely deterine the otion of the syste. The syste therefore has one degree of freedo, and one vibration frequency. Section 5.6 provides several ore exaples where it is fairly obvious that the syste has one degree of freedo. For a D syste, the nuber of degrees of freedo can be calculated fro the equation n 3r p Nc where: r is the nuber of rigid bodies in the syste p is the nuber of particles in the syste N is the nuber of constraints (or, if you prefer, independent reaction forces) in the syste. c To be able to apply this forula you need to know how any constraints appear in the proble. Constraints are iposed by things like rigid links, or contacts with rigid walls, which force the syste to ove in a particular way. The nubers of constraints associated with various types of D connections are listed in the table below. Notice that the nuber of constraints is always equal to the nuber of reaction forces you need to draw on an FBD to represent the joint Roller joint 1 constraint (prevents otion in one direction) or A A A A A () R (1/) Ay (1) R (1/) Ay R Ay Rigid (assless) link (if the link has ass, it should be represented as a rigid body) T T 1 constraint (prevents relative otion parallel to link) T T

12 1 Nonconforal contact (two bodies eet at a point) N N T No friction or slipping: 1 constraint (prevents interpenetration) T Sticking friction constraints (prevents relative otion Conforal contact (two rigid bodies eet along a line) M N N T M No friction or slipping: constraint (prevents interpenetration and rotation) T Sticking friction 3 constraints (prevents relative otion) Pinned joint (generally only applied to a rigid body, as it would stop a particle oving copletely) R Ax A R (1/) Ay A R (1/) Ax A () constraints (prevents otion horizontally and vertically) R Ay (1) A R (1/) Ax R (1/) Ay Claped joint (rare in dynaics probles, as it prevents otion copletely) A j A Can only be applied to a rigid body, not a particle R Ax M Az i R (1/) Ay R (1/) Ax M (1/) Az () 3 constraints (prevents otion horizontally, vertically and prevents rotation) R Ay (1) A M (1/) Az R (1/) Ax R (1/) Ay For a 3D syste, the nuber of degrees of freedo can be calculated fro the equation n 6r 3p Nc where the sybols have the sae eaning as for a D syste. A table of various constraints for 3D probles is given below.

13 13 Pinned joint (5 constraints prevents all otion, and prevents rotation about two axes) Roller bearing (5 constraints prevents all otion, and prevents rotation about two axes) Sleeve (4 constraints prevents otion in two directions, and prevents rotation about two axes) Swivel joint 4 constraints (prevents all otion, prevents rotation about 1 axis) Ball and socket joint 3 constraints prevents all otion.

14 14 Nonconforal contact (two rigid bodies eet at a point) No friction or slipping: 1 constraint (prevents interpenetration) Sticking friction 3 constraints, possibly 4 if friction is sufficient to prevent spin at contact) Conforal contact (two rigid bodies eet over a surface) No friction or slipping: 3 constraints: prevents interpenetration and rotation about two axes. Sticking: 6 constraints: prevents all relative otion and rotation. T A N A T A1 M A3 M A3 N A T A1 T A Claped joint (rare in dynaics probles, as it prevents all otion) 6 constraints (prevents all otion and rotation) 5..4 Calculating natural frequencies for 1DOF conservative systes In light of the discussion in the preceding section, we clearly need soe way to calculate natural frequencies for echanical systes. We do not have tie in this course to discuss ore than the very siplest echanical systes. We will therefore show you soe tricks for calculating natural frequencies of 1DOF, conservative, systes. It is best to do this by eans of exaples. Exaple 1: The spring-ass syste revisited Calculate the natural frequency of vibration for the syste shown in the figure. Assue that the contact between the block and wedge is frictionless. The spring has stiffness k and unstretched length L 0 k, L 0 s Our first objective is to get an equation of otion for s. We could do this by drawing a FBD, writing down Newton s law, and

15 15 looking at its coponents. However, for 1DOF systes it turns out that we can derive the EOM very quickly using the kinetic and potential energy of the syste. The potential energy and kinetic energy can be written down as: 1 1 ds V k s L0 gs sin T (The second ter in V is the gravitational potential energy it is negative because the height of the ass decreases with increasing s). Now, note that since our syste is conservative T V constant d T V 0 Differentiate our expressions for T and V (use the chain rule) to see that ds d s ds ds k( s L 0) g sin 0 d s g ( s L 0 sin ) 0 k k Finally, we ust turn this equation of otion into one of the standard solutions to vibration equations. Our equation looks very siilar to Thus let and substitute into the equation of otion: 1 d x x 0 n g g x s L0 sin s x L0 sin k k d x x 0 k By coparing this with our equation we see that the natural frequency of vibration is k n (rad/s) 3 1 k = Hz 3 Suary of procedure for calculating natural frequencies: (1) Describe the otion of the syste, using a single scalar variable (In the exaple, we chose to describe otion using the distance s); () Write down the potential energy V and kinetic energy T of the syste in ters of the scalar variable; d (3) Use T V 0 to get an equation of otion for your scalar variable; (4) Arrange the equation of otion in standard for; (5) Read off the natural frequency by coparing your equation to the standard for.

16 16 Exaple : A nonlinear syste. We will illustrate the procedure with a second exaple, which will deonstrate another useful trick. Find the natural frequency of vibration for a pendulu, shown in the figure. We will idealize the ass as a particle, to keep things siple. L We will follow the steps outlined earlier: (1) We describe the otion using the angle () We write down T and V: V glcos 1 d T L (if you don t see the forula for the kinetic energy, you can write down the position vector of the d d ass as r Lsini Lcos j, differentiate to find the velocity: v Lcos i Lsin j, and d then copute T ( v v) / L (sin cos ) and use a trig identity. You can also use the circular otion forulas, if you prefer). (3) Differentiate with respect to tie: d d d L glsin 0 L d sin 0 g (4) Arrange the EOM into standard for. Houston, we have a proble. There is no way this equation can be arranged into standard for. This is because the equation is nonlinear ( sin is a nonlinear function of ). There is, however, a way to deal with this proble. We will show what needs to be done, suarizing the general steps as we go along. (i) Find the static equilibriu configuration(s) for the syste. If the syste is in static equilibriu, it does not ove. We can find values of for which the syste is in static equilibriu by setting all tie derivatives of in the equation of otion to zero, and then solving the equation. Here, sin 0 0,,... o Here, we have used 0 to denote the special values of for which the syste happens to be in static equilibriu. Note that 0 is always a constant. (ii) Assue that the syste vibrates with sall aplitude about a static equilibriu configuration of interest. To do this, we let 0 x, where x 1. 0

17 17 Here, x represents a sall change in angle fro an equilibriu configuration.. Note that x will vary with tie as the syste vibrates. Instead of solving for, we will solve for x. Before going on, ake sure that you are cofortable with the physical significance of both x and 0. (iii) Linearize the equation of otion, by expanding all nonlinear ters as Taylor Maclaurin series about the equilibriu configuration. We substitute for in the equation of otion, to see that (Recall that 0 L d x sin( 0 x) 0 g is constant, so its tie derivatives vanish) Now, recall the Taylor-Maclaurin series expansion of a function f(x) has the for 1 f ( x0 x) f ( x0 ) xf ( x0 ) x f ( x0 )... where df d f f ( x0 ) f ( x0 ) dx xx dx 0 xx Apply this to the nonlinear ter in our equation of otion sin 0 x sin 1 0 xcos0 x sin 0... Now, since x<<1, we can assue that x x, and so sin x sin xcos n Finally, we can substitute back into our equation of otion, to obtain L d x cos 0x sin0 g (iv) Copare the linear equation with the standard for to deduce the natural frequency. We can do this for each equilibriu configuration. whence L d x 0 0,, 4... x 0 g n g L (rad/sec) 1 g fn (Hz) L Note that all these values of 0 really represent the sae configuration: the ass is hanging below the pivot. We have rediscovered the well-known expression for the natural frequency of a freely swinging pendulu. Next, try the reaining static equilibriu configuration 0

18 18 L d x 0, 3, 5... x 0 g If we look up this equation in our list of standard solutions, we find it does not have a haronic solution. Instead, the solution is 1 v0 t 1 v ( ) 0 t x t x0 e x0 e dx where x 0 x( t 0) and v 0 t0 g L Thus, except for soe rather special initial conditions, x increases without bound as tie increases. This is a characteristic of an unstable echanical syste. If we visualize the syste with 0, we can see what is happening. This equilibriu configuration has the pendulu 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/ becoe infinitely large. We know that a real pendulu would never rotate with infinite angular velocity. What has gone wrong? L Exaple 3: We will look at one ore nonlinear syste, to ake sure that k, L you are cofortable with this procedure. Calculate the resonant frequency of 0 sall oscillations about the equilibriu configuration 0 for the syste shown. The spring has stiffness k and unstretched length L 0. We follow the sae procedure as before. The potential and kinetic energies of the syste are 1 1 V k Lsin glcos Hence 1 L d T 3 d L d d d 1 d ( T V ) kl sin cos glsin 0 3 L d gl kl cos sin 0 3 L 0 L

19 19 Once again, we have found a nonlinear equation of otion. This tie we know what to do. We are told to find natural frequency of oscillation about 0, so we don t need to solve for the equilibriu configurations this tie. We set 0 x, with x 1 and substitute back into the equation of otion: L d x gl kl cos x sin x 0 3 Now, expand all the nonlinear ters (it is OK to do the one at a tie and then ultiply everything out. You can always throw away all powers of x greater than one as you do so) cos x 1 sin x x L d x gl 0 3 kl x d x x 0 3k 1 g / kl If you prefer, you can use Matheatica to do the Taylor series expansion: (The nuber 4 in the series coand specifies the highest power of x that should appear in the expansion). We now have an equation in standard for, and can read off the natural frequency n 3k g 1 (rad/sec) kl 1 3k g fn 1 (Hz) kl Question: what happens for g kl? Exaple 3: A syste with a rigid body (the KE of a rigid body will be defined in the next section of the course just live with it for now!). s Calculate the natural frequency of vibration for the syste k, L shown in the figure. Assue that the cylinder rolls without 0 slip on the wedge. The spring has stiffness k and unstretched length L 0 R Our first objective is to get an equation of otion for s. We do this by writing down the potential and kinetic energies of the syste in ters of s. The potential energy is easy:

20 0 1 V k s L0 gs sin The first ter represents the energy in the spring, while second ter accounts for the gravitational potential energy. The kinetic energy is slightly ore tricky. Note that the agnitude of the angular velocity of the disk is related to the agnitude of its translational velocity by ds R Thus, the cobined rotational and translational kinetic energy follows as 1 R 1 ds T 1 3 ds Now, note that since our syste is conservative T V constant d T V 0 Differentiate our expressions for T and V to see that 3 d s ds ds ds k( s L 0) g sin 0 3 d s g s L 0 sin k k The last equation is alost in one of the standard fors given on the handout, except that the right hand side is not zero. There is a trick to dealing with this proble siply subtract the constant right hand side fro s, and call the result x. (This only works if the right hand side is a constant, of course). Thus let g x s L0 sin k and substitute into the equation of otion: 3 d x g g x L 0 sin L0 sin k k k 3 d x x 0 k This is now in the for 1 d x x 0 n and by coparing this with our equation we see that the natural frequency of vibration is k n (rad/s) 3 1 k = Hz 3

21 1 5.3 Free vibration of a daped, single degree of freedo, linear spring ass syste. We analyzed vibration of several conservative systes in the preceding section. In each case, we found that if the syste was set in otion, it continued to ove indefinitely. This is counter to our everyday experience. Usually, if you start soething vibrating, it will vibrate with a progressively decreasing aplitude and eventually stop oving. The reason our siple odels 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-ass syste. As before, although we odel a very siple syste, the behavior we predict turns out to be representative of a wide range of real engineering systes Vibration of a daped spring-ass syste The spring ass dashpot syste shown is released with velocity u0 fro position s 0 at tie t 0. Find s( t ). Once again, we follow the standard approach to solving probles like this (i) Get a differential equation for s using F=a (ii) Solve the differential equation. s k, L 0 You ay have forgotten what a dashpot (or daper) does. Suppose we apply a force F to a dashpot, as shown in the figure. We would observe that the dashpot stretched at a F F rate proportional to the force dl L F One can buy dapers (the shock absorbers in your car contain dapers): a daper generally consists of a plunger inside an oil filled cylinder, which dissipates energy by churning the oil. Thus, it is possible to ake a spring-ass-daper syste that looks very uch like the one in the picture. More generally, however, the spring ass syste is used to represent a coplex echanical syste. In this case, the daper represents the cobined effects of all the various echaniss for dissipating energy in the syste, including friction, air resistance, deforation losses, and so on. To proceed, we draw a free body diagra, showing the forces exerted by the spring and daper on the ass. Newton s law then states that This is our equation of otion for s. ds d s k( s L0 ) a d s ds s L 0 0 k k k(s-l 0 ) ds Now, we check our list of solutions to differential equations, and see that we have a solution to:

22 We can get our equation into this for by setting 1 d x dx x 0 n n k s L0 x n As before, n is known as the natural frequency of the syste. We have discovered a new paraeter,, which is called the daping coefficient. It plays a very iportant role, as we shall see below. Now, we can write down the solution for x: Overdaped Syste 1 where d v0 ( ) 0 0 ( ) ( ) exp( ) n d x v exp( ) n d x x t 0 nt exp( d t) d d n 1 Critically Daped Syste 1 Underdaped Syste n 0 x( t) x v x t exp( t) n k v ( ) exp( ) cos nx x t nt x sind t d where 1 is known as the daped natural frequency of the syste. d n In all the preceding equations, are the values of x and its tie derivative at tie t=0. x0 s0 L0 v0 u0 These expressions are rather too coplicated to visualize what the syste is doing for any given set of paraeters. Instead, contains a Java Applet that can be used to show aniations along with graphs of the displaceent. You can use the sliders to set the values of either, k, and (in this case the progra will calculate the values of and n for you, and display the results), or alternatively, you can set the values of and n directly. You can also choose values for the initial conditions x0 and v 0. When you press `start, the applet will aniate the behavior of the syste, and will draw a graph of the position of the ass as a function of tie. You can also choose to display the phase plane, which shows the velocity of the ass as a function of its position, if you wish. You can stop the aniation at any tie, change the paraeters, 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 failiarize yourself with the behavior of the syste Set the dashpot coefficient to a low value, so that the daping coefficient 1. Make sure the graph is set to display position versus tie, and press `start. You should see the syste vibrate. The vibration looks very siilar to the behavior of the conservative syste we analyzed in the

23 3 preceding section, except that the aplitude decays with tie. Note that the syste vibrates at a frequency very slightly lower than the natural frequency of the syste. Keeping the value of fixed, vary the values of spring constant and ass 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 fixed, and vary. You should see that, as you increase, the vibration dies away ore and ore quickly. What happens to the frequency of oscillations as is increased? Is this behavior consistent with the predictions of the theory? Now, set the daping coefficient (not the dashpot coefficient this tie) to 1. For this value, the syste no longer vibrates; instead, the ass soothly returns to its equilibriu position x=0. If you need to design a syste that returns to its equilibriu position in the shortest possible tie, then it is custoary to select syste paraeters so that 1. A syste of this kind is said to be critically daped. Set to a value greater than 1. Under these conditions, the syste decays ore slowly towards its equilibriu configuration. Keeping >1, experient with the effects of changing the stiffness of the spring and the value of the ass. Can you explain what is happening atheatically, using the equations of otion and their solution? Finally, you ight like to look at the behavior of the syste on its phase plane. In this course, we will not ake uch use of the phase plane, but it is a powerful tool for visualizing the behavior of nonlinear systes. By looking at the patterns traced by the syste on the phase plane, you can often work out what it is doing. For exaple, if the trajectory encircles the origin, then the syste is vibrating. If the trajectory approaches the origin, the syste is decaying to its equilibriu configuration. We now know the effects of energy dissipation on a vibrating syste. One iportant conclusion is that if the energy dissipation is low, the syste will vibrate. Furtherore, the frequency of vibration is very close to that of an undaped syste. Consequently, if you want to predict the frequency of vibration of a syste, you can siplify the calculation by neglecting daping Using Free Vibrations to Measure Properties of a Syste We will describe one very iportant application of the results developed in the preceding section. It often happens that we need to easure the dynaical properties of an engineering syste. For exaple, we ight want to easure the natural frequency and daping coefficient for a structure after it has been built, to ake sure that design predictions were correct, and to use in future odels of the syste. You can use the free vibration response to do this, as follows. First, you instruent your design by attaching acceleroeters to appropriate points. You then use an ipulse haer to excite a particular ode of vibration, as discussed in Section You use your acceleroeter readings to deterine the displaceent at the point where the structure was excited: the results will be a graph siilar to the one shown below.

24 4 Displaceent x(t 0 ) x(t 1 ) x(t ) x(t 3 ) tie t 0 t 1 t t 3 t 4 T We then identify a nice looking peak, and call the tie there t 0, as shown. The following quantities are then easured fro the graph: 1. The period of oscillation. The period of oscillation was defined in Section 5.1.: it is the tie between two peaks, as shown. Since the signal is (supposedly) periodic, it is often best to estiate T as follows where t n T tn t0 n is the tie at which the nth peak occurs, as shown in the picture.. The Logarithic Decreent. This is a new quantity, defined as follows x( t ) log n x( tn1) where x( tn ) is the displaceent at the nth peak, as shown. In principle, you should be able to pick any two neighboring peaks, and calculate. You should get the sae answer, whichever peaks you choose. It is often ore accurate to estiate using the following forula 1 x( t0) log n x( tn ) This expression should give the sae answer as the earlier definition. Now, it turns out that we can deduce n and fro T and, as follows. 4 n 4 T Why does this work? Let us calculate T and using the exact solution to the equation of otion for a daped spring-ass syste. Recall that, for an underdaped syste, the solution has the for v ( ) exp( ) 0 0 cos nxn x t nt x sin d where 1. Hence, the period of oscillation is d n

25 5 Siilarly, where we have noted that 1 T d n 1 v0 exp( ) 0 cos nxn ntn x n sinn v0 log v exp( ( )) 0 0 cos ( ) nxn n tn T x d tn T sin d ( tn T ) v0 tn tn T. Fortunately, this horrendous equation can be siplified greatly: substitute for T in ters of, then cancel everything you possibly can to see that Finally, we can solve for as proised. n and to see that: 1 4 n 4 T Note that this procedure can never give us values for k, or. However, if we wanted to find these, we could perfor a static test on the structure. If we easure the deflection d under a static load F, then we know that F k d Once k had been found, and are easily deduced fro the relations n k k n and 5.4 Forced vibration of daped, single degree of freedo, linear spring ass systes. Finally, we solve the ost iportant vibration probles of all. In engineering practice, we are alost invariably interested in predicting the response of a structure or echanical syste to external forcing. For exaple, we ay need to predict the response of a bridge or tall building to wind loading, earthquakes, or ground vibrations due to traffic. Another typical proble you are likely to encounter is to isolate a sensitive syste fro vibrations. For exaple, the suspension of your car is designed to isolate a sensitive syste (you) fro bups in the road. Electron icroscopes are another exaple of sensitive instruents that ust be isolated fro vibrations. Electron icroscopes are designed to resolve features a few nanoeters in size. If the specien vibrates with aplitude of only a few nanoeters, it will be ipossible to see! Great care is taken to isolate this kind of instruent fro vibrations. That is one reason they are alost always in the baseent of a building: the baseent vibrates uch less than the floors above.

26 6 We will again use a spring-ass syste as a odel of a real engineering syste. As before, the springass syste can be thought of as representing a single ode of vibration in a real syste, whose natural frequency and daping coefficient coincide with that of our spring-ass syste. We will consider three types of forcing applied to the spring-ass syste, as shown below: External Forcing odels the behavior of a syste which has a tie varying force acting on it. An exaple ight be an offshore structure subjected to wave loading. Base Excitation odels the behavior of a vibration isolation syste. The base of the spring is given a prescribed otion, causing the ass to vibrate. This syste can be used to odel a vehicle suspension syste, or the earthquake response of a structure. Rotor Excitation odels the effect of a rotating achine ounted on a flexible floor. The crank with sall ass 0 rotates at constant angular velocity, causing the ass to vibrate. L 0 x(t) k, L 0 L 0 x(t) k, L 0 y(t)=y 0 sint L 0 x(t) k, L 0 F(t)=F 0 sin t External forcing Base Excitation y(t)=y 0 sint 0 Rotor Excitation Of course, vibrating systes can be excited in other ways as well, but the equations of otion will always reduce to one of the three cases we consider here. Notice that in each case, we will restrict our analysis to haronic excitation. For exaple, the external force applied to the first syste is given by F( t) F sin t The force varies haronically, with aplitude F0 and frequency. Siilarly, the base otion for the second syste is y( t) Y sin t and the distance between the sall ass and the large ass for the third syste has the sae for. We assue that at tie t=0, the initial position and velocity of each syste is dx x x0 v0 In each case, we wish to calculate the displaceent of the ass x fro its static equilibriu configuration, as a function of tie t. It is of particular interest to deterine the influence of forcing aplitude and frequency on the otion of the ass. We follow the sae approach to analyze each syste: we set up, and solve the equation of otion.

27 Equations of Motion for Forced Spring Mass Systes Equation of Motion for External Forcing We have no proble setting up and solving equations of otion by now. First draw a free body diagra for the syste, as show on the right Newton s law of otion gives Rearrange and susbstitute for F(t) d x dx F( t) kx d x dx 1 x F 0 sin t k k k Check out our list of solutions to standard ODEs. We find that if we set k 1 n,, K, k k our equation can be reduced to the for which is on the list. 1 d x dx x KF 0 sin t n n kx dx F(t) 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 systes. Equation of Motion for Base Excitation Exactly the sae approach works for this syste. The free body diagra is shown in the figure. Note that the force in the spring is now k(x-y) because the length of the spring is L0 x y. Siilarly, the rate of change of length of the dashpot is d(x-y)/. k(x-y) d(x-y) Newton s second law then tells us that d x ( ) dx dy k x y d x dx dy x y k k k Make the following substitutions k n,, K 1 k and the equation reduces to the standard for 1 d x dx dy x K y n n n

28 8 Given the initial conditions and the base otion dx x x0 v0 y( t) Y0 sin t we can look up the solution in our handy list of solutions to ODEs. Equation of otion for Rotor Excitation Finally, we will derive the equation of otion for the third case. Free body diagras are shown below for both the rotor and the ass Note that the horizontal acceleration of the ass 0 is d d x d y a ( L 0 x y) Hence, applying Newton s second law in the horizontal direction for both asses: Add these two equations to eliinate H and rearrange d x dx H kx d x d y 0 H 0 d x dx 0 d y x k k k To arrange this into standard for, ake the following substitutions k 0 n K ( ) k( ) whereupon the equation of otion reduces to d x dx K d y x n n n Finally, look at the picture to convince yourself that if the crank rotates with angular velocity, then y( t) Y sin t where Y 0 is the length of the crank. The solution can once again be found in the list of solutions to ODEs. 0 kx dx R 1 R V H H t V 0 Y 0 0 y=y 0 sin

29 Definition of Transient and Steady State Response. If you have looked at the list of solutions to the equations of otion 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 extreely difficult to work out what the equations are telling A Java applet posted at should help to visualize the otion. The applet will open in a new window so you can see it and read the text at the sae tie. The applet siply calculates the solution to the equations of otion using the forulae given in the list of solutions, and plots graphs showing features of the otion. You can use the sliders to set various paraeters in the syste, including the type of forcing, its aplitude and frequency; spring constant, daping coefficient and ass; as well as the position and velocity of the ass at tie t=0. Note that you can control the properties of the spring-ass syste in two ways: you can either set values for k, and using the sliders, or you can set, K and instead. n We will use the applet to deonstrate a nuber of iportant features of forced vibrations, including the following: The steady state response of a forced, daped, spring ass syste is independent of the initial conditions. To convince yourself of this, run the applet (click on `start and let the syste run for a while). Now, press `stop ; change the initial position of the ass, and press `start again. You will see that, after a while, the solution with the new initial conditions is exactly the sae 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 syste as tie gets very large the `steady state response; and as you see, it is independent of the initial position and velocity of the ass. The behavior of the syste while it is approaching the steady state is called the `transient response. The transient response depends on everything Now, reduce the daping coefficient and repeat the test. You will find that the syste takes longer to reach steady state. Thus, the length of tie to reach steady state depends on the properties of the syste (and also the initial conditions). The observation that the syste always settles to a steady state has two iportant consequences. Firstly, we rarely know the initial conditions for a real engineering syste (who knows what the position and velocity of a bridge is at tie t=0?). Now we know this doesn t atter 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 siplify the horrible solutions to our equations of otion. When analyzing forced vibrations, we (alost) always neglect the transient response of the syste, and calculate only the steady state behavior. If you look at the solutions to the equations of otion we calculated in the preceding sections, you will see that each solution has the for x( t) x ( t) x ( t) h p

30 30 The ter xh ( t) accounts for the transient response, and is always zero for large tie. The second ter gives the steady state response of the syste. Following standard convention, we will list only the steady state solutions below. You should bear in ind, however, that the steady state is only part of the solution, and is only valid if the tie is large enough that the transient ter can be neglected Suary of Steady-State Response of Forced Spring Mass Systes. This section suarizes all the forulas you will need to solve probles involving forced vibrations. Solution for External Forcing Equation of Motion with 1 d x dx x KF( t) n n Steady State Solution: k 1 n,, K k k x( t) X0 sin t KF0 1 / X0 tan n 1/ 1 / n 1 / n / n The expressions for X0 and are graphed below, as a function of / n L 0 x(t) k, L 0 F(t)=F 0 sin t (a) (b) Steady state vibration of a force spring-ass syste (a) aplitude (b) phase.