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2.6 Forced oscillations and resonance

Note: 2 lectures, §3.6 in [EP], §3.8 in [BD]

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Let us return back to the example of a mass on a spring. We now examine the case of forced oscillations, which we did not yet handle. That is, we consider the equation

mx ′′ + cx′ + kx = F(t),

for some nonzero F(t) . The setup is again: m is mass, c is friction, k is the spring constant and F(t) is an external force acting on the mass.

We are interested in periodic forcing, such as noncentered rotating parts, or perhaps loud sounds, or other sources of periodic force. Once we learn about Fourier series in chapter 4, we will see that we cover all periodic functions by simply considering F (t) = F0cos(ωt) (or sine instead of cosine, the calculations are essentially the same).

2.6.1 Undamped forced motion and resonance

First let us consider undamped (c = 0 ) motion for simplicity. We have the equation

mx ′′ + kx = F 0cos(ωt).

This equation has the complementary solution (solution to the associated homogeneous equation)

xc = C 1cos(ω0t) + C 2sin(ω 0t),

where  √ --- ω 0 = k∕m is the natural frequency (angular). It is the frequency at which the system “wants to oscillate” without external interference.

Let us suppose that ω0 ⇔ ω . We try the solution xp = Ac os(ωt ) and solve for A . Note that we do not need a sine in our trial solution as after plugging in we only have cosines. If you include a sine, it is fine; you will find that its coefficient is zero (I could not find a second rhyme).

We solve using the method of undetermined coefficients. We find that

 F xp = -----0-----cos(ωt). m(ω 20 - ω 2)

We leave it as an exercise to do the algebra required.

The general solution is

|--------------------------------------------------| | F 0 | | x = C 1cos(ω 0t) + C 2sin(ω 0t) +--2----2-cos(ωt). | -------------------------------m-(ω-0 --ω-)----------

Written another way

x = C cos(ω t - γ) +----F-0----co s(ωt). 0 m (ω20 - ω2)

Hence it is a superposition of two cosine waves at different frequencies.

Example 2.6.1: Take

 ′′ ′ 0.5x + 8x = 10cos(πt), x(0) = 0, x (0) = 0.

Let us compute. First we read off the parameters: ω = π ,  ---- ω0 = √ 8∕0.5 = 4 , F 0 = 10 , m = 0.5 . The general solution is

x = C1c os(4t) + C 2sin(4t) +--20---cos(πt). 16 - π2

Solve for C1 and C2 using the initial conditions. It is easy to see that C1 = --202 16-π and C2 = 0 . Hence

x = ---20--(cos(πt) - cos(4t)). 1 6 - π2


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Figure 2.5: Graph of -20-( ) 16-π2 cos(πt) - cos(4t) .


Notice the “beating” behavior in Figure 2.5. First use the trigonometric identity

 ( ) ( ) 2 sin A---B- sin A-+-B- = cosB - cosA 2 2

to get that

 ( ( ) ( )) --20--- 4 --π 4-+-π x = 16 - π 2 2sin 2 t sin 2 t .

Notice that x is a high frequency wave modulated by a low frequency wave.

Now suppose that ω0 = ω . Obviously, we cannot try the solution A cos(ωt ) and then use the method of undetermined coefficients. We notice that cos(ωt) solves the associated homogeneous equation. Therefore, we need to try xp = At cos(ωt ) + Btsin(ωt) . This time we do need the sine term since the second derivative of tco s(ωt ) does contain sines. We write the equation

x′′ + ω 2x = F-0cos(ωt). m

Plugging xp into the left hand side we get

2B ω cos(ωt) - 2A ω sin (ωt ) = F-0cos(ωt). m

Hence A = 0 and  F B = 2m0ω . Our particular solution is  F 2m0ωt sin(ωt) and our general solution is

 -F-0- x = C 1cos(ωt) + C 2sin(ωt ) + 2m ω tsin(ωt).


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Figure 2.6: Graph of 1πtsin(πt) .


The important term is the last one (the particular solution we found). This term grows without bound as t → ∞ . In fact it oscillates between F0t 2mω and -F0t- 2mω . The first two terms only oscillate between  ∘ -------- ± C 21 + C22 , which becomes smaller and smaller in proportion to the oscillations of the last term as t gets larger. In Figure 2.6 we see the graph with C 1 = C2 = 0 , F 0 = 2 , m = 1 , ω = π .

By forcing the system in just the right frequency we produce very wild oscillations. This kind of behavior is called resonance or perhaps pure resonance. Sometimes resonance is desired. For example, remember when as a kid you could start swinging by just moving back and forth on the swing seat in the “correct frequency”? You were trying to achieve resonance. The force of each one of your moves was small, but after a while it produced large swings.

On the other hand resonance can be destructive. In an earthquake some buildings collapse while others may be relatively undamaged. This is due to different buildings having different resonance frequencies. So figuring out the resonance frequency can be very important.

A common (but wrong) example of destructive force of resonance is the Tacoma Narrows bridge failure. It turns out there was a different phenomenon at play1.

2.6.2 Damped forced motion and practical resonance

In real life things are not as simple as they were above. There is, of course, some damping. Our equation becomes

 ′′ ′ mx + cx + kx = F0 cos(ωt ),
(2.8)

for some c > 0 . We have solved the homogeneous problem before. We let

 ∘ --- c k p = --- ω 0 = --. 2m m

We replace equation (2.8) with

 ′′ ′ 2 F-0 x + 2px + ω0x = m cos(ωt).

The roots of the characteristic equation of the associated homogeneous problem are  ∘-------- 2 2 r1,r2 = -p ± p - ω 0 . The form of the general solution of the associated homogeneous equation depends on the sign of p 2 - ω20 , or equivalently on the sign of c2 - 4km , as we have seen before. That is,

 (| r1t r2t 2 ||||C 1e + C 2e if c > 4km, xc = {||C 1e-pt + C2te-pt if c2 = 4km, |||( -pt( ) 2 e C1co s(ω 1t) + C2sin(ω1t) if c < 4km ,

where  ∘ -------- ω 1 = ω20 - p2 . In any case, we see that xc(t) → 0 as t → ∞ . Furthermore, there can be no conflicts when trying to solve for the undetermined coefficients by trying x = A cos(ωt) + B sin(ωt ) p . Let us plug in and solve for A and B . We get (the tedious details are left to reader)

((ω 2- ω 2)B - 2 ωpA )sin(ωt ) + ((ω 2- ω 2)A + 2ωpB )cos(ωt) = F0-cos(ωt). 0 0 m

We solve for A and B :

 (ω20 - ω 2)F0 A = --------2------2-----2 , m (2ωp ) + m (ω 0 - ω 2) 2ωpF 0 B = --------2------2----22 . m (2ωp ) + m (ω 0 - ω )
We also compute C = √A-2 +-B2 to be
 F C = --∘---------0----------. m (2ωp )2 + (ω2 - ω2)2 0

Thus our particular solution is

 (ω 2- ω 2)F0 2ωpF 0 xp = --------02------------2 cos(ωt) +--------2------------2 sin (ωt ). m (2ωp ) + m(ω 20 - ω 2) m (2ωp ) + m (ω 20 - ω 2)

Or in the alternative notation we have amplitude C and phase shift γ where (if ω ⇔ ω 0 )

 B 2ωp tan γ = --= --2-----. A ω 0 - ω 2

Hence we have

|------------------------------------------| | x = ----------F-0----------cos(ωt - γ). | | p ∘ 2 2 2 2 | | m (2ωp ) + (ω0 - ω ) | -------------------------------------------

If ω = ω0 we see that A = 0 ,  -F0- B = C = 2mωp , and  π γ = ∕2 .

The exact formula is not as important as the idea. Do not memorize the above formula, you should instead remember the ideas involved. For a different forcing function F , you will get a different formula for xp . So there is no point in memorizing this specific formula. You can always recompute it later or look it up if you really need it.

For reasons we will explain in a moment, we call xc the transient solution and denote it by xtr . We call the x p we found above the steady periodic solution and denote it by x sp . The general solution to our problem is

x = xc + xp = xtr + xsp.


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Figure 2.7: Solutions with different initial conditions for parameters k = 1 , m = 1 , F0 = 1 , c = 0.7 , and ω = 1.1 .


We note that xc = xtr goes to zero as t → ∞ , as all the terms involve an exponential with a negative exponent. So for large t , the effect of xtr is negligible and we see essentially only xsp . Hence the name transient. Notice that xsp involves no arbitrary constants, and the initial conditions only affect x tr . This means that the effect of the initial conditions is negligible after some period of time. Because of this behavior, we might as well focus on the steady periodic solution and ignore the transient solution. See Figure 2.7 for a graph given several different initial conditions.

The speed at which xtr goes to zero depends on p (and hence c ). The bigger p is (the bigger c is), the “faster” xtr becomes negligible. So the smaller the damping, the longer the “transient region.” This agrees with the observation that when c = 0 , the initial conditions affect the behavior for all time (i.e. an infinite “transient region”).

Let us describe what we mean by resonance when damping is present. Since there were no conflicts when solving with undetermined coefficient, there is no term that goes to infinity. We look at the maximum value of the amplitude of the steady periodic solution. Let C be the amplitude of xsp . If we plot C as a function of ω (with all other parameters fixed) we can find its maximum. We call the ω that achieves this maximum the practical resonance frequency. We call the maximal amplitude C(ω ) the practical resonance amplitude. Thus when damping is present we talk of practical resonance rather than pure resonance. A sample plot for three different values of c is given in Figure 2.8. As you can see the practical resonance amplitude grows as damping gets smaller, and any practical resonance can disappear when damping is large.


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Figure 2.8: Graph of C (ω ) showing practical resonance with parameters k = 1 , m = 1 , F 0 = 1 . The top line is with c = 0.4 , the middle line with c = 0.8 , and the bottom line with c = 1.6 .


To find the maximum we need to find the derivative C ′(ω ) . Computation shows

 2 2 2 C ′(ω ) =---2ω-(2p-+-ω----ω-0)F0--. ( 2 2 22)3∕2 m (2 ωp) + (ω 0 - ω )

This is zero either when ω = 0 or when 2p 2 + ω 2 - ω 20 = 0 . In other words, C ′(ω ) = 0 when

|------------------------------| | ∘--2-----2 | | ω = ω 0 - 2p or ω = 0. | -------------------------------

It can be shown that if  2 2 ω 0 - 2p is positive, then ∘ --2------ ω 0 - 2p 2 is the practical resonance frequency (that is the point where C (ω ) is maximal, note that in this case C ′(ω ) > 0 for small ω ). If ω = 0 is the maximum, then essentially there is no practical resonance since we assume that ω > 0 in our system. In this case the amplitude gets larger as the forcing frequency gets smaller.

If practical resonance occurs, the frequency is smaller than ω0 . As the damping c (and hence p ) becomes smaller, the practical resonance frequency goes to ω0 . So when damping is very small, ω 0 is a good estimate of the resonance frequency. This behavior agrees with the observation that when c = 0 , then ω0 is the resonance frequency.

The behavior is more complicated if the forcing function is not an exact cosine wave, but for example a square wave. It will be good to come back to this section once we have learned about the Fourier series.

2.6.3 Exercises

Exercise 2.6.1: Derive a formula for xsp if the equation is mx ′′ + cx′ + kx = F 0sin(ωt) . Assume c > 0 .

Exercise 2.6.2: Derive a formula for xsp if the equation is  ′′ ′ mx + cx + kx = F0 cos(ωt ) + F1 cos(3 ωt) . Assume c > 0 .

Exercise 2.6.3: Take  ′′ ′ mx + cx + kx = F 0cos(ωt) . Fix m > 0 , k > 0 , and F0 > 0 . Consider the function C (ω ) . For what values of c (solve in terms of m , k , and F0 ) will there be no practical resonance (that is, for what values of c is there no maximum of C (ω) for ω > 0 )?

Exercise 2.6.4: Take mx ′′ + cx ′ + kx = F0co s(ωt) . Fix c > 0 , k > 0 , and F0 > 0 . Consider the function C (ω) . For what values of m (solve in terms of c , k , and F 0 ) will there be no practical resonance (that is, for what values of m is there no maximum of C (ω ) for ω > 0 )?

Exercise 2.6.5: Suppose a water tower in an earthquake acts as a mass-spring system. Assume that the container on top is full and the water does not move around. The container then acts as a mass and the support acts as the spring, where the induced vibrations are horizontal. Suppose that the container with water has a mass of m = 10,000 kg . It takes a force of 1000 newtons to displace the container 1 meter. For simplicity assume no friction. When the earthquake hits the water tower is at rest (it is not moving).

Suppose that an earthquake induces an external force  2 F (t) = mA ω cos(ωt ) .

a) What is the natural frequency of the water tower?

b) If ω is not the natural frequency, find a formula for the maximal amplitude of the resulting oscillations of the water container (the maximal deviation from the rest position). The motion will be a high frequency wave modulated by a low frequency wave, so simply find the constant in front of the sines.

c) Suppose A = 1 and an earthquake with frequency 0.5 cycles per second comes. What is the amplitude of the oscillations? Suppose that if the water tower moves more than 1.5 meter from the rest position, the tower collapses. Will the tower collapse?

Exercise 2.6.101: A mass of 4 kg on a spring with k = 4 and a damping constant c = 1 . Suppose that F = 2 0 . Using forcing function F cos(ωt) 0 , find the ω that causes practical resonance and find the amplitude.

Exercise 2.6.102: Derive a formula for x sp for mx ′′ + cx ′ + kx = F co s(ωt) + A 0 , where A is some constant. Assume c > 0 .

Exercise 2.6.103: Suppose there is no damping in a mass and spring system with m = 5 , k = 20 , and F 0 = 5 . Suppose that ω is chosen to be precisely the resonance frequency. a) Find ω . b) Find the amplitude of the oscillations at time t = 100 .

1K. Billah and R. Scanlan, Resonance, Tacoma Narrows Bridge Failure, and Undergraduate Physics Textbooks, American Journal of Physics, 59(2), 1991, 118–124, http://www.ketchum.org/billah/Billah-Scanlan.pdf