Forced Oscillations and Resonance — Question 7

Question 7

Detuned undamped driving retains a free oscillation

An undamped oscillator starts at rest at equilibrium and obeys $\ddot x+\omega_0^2x=(F_0/m)\cos\Omega t$, with $\Omega\ne\omega_0$.

1. Derive its complete motion and rewrite it to expose the slow modulation when the frequencies are close.

2. For $\omega_0=5\,\mathrm{rad/s}$, $\Omega=4\,\mathrm{rad/s}$ and $F_0/m=.090\,\mathrm{m/s^2}$, find the envelope bound and the spacing of its slow envelope zeros. Explain why the forced term alone fails the initial conditions.

Question 7 - Solution

Detuned undamped driving retains a free oscillation

(a) Derivation and calculation

A particular solution is $C\cos\Omega t$, where $C=F_0/[m(\omega_0^2-\Omega^2)]$. Initial rest fixes the free term:$$\boxed{x=C(\cos\Omega t-\cos\omega_0t)}.$$$$\boxed{x=\frac{2F_0}{m(\omega_0^2-\Omega^2)}\sin\frac{(\omega_0+\Omega)t}{2}\sin\frac{(\omega_0-\Omega)t}{2}}.$$

(b) Result and interpretation

$$C=.090/(25-16)=.0100\,\mathrm m,\quad\boxed{x=.0200\sin4.5t\sin.5t\,\mathrm m}.$$$$|x|\le.0200|\sin.5t|\,\mathrm m,\quad\boxed{\Delta t_{\mathrm{envelope\ zeros}}=2\pi\,\mathrm s}.$$The forced term alone has $x(0)=C\ne0$; the nondecaying free cosine cancels it at release.

Physical check

As $\Omega\to\omega_0$, the cosine difference divided by $\omega_0^2-\Omega^2$ tends to the resonant growth solution.