# Driven oscillators

#### Prerequisites

How do you like to go up in a swing, Up in the air so blue?
Oh, I do think it the pleasantest thing Ever a child can do!
Up in the air and over the wall, Till I can see so wide,
River and trees and cattle and all Over the countryside--
Till I look down on the garden green, Down on the roof so brown--
Up in the air I go flying again, Up in the air and down!

R.L. Stevenson, A Child’s Garden of Verses

## Everyday experiences

Most of us have had the experience of swinging in a child’s swing in a playground. One of the things you may remember is being pushed — or you might have had the experience of pushing a friend, cousin, or younger sibling. What you might recall is that if you wanted to swing high, you had to be pushed (or push) just at the right time.

A swing is a pendulum and a pendulum is an oscillator. This phenomenon of pushing a swing is our base personal-experience example of the phenomenon of a driven oscillator and resonance. The swing is a (damped) oscillator because it will oscillate naturally — for small amplitudes at a frequency (cycles per second) that we have found to be

$$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$$

(See The pendulum.)

When you push it, it is a driven oscillator because you are periodically adding energy to it through the use of an outside force. And it demonstrates resonance because the effect depends on how well the frequency of the outside force (pusher) matches with the natural frequency of the oscillator (the frequency with which you would swing naturally. This makes sense; if you push the swinger to speed her up just when she is beginning to swing down, her speed will increase a bit each time and her amplitude grow.

This phenomenon is what is responsible for the tuning of the hairs in the ear to distinguish different tones.

## Setting up the model and its equation of motion

To study this important phenomenon, let’s turn to (as usual) the simplest example we can think of that has all the parts needed. Consider a mass on a spring where the base of the spring is being oscillated back and forth at some frequency.

Without the hand, if the spring’s rest length is $L_0$ and its length is $L$ then the force on the cart is due to the stretch of the spring. We’ll choose to take the origin of our coordinate system ($x = 0$) when the cart is at its equilibrium position — the spring is at its rest length.

$$ma = -k(L-L_0) - bv$$

$$L = x + L_0\quad \mathrm{(not\;driven)}$$

Now let's let the hand holding the back end of the spring shift its position by an amount $Δx_h$. If this is positive, the spring is shorter — scrunched up more than it would otherwise be. So the equation that corrects for the shift of the hand’s position is

$$L = x - \Delta x_h +L_0\quad \mathrm{(driven)}$$

$$ma = -k(x-\Delta x_h) - bv$$

To see how this looks compared to what we did before, we’ll put the terms that include an “$x$” to the left and make it into a differential equation. Then we’ll oscillate the hand back and forth with an amplitude $A_h$ and an angular frequency $ω$. Here’s the resulting equation:

$$ma +bv +kx = \Delta x_h$$

Putting this in differential equation form by using the derivative definitions of $a$ and $v$ given

$$\frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2 x = \omega_0^2 \Delta x_h$$

If we now put in the way the hand is oscillating, we get

$$\frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2 x = \omega_0^2 A_0 \cos{(\omega t)}$$

Note that the driving frequency, $\omega$, is NOT necessarily equal to the natural frequency of the oscillator, $\omega_0$. We can vary $\omega$ as we like and see what happens.

This is a differential equation — an equation that relates an unknown function $x(t)$ and its derivatives. But it differs from the plain vanilla harmonic oscillator differential equation in that there is a term on the right that does NOT depend on $x(t)$ at all. This makes the differential equation inhomogeneous. We will not attempt to solve this equation here. (If you want to see how to do it, go to a text or website on ordinary differential equations and look up inhomogeneous linear equations.) But we will explore the results qualitatively.

## Exploring the driven oscillator qualitatively: resonance

Our “simplest possible model” of a mass on a spring with a “hand” driving the other end of the spring is nicely modeled in the on-line “Resonance” simulation from the University of Colorado’s PhET group. The screen is shown below.

Since “gravity” is turned off in the control panel, we can visualize it as a horizontal oscillation viewed from above. We’ve turned the ruler on so we can measure the amplitude of the oscillation. We have set the mass to be $m$ = 1 kg and the spring constant to be $k$ = 25 N/m. This makes the undamped angular frequency $ω_0$ (equal $\sqrt{k/m}$) to be 5 radians/s. The frequency is then$f = ω_0/2π$ (since 2π radians = 1 cycle so 1 = (1 cycle)/($2π$ radians)) about 0.8 Hertz. We’ve chosen a damping constant b = 0.5, so we have $γ = b/m$ = 0.5 s-1, a time constant $τ = 2/γ$ = 2 seconds.

Run this program and test it out by first grabbing the mass with your mouse, pulling the mass up a bit and releasing it. Watch the mass oscillate, losing amplitude as it goes. The time constants should agree reasonably with a 1.2 second period and a 2 second time scale for losing a factor of 3 of amplitude. Play with the parameters and see how the oscillation responds.

Now turn on the driver. This oscillates the other end of the spring. You should find that the driver creates the largest oscillation when its angular frequency of the driver matches the shifted natural frequency of the damped oscillator, $ω_1$

$$\omega_1 = \sqrt{\omega_0^2 - \frac{\gamma^2}{4}}.$$

(Recall from our damped oscillator discussion that the undamped frequency gets shifted down a bit by the damping.)

The PhET program allows you to illustrate this by driving multiple masses at the same time, changing either their masses or spring constants or both. Here's the result of driving 5 different masses with the same spring constant. The one that has its value of $ω_1$ best matched to the driving frequency gets kicked to the largest amplitude, just like the child in the swing, pushed at the right frequency (or an ear-hair of the right length being driven by a particular sound frequency). This is called a resonance and the natural frequency of the oscillator, $ω_1$, is called the resonant frequency.

## Q value

How big an amplitude you get when you drive an oscillator at the resonant frequency is determined by how strong the damping is. To make a dimensionless constant, we can take the ratio of the undamped angular frequency, $ω_0$, to the damping constant, γ, since both have dimensions of 1/time. This ratio is called the Q-value of the oscillator.

$$Q = \frac{ω_0}{γ}.$$

Since these are related to time constants, $ω_0$ to the undamped period, $T$; and $γ$ to the decay time, $τ$, we can also write $Q$ as a ratio of these two times:

$$Q = \frac{ω_0}{γ} =(\frac{2π}{T})/(\frac{2}{τ}) = π\frac{τ}{T}.$$

How high, medium, and low $Q$ oscillators respond to driving is shown in the figure at the right. The narrower the curve (the higher $Q$), the better "tuned" o the particular driving frequency is to the oscillator.

Joe Redish 3/24/12

Article 671