Simple Harmonic Motion — Question 7

Question 7

Periodic does not necessarily mean simple harmonic

A periodic signal is $x(t)=A\cos\omega t+\epsilon A\cos2\omega t$, where $A>0$, $\omega>0$ and $\epsilon=0.10$. The equilibrium reference is zero.

1. Compute $a+\omega^2x$ and decide whether this signal satisfies the SHM equation with frequency $\omega$.

2. Could it instead be exactly SHM at some other single frequency? Use the two independent harmonics to justify the answer.

Question 7 - Solution

Periodic does not necessarily mean simple harmonic

(a) Derivation and calculation

$$x=A\cos\omega t+\epsilon A\cos2\omega t,$$$$a=-\omega^2A\cos\omega t-4\epsilon\omega^2A\cos2\omega t.$$$$\boxed{a+\omega^2x=-3\epsilon\omega^2A\cos2\omega t=-0.30\omega^2A\cos2\omega t\ne0}.$$The original SHM equation therefore fails, even though $x(t+2\pi/\omega)=x(t)$.

(b) Result and interpretation

For an arbitrary candidate frequency $\Omega$,$$a+\Omega^2x=A(\Omega^2-\omega^2)\cos\omega t$$$$\hspace{2.0cm}+\epsilon A(\Omega^2-4\omega^2)\cos2\omega t.$$Independent nonzero harmonics require both coefficients to vanish:$$\Omega^2=\omega^2\quad\text{and}\quad\Omega^2=4\omega^2,$$which is impossible for $\omega>0$.

Physical check

The limiting case $\epsilon=0$ removes the second constraint and restores exact SHM.