Simple Harmonic Motion — Question 8

Question 8

A position alone does not specify the oscillator state

An oscillator has $A=0.050\,\mathrm{m}$ and period $T=\pi/2\,\mathrm{s}$. Consider its two passages through $x=0.030\,\mathrm{m}$ during a cycle.

1. Find both allowed velocities and the acceleration at that position.

2. For each passage, determine whether the speed is increasing or decreasing, and explain why identical acceleration does not imply identical subsequent motion.

Question 8 - Solution

A position alone does not specify the oscillator state

(a) Derivation and calculation

$$\omega=\frac{2\pi}{\pi/2}=4\,\mathrm{rad/s}.$$$$v=\pm\omega\sqrt{A^2-x^2}=\pm4\sqrt{0.050^2-0.030^2}=\boxed{\pm0.160\,\mathrm{m/s}}.$$$$a=-\omega^2x=-16(0.030)=\boxed{-0.480\,\mathrm{m/s^2}}.$$

(b) Result and interpretation

Speed changes according to$$\frac{d|v|}{dt}=\frac{va}{|v|}.$$$$v=+0.160:\quad d|v|/dt=-0.480\,\mathrm{m/s^2},$$$$v=-0.160:\quad d|v|/dt=+0.480\,\mathrm{m/s^2}.$$The two allowed states have equal position and acceleration but opposite velocities; the outward passage slows and the inward passage speeds up.

Physical check

A second-order evolution law needs the pair $(x,v)$, not position alone, to determine subsequent motion.