Driven harmonic oscillator: Why does the phase of the driver have such a big impact on the solution?

Question

Consider the equation for a driven harmonic oscillator, $$\frac{d^2 x}{d t^2} + \omega_0^2x(t) = f(t),$$ with initial conditions $$x(0)=\left.\frac{dx}{dt}\right|_{t=0}=0.$$ If we assume $\omega_0\ne\omega$ then $$x(t)=x_0\omega^2\begin{cases} \frac{\omega_0\sin(\omega t) - \omega \sin(\omega_0 t)}{\omega_0(\omega_0^2 - \omega^2)}, & f(t)=x_0\omega^2\sin(\omega t), \\ \frac{\cos(\omega t) - \cos(\omega_0 t)}{\omega_0^2 - \omega^2}, & f(t)=x_0\omega^2\cos(\omega t). \end{cases}$$ In the limit $\omega\rightarrow \infty$, $$x(t)\rightarrow x_0\begin{cases} \frac{\omega}{\omega_0}\sin(\omega_0 t), & f(t)=x_0\omega^2\sin(\omega t), \\ \cos(\omega_0 t) - \cos(\omega t), & f(t)=x_0\omega^2\cos(\omega t). \end{cases}$$ This shows that depending on the phase of the driver the amplitude can be infinite or finite. Do you know a physical reason why this might be, or have I made a mistake in the maths somewhere?

Answer 1

Consider the easier differential equation $$ \frac{d^2x}{dt^2} = f(t) $$ with $x(0) = \frac{dx}{dt}\Big|_{t=0} = 0$. If $f(t) = x_0 \omega^2 \sin(\omega t)$ we get $$ x(t) = x_0 \omega t - x_0 \sin(\omega t) .$$ If $f(t) = x_0 \omega^2 \cos(\omega t)$ we get $$ x(t) = x_0 - x_0 \cos(\omega t) .$$ Again, the phase of the driving term determines whether the amplitude blows up or stays bounded.

It basically boils down to the constant terms. In the first case, there is a net velocity in the motion, which causes the solution to drift to infinity. In the second case, you 'get lucky' and the drift is exactly zero.