Oscillations and Waves
Beginner
Simple harmonic oscillation
Source: High school physics (Chinese)
Problem
A 3.0 kg particle has a displacement given by
$$x = 5.0\cos\left(\frac{\pi}{3}t - \frac{\pi}{4}\right) \text{ m}.$$- For what value of $x$ is the potential energy equal to half the total energy?
- How long does it take the particle to travel from the equilibrium position to that position?
(1) $x = \pm A/\sqrt{2} \approx \pm 3.5$ m. (2) $t = 0.75$ s.
The motion has amplitude $A = 5.0$ m and angular frequency $\omega = \frac{\pi}{3}$ rad/s, so the period is $T = \frac{2\pi}{\omega} = 6.0$ s.
(1) The total energy is $E = \frac{1}{2}kA^2$ and the potential energy is $U = \frac{1}{2}kx^2$. Setting $U = \frac{1}{2}E$:
$$\frac{1}{2}kx^2 = \frac{1}{2}\cdot\frac{1}{2}kA^2 \implies x = \pm\frac{A}{\sqrt{2}} = \pm\frac{5.0}{\sqrt{2}} \approx \pm 3.5 \text{ m}.$$(2) Starting from equilibrium ($x=0$) the phase is $\frac{\pi}{2}$ from maximum displacement; at $x = A/\sqrt{2}$ it is $\frac{\pi}{4}$ from maximum. The phase swept is $\Delta\phi = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}$, so
$$t = \frac{\Delta\phi}{\omega} = \frac{\pi/4}{\pi/3} = 0.75 \text{ s} = \frac{T}{8}.$$