Simple Harmonic Motion — Question 10

Question 10

Average speed and root-mean-square speed answer different questions

An SHM oscillator has amplitude $A$ and period $T$. Define the rms speed as $v_{\mathrm{rms}}=\sqrt{T^{-1}\int_0^T v^2dt}$.

1. Find the mean velocity and mean speed over a complete cycle.

2. Find the rms speed and compare it with mean speed and maximum speed. Explain the strict ordering for nonzero amplitude.

Question 10 - Solution

Average speed and root-mean-square speed answer different questions

(a) Derivation and calculation

For one closed cycle, the signed displacement vanishes but the distance traveled is $4A$:$$\overline v=\frac{x(T)-x(0)}T=\boxed0,\qquad\overline{|v|}=\frac{4A}{T}=\boxed{\frac{2A\omega}{\pi}}.$$

(b) Result and interpretation

$$v_{\mathrm{rms}}^2=\frac{A^2\omega^2}{T}\int_0^T\sin^2(\omega t+\phi)\,dt=\frac{A^2\omega^2}{2}.$$$$\boxed{v_{\mathrm{rms}}=\frac{A\omega}{\sqrt2}},\qquad v_{\max}=A\omega.$$$$\frac{2}{\pi}=0.63662<\frac1{\sqrt2}=0.70711<1\quad\Rightarrow\quad\overline{|v|}<v_{\mathrm{rms}}<v_{\max}.$$RMS averaging weights larger speeds more strongly; equality would require constant speed.

Physical check

The inequalities are strict for $A>0$ and $\omega>0$. Mean velocity remains zero despite nonzero mean speed.