Chapter 5

Vibrations

## 5.1 Overview of Vibrations

5.1.1 Examples of practical vibration problems

### 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 1940s, 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/05083/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 2003 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 2005 (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  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  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  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, , is defined as . We specify angular frequency in radians per second.  Thus, a signal that oscillates at 100 Hz has angular frequency  radians per second.

Period, frequency and angular frequency are related by

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

## 5.1.4  Harmonic Oscillations

Harmonic oscillations are a particularly simple form of vibration response.  A conservative spring-mass system will exhibit harmonic motion  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, , has the form

Here,  is the amplitude of the displacement,  is the frequency of oscillations in radians per second, and  (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

 Frequency /Hz Amplitude/mm Atomic Vibration Threshold of human perception 1-8 Machinery and building vibes 10-100 Swaying of tall buildings 1-5 10-1000

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

The velocity  and acceleration  of the mass follow as

Here,  is the amplitude of the velocity, and  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 .  The mass is released with velocity  from position  at time .  Find .

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

The spring force is related to the length of the spring by .  The i component of the equation of motion and this equation then shows that

This is our equation of motion for s.

Now, we need to solve this equation.  We could, of course, use Matlab to do this  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  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

with initial conditions   from position  at time .

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

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

We see that if we define

then our equation is equivalent to the standard one.

HEALTH WARNING: it is important to note that this substitution only works if  is constant, so its time derivative is zero.

The solution for x is

Here,  and  are the initial value of x and  its time derivative, which must be computed from the initial values of s and its time derivative

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):

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

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, , 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 springmass 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

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

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

with

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:

then the response is simply

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

Similarly, if we choose

then

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  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

where:

is the number of rigid bodies in the system

p is the number of particles in the system

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

 Roller joint   1 constraint (prevents motion in one direction) Rigid (massless) link (if the link has mass, it should be represented as a rigid body)   1 constraint (prevents relative motion parallel to link) Nonconformal contact (two bodies meet at a point)   No friction or slipping: 1 constraint (prevents interpenetration)   Sticking friction 2 constraints (prevents relative motion Conformal contact (two rigid bodies meet along a line)   No friction or slipping: 2 constraint (prevents interpenetration and rotation)   Sticking friction 3 constraints (prevents relative motion) Pinned joint (generally only applied to a rigid body, as it would stop a particle moving completely)   2 constraints (prevents motion horizontally and vertically) Clamped joint (rare in dynamics problems, as it prevents motion completely)   Can only be applied to a rigid body, not a particle   3 constraints (prevents motion horizontally, vertically and prevents rotation)

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

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

 Pinned joint   (5 constraints  prevents all motion, and prevents rotation about two axes) Roller bearing   (5 constraints  prevents all motion, and prevents rotation about two axes) Sleeve   (4 constraints  prevents motion in two directions, and prevents rotation about two axes) Swivel joint   4 constraints (prevents all motion, prevents rotation about 1 axis) Ball and socket joint   3 constraints  prevents all motion. Nonconformal contact (two rigid bodies meet 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) Conformal contact (two rigid bodies meet over a surface)   No friction or slipping: 3 constraints: prevents interpenetration and rotation about two axes.   Sticking: 6 constraints: prevents all relative motion and rotation. Clamped joint (rare in dynamics problems, as it prevents all motion)   6 constraints (prevents all motion and rotation)

## 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

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:

(The second term in V is the gravitational potential energy  it is negative because the height of the mass decreases with increasing s).  Now, note  that since our system is conservative

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

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

Our equation looks very similar to

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

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  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

(2)  We write down T and V:

(if you don’t see the formula for the kinetic energy, you can write down the position vector of the mass as , differentiate to find the velocity: , and then compute  and use a trig identity.  You can also use the circular motion formulas, if you prefer).

(3) Differentiate with respect to time:

(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 (  is a nonlinear function of  ).  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  for which the system is in static equilibrium by setting all time derivatives of  in the equation of motion to zero, and then solving the equation.  Here,

Here, we have used  to denote the special values of  for which the system happens to be in static equilibrium.  Note that  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 , where .

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 , we will solve for x.  Before going on, make sure that you are comfortable with the physical significance of  both x and .

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

We substitute for  in the equation of motion, to see that

(Recall that  is constant, so its time derivatives vanish)

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

where

Apply this to the nonlinear term in our equation of motion

Now, since x<<1, we can assume that , and so

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

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

We can do this for each equilibrium configuration.

whence

Note that all these values of  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

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

where  and

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 , 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/dt become 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  for the system shown. The spring has stiffness k and unstretched length .

We follow the same procedure as before.

The potential and kinetic energies of the system are

Hence

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 , so we don’t need to solve for the equilibrium configurations this time.  We set , with  and substitute back into the equation of motion:

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)

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

Question: what happens for ?

Example 3: A system 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!).

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

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:

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

Thus, the combined rotational and translational kinetic energy follows as

Now, note that since our system is conservative

Differentiate our expressions for T and V to see that

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  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

and substitute into the equation of motion:

This is now in the form

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

## 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  from position  at time .  Find .

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

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

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:

We can get our equation into this form by setting

As before,  is known as the natural frequency of the system.  We have discovered a new parameter, , 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

where

Critically Damped System

Underdamped System

where  is known as the damped natural frequency of the system.

In all the preceding equations,  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.   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  (in this case the program will calculate the values of  and  for you, and display the results), or alternatively, you can set the values of  and  directly.  You can also choose values for the initial conditions  and .  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  to a low value, so that the damping coefficient .   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  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 .  You should see that, as you increase , the vibration dies away more and more quickly.  What happens to the frequency of oscillations as  is increased?  Is this behavior consistent with the predictions of the theory?

Now, set the damping coefficient (not the dashpot coefficient this time) to .  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 .  A system of this kind is said to be critically damped.

Set  to a value greater than 1.  Under these conditions, the system decays more slowly towards its equilibrium configuration.

Keeping  >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 , 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

where  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

where  is the displacement at the nth peak, as shown.  In principle, you should be able to pick any two neighboring peaks, and calculate .  You should get the same answer, whichever peaks you choose.   It is often more accurate to estimate  using the following formula

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

Now, it turns out that we can deduce  and  from T and , as follows.

Why does this work? Let us calculate T and  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

where . Hence, the period of oscillation is

Similarly,

where we have noted that .

Fortunately, this horrendous equation can be simplified greatly:  substitute for T in terms of  and , then cancel everything you possibly can to see that

Finally, we can solve for  and  to see that:

as promised.

Note that this procedure can never give us values for k, m or .  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

Once k had been found, m and  are easily deduced from the relations

## 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  and mass  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

The force varies harmonically, with amplitude  and frequency . Similarly, the base motion for the second system is

and the distance between the small mass  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

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

Rearrange and susbstitute for F(t)

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

,

our equation can be reduced to the form

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 .  Similarly, the rate of change of length of the dashpot is d(x-y)/dt.

Newton’s second law then tells us that

Make the following substitutions

and the equation reduces to the standard form

Given the initial conditions

and the base motion

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  is

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

Add these two equations to eliminate H and rearrange

To arrange this into standard form, make the following substitutions

whereupon the equation of motion reduces to

Finally, look at the picture to convince yourself that if the crank rotates with angular velocity , then

where  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  using the sliders, or you can set , K and  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  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

The term  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

Equation of Motion

with

Here, the function M is called the ‘magnification’ for the system.  M and  are graphed below, as a function of

(a)                                                                                (b)

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

Solution for Base Excitation

Equation of Motion

with

The expressions for  and  are graphed below, as a function of

(a)                                                                           (b)

Steady state vibration of a base excited springmass system (a) Amplitude and (b) phase

#### Solution for Rotor Excitation

Equation of Motion

with

The expressions for  and  are graphed below, as a function of

Steady state vibration of a rotor excited springmass 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 .  Recall that  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 , 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 springmass system.

To see this mathematically, note that the solution has the form .  Observe that  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  are quite complicated, but you will learn a great deal if you are able to sketch graphs of  as a function of  for various values of .

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 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  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 .  Expressions for  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  e.g. to stop a tall building from swaying.   Structures are always deformable to some extent  this is represented qualitatively by the spring in a spring-mass system.  They always have mass  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  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

## 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 . Our goal is to design the system to minimize the vibration of the mass.

Our vibration solution predicts that the mass vibrates with displacement

Again, the graph is helpful to understand how the vibration amplitude  varies with system parameters.

Clearly, we can minimize the vibration amplitude of the mass by making .   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  small.  This is counter-intuitive  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  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  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 , and draw a horizontal line at amplitude .  Finally, we measure the frequencies ,  and  as shown in the picture.

We define the bandwidth of the response  as

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

The formulae are accurate for small  - say .

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

We can find the frequency at which the amplitude is a maximum by differentiating with respect to , setting the derivative equal to zero and solving the resulting equation for frequency.  It turns out that the maximum amplitude occurs at a frequency

For small , we see that

Next, to get an expression relating the bandwidth  to , we first calculate the frequencies  and .  Note that the maximum amplitude of vibration can be calculated by setting , which gives

Now, at the two frequencies of interest, we know , so that  and  must be solutions of the equation

Rearrange this equation to see that

This is a quadratic equation for  and has solutions

Expand both expressions in a Taylor series about  to see that

so, finally, we confirm that

## 5.4.7 Example Problems in Forced Vibrations

Example 1: A structure is idealized as a damped springmass 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:

Now, the list of solutions to forced vibration problems gives

For the present problem:

Substituting numbers into the expression for the vibration amplitude shows that

Example 2: A car and its suspension system are idealized as a damped springmass 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

Noting that , we have that

i.e., the wheel oscillates vertically with harmonic motion, at frequency .

Now, the suspension has been idealized as a springmass system subjected to base excitation.  The steady state vibration is

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

Note that K=1 for base excitation, so that the amplitude of vibration at  is approximately

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 3000lb.  Select values for the spring stiffness and the dashpot coefficient.

We must first determine values for  and  that will satisfy the design specifications. To this end:

(i)                 The specification requires that

for any value of  (remember  ).  Recall that  and that K=1 for a base excited springmass system. This tells us that the magnification  has to be below 1.75 for any frequency. The graph shows that if , the magnification never exceeds 1.75.  We also see that smaller values of  make the suspension more effective (M is smaller) at high frequencies.   So  is a good choice.

If you prefer not to use the graph, you can use the approximation  which suggests that  which gives  - but the approximation is not very accurate for such large values of  (to get a better estimate you’d have to maximize the formula for magnification with respect to  but that’s very messy).

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

We must choose system parameters so that, at this excitation frequency, .

This tells us that M must be less than  when  is 15.45 rad/s or greater.  We already know that , and following the curve for this value of  we see that M<1/2 if  . Therefore, we must pick .

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

for  , 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

Similarly

## 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

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

where a and b are two real numbers.

Define the complex conjugate as

It follows that

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

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

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

Note that

Solution to the equation of motion for an undamped harmonic oscillator

Solve  with initial conditions

Guess a solution of the form  where A and  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

We can satisfy this for any A by choosing  .   This gives us two families of solutions to the equations, one with  and another with .  The most general solution is the sum of these, with different coefficients

We need to find : we can do this by substituting t=0 into x and using the given values of x at t=0

Add and subtract these two equations to see that

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

(to see this just substitute  into the formulas and use Euler’s formula to show   are correct).  Finally substitute  into the general solution for x to see that

This agrees with the answer on the formula sheet.

Solution to the equation of motion for a free damped system

Solve  with initial conditions

As before we guess a solution  where A and  are two complex numbers to be determined.  Substituting into the equation:

This gives a quadratic equation for  (it is called the ‘characteristic equation’ for the differential equation).  It has solutions

Depending on the value of  we find

·           (overdamped)  two real values of

·           (critical damping):

·          (underdamped)  two complex values of

where we have defined   To write the answers in terms of real valued functions we need to treat these cases separately.

Overdamped solution: We have that

We can use the initial conditions to determine  :

Hence

Critically damped solution: our guess for the critically damped solution gives only  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  it turns out that

will also satisfy the differential equation (this is a standard trick in situations where the characteristic equation has repeated roots).   We can solve for  using the initial conditions:

It follows that  so the solution is

Underdamped solution: For this case

We can use the initial conditions to determine  (which are now complex):

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

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

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

Solve  with initial conditions

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

We guess a solution of the form

where   are complex numbers to be determined.   Substituting into the ODE:

We can satisfy this by setting

The first two equations show that

(we introduced M and  to re-write  in polar form).  Finally substitute back for  into the guess for  and simplify the solution to see that

## Finally, we must determine  .  By construction, our guess for  satisfies

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

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

so we only need to determine the magnification  and the phase  .   We can do this quickly by

(i)     Replacing the harmonic function  by a complex exponential

(ii)   Substituting  into the solution

This gives

Hence

Finally, write the complex numbers on the right hand side in polar form and read off M and

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

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

Re-write the right hand side in polar form

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  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  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  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  draw a FBD, use Newton’s law and all that tedious stuff), but here is the final answer:

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

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