Why does the amplitude of an undamped driven oscillator not vary with time?

[adamjts]

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/(ω²-ω²))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?

 

[BvU]

Which is to say, that the exact moment that we introduce this driving force, the amplitude of the oscillator instantaneously becomes enormous

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.

 

[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 $x''+\omega_0^2 x=\cos(\omega_0 t)$

 

[JJacquelin]

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