Energy in Simple Harmonic Motion — Question 7

Question 7

Peak power occurs between equilibrium and the turning point

For an ideal spring oscillator, the instantaneous power delivered by the spring is $P=Fv=-kxv$. Let $x=A\cos\omega t$ and $\omega^2=k/m$.

1. Find the maximum magnitude of spring power and the positions where it occurs.

2. Find the net spring work over a full cycle and the work from $x=A$ to $x=0$. Explain how zero net work can coexist with large instantaneous power.

Question 7 - Solution

Peak power occurs between equilibrium and the turning point

(a) Derivation and calculation

$$x=A\cos\psi,\quad v=-A\omega\sin\psi,\quad F=-kA\cos\psi.$$$$\boxed{P=Fv=kA^2\omega\sin\psi\cos\psi=E\omega\sin2\psi}.$$$$\boxed{|P|_{\max}=E\omega},\qquad|\sin2\psi|=1\Rightarrow |x|=A/\sqrt2.$$

(b) Result and interpretation

$$W_{\mathrm{cycle}}=\oint(-kx)\,dx=0.$$$$W_{A\to0}=\int_A^0(-kx)\,dx=\tfrac12kA^2=\boxed E.$$$$P=\frac{dK}{dt}=-\frac{dU}{dt}.$$Power is positive while moving toward equilibrium and negative while moving away.

Physical check

Power vanishes at both equilibrium ($F=0$) and a turning point ($v=0$).