Source: High school physics (Chinese)
Problem
A particle undergoes simple harmonic motion about $x = 0$. At $t = 0$ its displacement is $x = 0.37$ cm and its velocity is zero. The vibration frequency is $0.25$ Hz.
- Find the period, the angular frequency, and the amplitude.
- Write the displacement and the velocity as functions of time, and find the maximum speed and the maximum displacement.
- Find the displacement and the velocity at $t = 3.0$ s.
Period and angular frequency. $T = \dfrac{1}{f} = \dfrac{1}{0.25} = 4.0$ s, and $\omega = 2\pi f = 0.5\pi \approx 1.57$ rad/s.
Amplitude. The velocity vanishes only at an extreme position, so at $t = 0$ the particle is at maximum displacement. Hence $A = 0.37$ cm and the motion is $x = A\cos\omega t$.
Displacement and velocity. $x = 0.37\cos(0.5\pi t)$ cm and $v = -A\omega\sin\omega t = -0.185\pi\sin(0.5\pi t) \approx -0.58\sin(0.5\pi t)$ cm/s. The maximum speed is $v_{\max} = A\omega = 0.185\pi \approx 0.58$ cm/s, and the maximum displacement equals the amplitude, $0.37$ cm.
At $t = 3.0$ s. Here $\omega t = 1.5\pi$, so $x = 0.37\cos(1.5\pi) = 0$ and $v = -0.185\pi\sin(1.5\pi) = +0.185\pi \approx +0.58$ cm/s; the particle passes through equilibrium at maximum speed moving in the $+x$ direction.