# Oscillations III: Damped Driven Oscillator

*Michael Fowler *

### A Driven Damped Oscillator: Equation of Motion

Now apply a periodic external driving force to the damped oscillator analyzed above: if the driving force has the same period as the oscillator, the amplitude can increase, perhaps to disastrous proportions, as in the famous case of the Tacoma Narrows Bridge.

The equation of motion for a driven damped oscillator is:

_{$$m\frac{{d}^{2}x}{d{t}^{2}}+b\frac{dx}{dt}+kx={F}_{0}\mathrm{cos}\omega t.$$}_{}

We shall be using $\omega $ for the *driving
*frequency, and ${\omega}_{0}$ for the *natural
*frequency of the oscillator (meaning that ignoring damping, so ${\omega}_{0}=\sqrt{k/m}.$ )

### The Driven Steady State Solution and Initial Transient Behavior

The solution to this differential equation is not unique: as with any second order differential equation, there are two constants of integration, which are determined by specifying the initial position and velocity.

To get a general idea of how a damped driven oscillator behaves under a wide variety of conditions, check out this spreadsheet for damped driven oscillator.

However, as we shall prove below using complex numbers, the
equation *does* have a *unique* steady state solution with $x$ oscillating at the same frequency as the
external drive. How can that be fitted
to arbitrary initial conditions? The key
is that we can add to the steady state solution any solution of the *undriven* equation $m\frac{{d}^{2}x}{d{t}^{2}}+b\frac{dx}{dt}+kx=0,$ and we’ll clearly still have a solution of the
full damped driven equation.

We know what those undriven solutions look like: they all
die away as time goes on. So, we can add
such a solution to fit the specified initial conditions, and after a while the
system will lose memory of those conditions and settle into the steady driven
solution. The initial deviations from the steady solution needed to satisfy
initial conditions are termed *transients*.

Here’s a pair of examples: the same driven damped oscillator, started with zero velocity, once from the origin and once from 0.5:

Notice that after about 70 seconds, the two curves are the same, both in amplitude and phase.

### Using Complex Numbers to Solve the Driven Steady State Equation Easily

We begin by writing:

external driving force = ${F}_{0}{e}^{i\omega t}$

with ${F}_{0}$ real, so the actual physical driving force is just the real part of this, that is, ${F}_{0}\mathrm{cos}\omega t.$

So now we’re trying to solve the equation

$$m\frac{{d}^{2}x}{d{t}^{2}}+b\frac{dx}{dt}+kx={F}_{0}{e}^{i\omega t}.$$

We’ll try the complex function, $x(t)=A{e}^{i(\omega t+\phi )}$,
with $A$ a real number, $x\left(t\right)$ cycling at the same frequency as the driving
force. We can always take the amplitude $A$ to be real: that is *not* a restriction, since we’ve added the adjustable phase factor ${e}^{i\phi}.$ Physically, this factor allows the solution
to differ from the driver in phase, which it usually does. If we succeed in finding an $x\left(t\right)$ that satisfies the equation, the real parts of
the two sides of the equation must be equal:

*If **$x(t)=A{e}^{i(\omega t+\phi )}$** is a solution to the
equation with the complex driving force, **${F}_{0}{e}^{i\omega t},$** its *real *part, **$A\mathrm{cos}\left(\omega t+\phi \right),$** will be a solution to
the equation with the *real physical* driving force, **${F}_{0}\mathrm{cos}\omega t$**.*

It’s very easy to check that $x\left(t\right)=A{e}^{i\left(\omega t+\phi \right)}$ *is* a
solution to the equation, with the right $A$ and $\phi $! Just plug it in and see what happens. The differentiations are simple, giving

$-m{\omega}^{2}A{e}^{i\left(\omega t+\phi \right)}+ib\omega A{e}^{i\left(\omega t+\phi \right)}+kA{e}^{i\left(\omega t+\phi \right)}={F}_{0}{e}^{i\omega t}.$

To nail down $A$ and $\phi ,$ we begin by cancelling out the common factor ${e}^{i\omega t},$ then shift the ${e}^{i\phi}$ to the other side to find

$$A=\frac{{F}_{0}{e}^{-i\phi}}{k-m{\omega}^{2}+ib\omega}.$$

Now $A$ is a real number, and the right hand side of this equation looks alarmingly complex, so what’s going on?

Let’s begin to untangle this by diagramming that complex number in the denominator,

$k-m{\omega}^{2}+ib\omega $.

It has real part $k-m{\omega}^{2}$ and imaginary part $ib\omega .$.

Like any complex number, it can be expressed in terms of its amplitude $r$ and its phase $\theta :$

$k-m{\omega}^{2}+ib\omega =r{e}^{i\theta}$.

Putting it in the equation in this $r,\theta $ notation gives

$$A=\frac{{F}_{0}{e}^{-i\phi}}{k-m{\omega}^{2}+ib\omega}=\frac{{F}_{0}{e}^{-i\phi}}{r{e}^{i\theta}}=\frac{{F}_{0}}{r}{e}^{-i\left(\phi +\theta \right)}.$$

Now, ${F}_{0}$ and $r$ are real, so $A$ *will*
be real (as it must be) *if* ${e}^{-i\left(\phi +\theta \right)}$ is real: so $\phi =-\theta ,$ and

$$A=\frac{{F}_{0}}{r}=\frac{{F}_{0}}{\sqrt{{m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}}},\text{\hspace{1em}}x\left(t\right)=A{e}^{i\left(\omega t-\theta \right)},$$

recalling $k=m{\omega}_{0}^{2}$, and here $$\theta ={\mathrm{tan}}^{-1}\left(\frac{b\omega}{k-m{\omega}^{2}}\right)$$ (see diagram).

So we’ve already solved the differential equation: the amplitude $A$ is proportional to the strength of the driving force, and that ratio is determined by the parameters of the undriven oscillator, and the driving oscillation frequency.

The important thing to note about the amplitude $A$ is that if the damping $b$ is small, $A$ gets *very large* when the frequency of
the driver approaches the natural frequency of the oscillator! This is called *resonance*, and is what
happened to the

The *phase lag* of
the oscillations behind the driver, $\theta ={\mathrm{tan}}^{-1}\left(b\omega /\left(k-m{\omega}^{2}\right)\right)$,
is completely determined by the frequency together with the physical constants
of the undriven oscillator: the mass, spring constant, and damping
strength. So, when the driving force *${F}_{0}{e}^{i\omega t}$** *generates
the motion $x\left(t\right)=A{e}^{i\left(\omega t+\phi \right)}=A{e}^{i\left(\omega t-\theta \right)}$, the lag angle $\theta $ is independent of the strength of the driving
force: a stronger force doesn’t get the oscillator more in sync, it just
increases the amplitude of the oscillations.

Note that at *low*
frequencies, $\omega \ll {\omega}_{0},$ the oscillator lags behind by a small angle,
but at resonance $\omega ={\omega}_{0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =\pi /2,$ and for driving frequencies above ${\omega}_{0},$ $\theta >\pi /2.$

### Back to Reality

** To summarize**: we’ve just established that $x\left(t\right)=A{e}^{i\left(\omega t-\theta \right)}$ with $A={F}_{0}/\sqrt{{m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}}$ and $\theta ={\mathrm{tan}}^{-1}\left(b\omega /\left(k-m{\omega}^{2}\right)\right)$ is a solution to the driven damped oscillator
equation $$m\frac{{d}^{2}x}{d{t}^{2}}+b\frac{dx}{dt}+kx={F}_{0}{e}^{i\omega t}$$with the complex driving force ${F}_{0}{e}^{i\omega t}.$

So, *equating the real parts
of the two sides of the equation*, since $m,b,k$ are all real,

*$x=A\mathrm{cos}\left(\omega t-\theta \right)$*

*is a solution of the equation with the real
driving force *

*${F}_{0}\mathrm{cos}\omega t$*

*.*We could have found this out without complex numbers, by using a trial solution $A\mathrm{cos}\left(\omega t+\phi \right).$ However, it’s not that easy$\u2014$the left hand side becomes a mix of sines and cosines, and one needs to use trig identities to sort it all out. With a little practice, the complex method is easier and is certainly more direct.

Now the total energy of the oscillator is

$$\begin{array}{c}E={\scriptscriptstyle \frac{1}{2}}m{v}^{2}+{\scriptscriptstyle \frac{1}{2}}k\text{}{x}^{2}\\ ={\scriptscriptstyle \frac{1}{2}}m{v}^{2}+{\scriptscriptstyle \frac{1}{2}}m{\omega}_{0}^{2}\text{}{x}^{2}.\end{array}$$

Putting in

$$x\left(t\right)=A\mathrm{cos}\left(\omega t-\theta \right),\text{\hspace{1em}}v\left(t\right)=-A\omega \mathrm{sin}\left(\omega t-\theta \right)$$

gives

$$E={\scriptscriptstyle \frac{1}{2}}m{A}^{2}\left({\omega}^{2}{\mathrm{sin}}^{2}\left(\omega t-\theta \right)+{\omega}_{0}^{2}{\mathrm{cos}}^{2}\left(\omega t-\theta \right)\right).$$

Note that this is *not*
constant through the cycle unless the oscillator is at resonance, $\omega ={\omega}_{0}.$

We can see from the above that at the resonant frequency $\omega ={\omega}_{0},$ $E={\scriptscriptstyle \frac{1}{2}}m{\omega}_{0}^{2}{A}^{2}$, and from the previous section

$$A=\frac{{F}_{0}}{\sqrt{{m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}}},$$

so the energy in the oscillator at the resonant frequency is

$${E}_{\text{resonance}}={\scriptscriptstyle \frac{1}{2}}m{\omega}_{0}^{2}{A}^{2}={\scriptscriptstyle \frac{1}{2}}m{\omega}_{0}^{2}\frac{{F}_{0}^{2}}{{b}^{2}{\omega}_{0}^{2}}={\scriptscriptstyle \frac{1}{2}}m\frac{{F}_{0}^{2}}{{b}^{2}}=\frac{{Q}^{2}}{2}\frac{{F}_{0}^{2}}{m{\omega}_{0}^{2}},$$

recalling that $Q={\omega}_{0}\tau ={\omega}_{0}m/b.$

So *Q*, the quality
factor, the measure of how long an oscillator keeps ringing, also *measures the strength of response* of the
oscillator to an external driver at the resonant frequency.

But what happens on going *away* from the resonant frequency?
Let’s assume that $Q$ is large, and the driving force is kept
constant. It won’t take much change in $\omega $ from ${\omega}_{0}$ for the denominator ${m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}$ in the expression for $E$ to double in size. In fact, for large $Q$ it’s a good approximation to replace $b\omega $ by $b{\omega}_{0}$ over that variation, and it is then
straightforward to check that the energy in the oscillator drops to one-half
its resonant value for $\omega -{\omega}_{0}\cong \pm {\omega}_{0}/2Q.$

*Exercise*: prove this.

The bottom line is that for increasing $Q,$ the response at the resonant frequency gets larger, but this large response takes place over a narrower and narrower range in driving frequencies.

### How Hard is the Driver Working to Keep this Thing Going?

It’s simplest to work with the real solution. Suppose the oscillator moves through $\Delta \text{\hspace{0.05em}}x$ in a time $\Delta t$ the driving force does work $\left({F}_{0}\mathrm{cos}\omega t\right)\Delta x$, so

$\text{rateofworkingattime}t\text{}=\text{}\left({F}_{0}\mathrm{cos}\omega t\right)\left(\Delta x/\Delta t\right)=\left({F}_{0}\mathrm{cos}\omega t\right)v\left(t\right)$

The important thing is the *average* rate of working of the driving force, the *mean power input*, found by averaging
over a complete cycle:

From $x(t)=A\mathrm{cos}(\omega t-\theta )$, $v\left(t\right)=-A\omega \mathrm{sin}\left(\omega t-\theta \right)$,

averaging the power input (the bar above means average over a complete cycle) and denoting average power by $P,$

$\begin{array}{l}P={F}_{0}\overline{\left(\mathrm{cos}\omega t\right)v\left(t\right)}\\ =-{F}_{0}A\omega \text{\hspace{0.17em}}\overline{\mathrm{cos}\omega t\mathrm{sin}\left(\omega t-\theta \right)}\\ =-{F}_{0}A\omega \text{\hspace{0.17em}}\overline{\mathrm{cos}\omega t\mathrm{sin}\omega t}\mathrm{cos}\theta +{F}_{0}A\omega \text{\hspace{0.17em}}\overline{{\mathrm{cos}}^{2}\omega t}\mathrm{sin}\theta \\ ={\scriptscriptstyle \frac{1}{2}}{F}_{0}A\omega \mathrm{sin}\theta \end{array}$

since over one cycle the average $\overline{{\mathrm{cos}}^{2}\omega t}={\scriptscriptstyle \frac{1}{2}}$ and $\overline{\mathrm{cos}\omega t\mathrm{sin}\omega t}={\scriptscriptstyle \frac{1}{2}}\overline{\mathrm{sin}2\omega t}=0$.

(Remembering ${\mathrm{cos}}^{2}\omega \text{}\text{\hspace{0.05em}}t+{\mathrm{sin}}^{2}\omega \text{\hspace{0.05em}}t=1$ at all times, and sine is just cosine moved over, so they must have the same average over a complete cycle.)

This can be expressed entirely in terms of the driving force and frequency. Since

$$A=\frac{{F}_{0}}{\sqrt{{m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}}},\text{\hspace{1em}}\mathrm{sin}\theta =\frac{b\omega}{\sqrt{{m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}}},$$

$$P={\scriptscriptstyle \frac{1}{2}}{F}_{0}A\omega \mathrm{sin}\theta & =\frac{1}{2}\frac{b{\omega}^{2}{F}_{0}^{2}}{{m}^{2}{({\omega}_{0}^{2}-{\omega}^{2})}^{2}+{(b\omega )}^{2}}.$$

*Exercise
1*: Prove that for a lightly damped oscillator, at resonance the
oscillator extracts the most work from the driving force.

*Exercise 2*: Prove that
any solution of the damped oscillator equation (with $F=0$ ) can be added to the driven oscillator
solution, and gives another solution to the driven oscillator. How do you pick
the “right solution”?