Simple Harmonic Motion — Question 4

Question 4

An oscillator spends more time near its turning points

A particle executes SHM with amplitude $A$. Over one complete cycle, compare time spent in the central region $|x|<A/2$ with time spent outside it. Endpoint instants do not affect the fractions.

1. Find the fraction of a period spent in the central region.

2. Find the fraction spent in the final ten percent of the amplitude near either end, $|x|>0.9A$. Why is it not $10\%$?

Question 4 - Solution

An oscillator spends more time near its turning points

(a) Derivation and calculation

Uniform phase advance gives a time fraction equal to phase length divided by $2\pi$:$$|x|<A/2\iff|\cos\psi|<\tfrac12.$$$$\psi\in(\pi/3,2\pi/3)\cup(4\pi/3,5\pi/3),\quad\Delta\psi=2\pi/3.$$$$\boxed{\Delta t_{\mathrm{central}}/T=1/3},\qquad\Delta t_{\mathrm{outer}}/T=2/3.$$

(b) Result and interpretation

For $|x|>qA$, the four end arcs have total length $4\arccos q$:$$\frac{\Delta t_{\mathrm{ends}}}T=\frac{4\arccos q}{2\pi}.$$$$q=0.9:\quad\boxed{\Delta t_{\mathrm{ends}}/T=\frac{2\arccos0.9}{\pi}=0.28713}.$$Equivalently, $dt=|dx|/[\omega\sqrt{A^2-x^2}]$, which weights positions near $|x|=A$ more strongly.

Physical check

The integrable divergence of $dt/dx$ at a turning point reflects zero speed, not an infinite period.