** Higher order linear ODEs**

**2.6. FORCED OSCILLATIONS AND RESONANCE 99 Hence we have**

xp=
F0
m
q
(2ωp)2+(ω2_{0}−ω2_{)}2
cos(ωt−γ).

Ifω =ω0we see thatA= 0,B=C = _{2}_{m}F_{ω}0_{p}, andγ= π/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 functionF, you will get a different formula forxp. 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 thexpwe found above thesteady periodic solutionand denote it byxsp. The general

solution to our problem is

x= xc+xp = xtr+ xsp.

The transient solution xc = xtr goes to zero

0 5 10 15 20 0 5 10 15 20 -5.0 -2.5 0.0 2.5 5.0 -5.0 -2.5 0.0 2.5 5.0

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

ω=1.1.

ast → ∞, as all the terms involve an exponential with a negative exponent. So for larget, the effect of xtr is negligible and we see essentially only

xsp. Hence the nametransient. Notice that xsp

involves no arbitrary constants, and the initial con- ditions only affectxtr. This means that the effect

of the initial conditions is negligible after some pe- riod of time. Because of this behavior, we might as well focus on the steady periodic solution and ignore the transient solution. See for a graph given several different initial conditions. The speed at which xtr goes to zero depends

on p(and hencec). The bigger pis (the bigger cis), the “faster”xtr becomes negligible. So the

smaller the damping, the longer the “transient region.” This agrees with the observation that

whenc= 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. LetCbe the amplitude of xsp. If we plotCas a function ofω(with all other parameters fixed) we can find its maximum. We

call theω that achieves this maximum thepractical resonance frequency. We call the maximal amplitude C(ω) the practical resonance amplitude. Thus when damping is present we talk of practical resonancerather than pure resonance. A sample plot for three different values ofcis given

in . As you can see the practical resonance amplitude grows as damping gets smaller, and practical resonance can disappear altogether when damping is large.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 0.0 0.5 1.0 1.5 2.0 2.5

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 derivativeC0(ω). Computation shows
C0(ω)= −2ω(2p
2+ω2_{−}ω2
0)F0
m (2ωp)2+(ω2
0−ω2)
23/2
.

This is zero either whenω =0 or when 2p2+ω2−ω2_{0} = 0. In other words,C0(ω)= 0 when

ω= qω2

0−2p2 or ω= 0.

It can be shown that ifω2_{0}−2p2is positive, then
q

ω2

0−2p2is the practical resonance frequency

(that is the point whereC(ω) is maximal, note that in this caseC0_{(}ω_{)}>_{0 for small}ω_{). If}ω=_{0 is}

the maximum, then there is no practical resonance since we assumeω >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 thanω0. As the dampingc(and hence p)

becomes smaller, the practical resonance frequency goes toω0. So when damping is very small,ω0

is a good estimate of the resonance frequency. This behavior agrees with the observation that when c=0, thenω0is the resonance frequency.

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

2.6. FORCED OSCILLATIONS AND RESONANCE 101

### 2.6.3

### Exercises

Exercise2.6.1: Derive a formula for xspif the equation is mx00+cx0+kx =F0sin(ωt). Assume

c>0.

Exercise2.6.2: Derive a formula for xspif the equation is mx00+cx0+kx= F0cos(ωt)+F1cos(3ωt).

Assume c> 0.

Exercise2.6.3: Take mx00 +cx0+kx = F0cos(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)forω >0)?

Exercise2.6.4: Take mx00+cx0 +kx = F0cos(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 F0) will there be no practical

resonance (that is, for what values of m is there no maximum of C(ω)forω >0)?

Exercise2.6.5: 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 F(t)= mAω2cos(ωt). a) What is the natural frequency of the water tower?

b) Ifωis 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=1and 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?

Exercise2.6.101: A mass of 4 kg on a spring with k =4N/_{m}_{and a damping constant c} = _{1}Ns/_{m}_{.}

Suppose that F0= 2 N. Using forcing function F0cos(ωt), find theωthat causes practical resonance

and find the amplitude.

Exercise2.6.102: Derive a formula for xspfor mx00+cx0+kx= F0cos(ωt)+A, where A is some

constant. Assume c>0.

Exercise2.6.103: Suppose there is no damping in a mass and spring system with m =5, k=20,

and F0=5. Supposeωis chosen to be precisely the resonance frequency. a) Findω. b) Find the