Simple harmonic motion

What you’ll learn

• Prove SHM by measuring displacement from equilibrium.
• Find period, amplitude, speed and phase.

Understand the idea

Motion is simple harmonic when acceleration is proportional to displacement from a fixed equilibrium point and directed towards that point. Derive the force equation using displacement from equilibrium so that any constant gravity and static-extension terms cancel. The resulting coefficient of $-x$ is $\omega^2$.

The amplitude is the largest distance from equilibrium, while the period is the time for a complete oscillation. Initial displacement and velocity determine the phase in a sine or cosine solution. Speed is greatest at equilibrium and zero at the extremes. For elastic-string oscillations, check that the string stays taut throughout the proposed motion.

Key formulae

$$\ddot x=-\omega^2x$$

$$x=A\cos\omega t+B\sin\omega t$$

$$T=\frac{2\pi}{\omega}$$

$$v^2=\omega^2(a^2-x^2),\qquad v_{\max}=a\omega$$

Worked example

A particle satisfies $\ddot x=-9x$ and has amplitude $0.4\,\mathrm m$. Find its period and maximum speed.

1. Compare with $\ddot x=-\omega^2x$ to get $\omega=3\,\mathrm{s^{-1}}$.
2. The period is $2\pi/3\,\mathrm s$.
3. Maximum speed is $a\omega=0.4(3)=1.2\,\mathrm{m\,s^{-1}}$.

Answer: $T=2\pi/3\,\mathrm s$ and $v_{\max}=1.2\,\mathrm{m\,s^{-1}}$.