Simple Harmonic Motion — Question 6

Question 6

An offset coordinate can hide the equilibrium

A laboratory coordinate is recorded as $y(t)=0.30+0.040\cos(6t)\,\mathrm{m}$, with $t$ in seconds. A student tests $a=-\omega^2y$ and concludes the motion cannot be SHM.

1. Identify the equilibrium coordinate, amplitude and correct acceleration law.

2. Find the maximum and minimum values of $y$, and explain why the particle never crossing $y=0$ is irrelevant to the classification.

Question 6 - Solution

An offset coordinate can hide the equilibrium

(a) Derivation and calculation

First remove the constant offset:$$y_e=0.30\,\mathrm m,\quad x=y-y_e=0.040\cos6t.$$$$\dot x=-0.240\sin6t,\qquad\ddot x=-1.440\cos6t=-36x.$$$$\boxed{A=0.040\,\mathrm m},\quad\boxed{\omega=6\,\mathrm{rad/s}},\quad\boxed{a=-36(y-0.30)}.$$

(b) Result and interpretation

$$y_{\max}=y_e+A=\boxed{0.340\,\mathrm m},\qquad y_{\min}=y_e-A=\boxed{0.260\,\mathrm m}.$$$$y=y_e\iff\cos6t=0\iff t=\frac{\pi/2+n\pi}{6}.$$There are two equilibrium crossings per period $T=2\pi/6$. The laboratory origin is not the equilibrium point.

Physical check

Changing coordinate origin adds a constant to position but leaves velocity, acceleration, amplitude and period unchanged.