Forced Oscillations and Resonance — Question 6

Question 6

Exact undamped resonance has growth instead of a settled amplitude

An undamped oscillator starts at rest at equilibrium. For $t\ge0$, it obeys $\ddot x+\omega_0^2x=(F_0/m)\cos\omega_0t$. The force remains exactly resonant.

1. Find a solution satisfying both initial conditions.

2. Find its energy at times $t_n=n\pi/\omega_0$. Why is taking $b\to0$ in a damped steady-state amplitude insufficient to describe this initial-value problem?

Question 6 - Solution

Exact undamped resonance has growth instead of a settled amplitude

(a) Derivation and calculation

Try $x=Ct\sin\omega_0t$. Then$$\dot x=C\sin\omega_0t+C\omega_0t\cos\omega_0t,$$$$\ddot x+\omega_0^2x=2C\omega_0\cos\omega_0t.$$$$C=\frac{F_0}{2m\omega_0}\Rightarrow\boxed{x=\frac{F_0t}{2m\omega_0}\sin\omega_0t}.$$Both $x(0)=0$ and $\dot x(0)=0$ are satisfied.

(b) Result and interpretation

At $t_n=n\pi/\omega_0$,$$x(t_n)=0,\quad v(t_n)=\frac{F_0t_n}{2m}(-1)^n.$$$$\boxed{E(t_n)=\tfrac12mv(t_n)^2=\frac{F_0^2t_n^2}{8m}}.$$There is no bounded periodic steady state. The limit $b\to0$ of a settled amplitude signals divergence but does not supply this initial-value solution or its time dependence.

Physical check

Without damping, the free transient never decays; the long-time and zero-damping limits cannot be interchanged casually.