# I Undamped Driven Oscillation

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1. Mar 24, 2016

The equations I'm getting when I solve the differential equations seem to imply that the amplitude of oscillation does not vary in time.

For example, if I have

x'' + ω02x = cos(ωt)

If we suppose that ω≠ω0,

then the general solution should look something like:

x(t) = c1cos(ω0t) + c2sin(ω0t) + (1/(ω22))cos(ωt)

This is okay with me mostly. But then thinking about what happens when ω→ω0 AND ω≠ω0, then obviously the amplitude of the oscillator should be huge. However, It would seem that the amplitude does not depend on time. Which is to say, that the exact moment that we introduce this driving force, the amplitude of the oscillator instantaneously becomes enormous. Which is hard to believe, because I would expect the object to start deviating from its simple oscillations more slowly and grow in time.

I know that when ω=ω0 that there is a factor of t in the amplitude, but that is not the case here.

Is it because the superposition of the two sinusoids makes it seem like the initial amplitudes are small. So when the driving force is introduced, the waves align such that the oscillating body does not seem to have a huge amplitude. But over some time, the waves will align such that the body does have evidently huge oscillations. This would imply, though, that the oscillations would become small again. In other words, we would expect long beats. Is this correct?

Or, maybe it's more likely that after the driver begins, the motion converges onto that of the driven oscillation? Why and how would one describe so mathematically?

2. Mar 24, 2016

### BvU

You do have initial conditions to determine c1 and c2 that determine the starting situation. Try to plot a simple case to see how it looks.
Without damping the $\omega_0$ oscillation goes on forever, indeed.

3. Mar 30, 2016

### JJacquelin

In order to understand what happen when $\omega$ tends to $\omega_0$ , let $\omega=\omega_0+\epsilon$
$\omega^2-\omega_0^2=(\omega+\omega_0)\epsilon \simeq 2\omega_0\epsilon$
$\sin(\omega t)=\sin(\omega_0 t+\epsilon t)=\sin(\omega_0 t)\cos(\epsilon t)+\cos(\omega_0 t)\sin(\epsilon t) \simeq \sin(\omega_0 t)+\cos(\omega_0 t)\epsilon t$
$\cos(\omega t)=\cos(\omega_0 t+\epsilon t)=\cos(\omega_0 t)\cos(\epsilon t)-\sin(\omega_0 t)\sin(\epsilon t) \simeq \cos(\omega_0 t)+\sin(\omega_0 t)\epsilon t$
The solution : $x(t)=c_1 \cos(\omega t)+c_2\sin(\omega t)+\frac{1}{\omega^2-\omega_0^2}\cos(\omega t)$ becomes :
$x(t) \simeq c_1 \cos(\omega_0 t)+c_2\sin(\omega_0 t)+c_2\cos(\omega_0 t) \epsilon t+\frac{1}{2\omega_0\epsilon}\left(\cos(\omega_0 t)+\sin(\omega_0 t)\epsilon t\right)$
$x(t) \simeq \left(c_1+\frac{1}{2\omega_0\epsilon} \right)\cos(\omega_0 t)+c_2\sin(\omega_0 t)+\frac{t}{2\omega_0}\sin(\omega_0 t)+c_2\cos(\omega_0 t)\epsilon t$
At $t=0$ the starting value is $x(0)=x_0=\left(c_1+\frac{1}{2\omega_0\epsilon} \right)$
In fact, $c_1$ depends on $\epsilon$ so that the initial condition be fulfilled. : $c_1=x_0-\frac{1}{2\omega_0\epsilon}$
$x(t) \simeq x_0\cos(\omega_0 t)+c_2\sin(\omega_0 t)+\frac{t}{2\omega_0}\sin(\omega_0 t)+c_2\epsilon t \cos(\omega_0 t)$
If $\epsilon=0$ the solution is : $x(t) = x_0\cos(\omega_0 t)+c_2\sin(\omega_0 t)+\frac{t}{2\omega_0}\sin(\omega_0 t)$
which is exacly the solution of the equation $x''+\omega_0^2 x=\cos(\omega_0 t)$

4. Mar 31, 2016

### JJacquelin

There are several typo in my first answer; I suppose that you can correct them.