Energy in Simple Harmonic Motion — Question 2

Question 2

Equal mean energies do not imply equal energies at every instant

An ideal spring oscillator has $x=A\cos\omega t$ and total energy $E=kA^2/2$.

1. Find $U(t)$ and $K(t)$, their cycle averages, and the frequency at which each energy oscillates.

2. What fraction of a cycle has $K>3U$? Compare this with the statement that $\langle K\rangle=\langle U\rangle$.

Question 2 - Solution

Equal mean energies do not imply equal energies at every instant

(a) Derivation and calculation

Put $\psi=\omega t+\phi$ and $E=kA^2/2$:$$U=\tfrac12kx^2=E\cos^2\psi=\tfrac E2(1+\cos2\psi),$$$$K=\tfrac12m\omega^2A^2\sin^2\psi=\tfrac E2(1-\cos2\psi).$$$$\boxed{\langle U\rangle=\langle K\rangle=E/2},\qquad \omega_{\mathrm{energy}}=2\omega.$$

(b) Result and interpretation

$$K>3U\iff\sin^2\psi>3\cos^2\psi\iff|\cos\psi|<\tfrac12.$$$$\psi\in(\pi/3,2\pi/3)\cup(4\pi/3,5\pi/3).$$$$\boxed{\frac{\Delta t}{T}=\frac{2(\pi/3)}{2\pi}=\frac13}.$$

Physical check

Energy repeats after $T/2$ because reversing both $x$ and $v$ leaves their squares unchanged.