Simple Harmonic Motion — Question 1

Question 1

Position and velocity fix the phase quadrant

An oscillator obeys $x=A\cos(\omega t+\phi)$ with $A>0$, $\omega=5.0\,\mathrm{rad/s}$ and $-\pi<\phi\le\pi$. At $t=0$, $x_0=0.060\,\mathrm{m}$ and $v_0=-0.40\,\mathrm{m/s}$.

1. Find amplitude and phase, using the velocity sign to choose the quadrant.

2. Find initial acceleration and the first positive time at which the oscillator reaches $x=0$.

Question 1 - Solution

Position and velocity fix the phase quadrant

(a) Derivation and calculation

Use both initial-state equations to fix the quadrant:$$x_0=A\cos\phi,\qquad v_0=-A\omega\sin\phi.$$$$A^2=x_0^2+\left(\frac{v_0}{\omega}\right)^2=(0.060)^2+(-0.080)^2=0.0100\,\mathrm{m^2}.$$$$\boxed{A=0.100\,\mathrm m},\quad (\cos\phi,\sin\phi)=(0.6,0.8),\quad\boxed{\phi=\operatorname{atan2}(0.8,0.6)=0.92730\,\mathrm{rad}}.$$

(b) Result and interpretation

$$a_0=-\omega^2x_0=-25(0.060)=\boxed{-1.50\,\mathrm{m/s^2}}.$$The next zero occurs at phase $\pi/2$, since the initial phase is in quadrant I:$$\omega t_1+\phi=\pi/2,\qquad t_1=\frac{\pi/2-0.92730}{5}=\boxed{0.12870\,\mathrm s}.$$

Physical check

$v_0=-A\omega\sin\phi=-0.40\,\mathrm{m/s}$ verifies motion toward equilibrium; an inverse cosine alone would leave the sign ambiguous.