C.1 Simple harmonic motion

Locate equilibrium

Measure $x$ from the equilibrium position, not from an arbitrary origin. At equilibrium $x=0$, so the restoring force and acceleration are zero; away from equilibrium, the acceleration points opposite to the displacement.

Check the model

A spring–mass system is an SHM model when the spring force is linear. A simple pendulum approximates SHM only for small angular displacements, where $\sin\theta\approx\theta$ in radians.

Interpret the minus sign

If the particle is displaced to positive $x$, acceleration is negative; if it is displaced to negative $x$, acceleration is positive. At equilibrium, $x=0$ and $a=0$, although the particle may have maximum speed there.

Connect frequency to acceleration

A larger $\omega$ gives a larger acceleration for the same displacement. The equation also shows why increasing amplitude does not change the period of ideal SHM: acceleration scales with the displacement.

Connect time measures

The period $T$ is the time for one complete cycle. Frequency $f$ is the number of cycles per second, so $T=1/f$. Angular frequency is $\omega=2\pi f=2\pi/T$, measured in radians per second.

Keep amplitude and displacement distinct

Amplitude is a non-negative fixed maximum for an ideal oscillation; displacement changes continuously between $-x_0$ and $+x_0$. The sign of displacement identifies the side of equilibrium.

Three equivalent measures

$$T=\frac1f=\frac{2\pi}{\omega},\qquad \omega=2\pi f$$

Use seconds for $T$, hertz for $f$, and radians per second for $\omega$.

Worked example from the mapped local textbook

A guitar-string point oscillates at $f=196\,\mathrm{Hz}$.

$$T=\frac1{196}=5.10\times10^{-3}\,\mathrm s$$

$$\omega=2\pi(196)=1.23\times10^3\,\mathrm{rad\,s^{-1}}$$

Mass–spring period

For an ideal mass $m$ attached to a linear spring of spring constant $k$,

$$T=2\pi\sqrt{\frac{m}{k}}$$

Use $m$ in kilograms and $k$ in $\mathrm{N\,m^{-1}}$ to obtain $T$ in seconds.

Read the dependence

$T\propto\sqrt m$: more mass increases the period. $T\propto1/\sqrt k$: a stiffer spring decreases the period. The ideal period is independent of amplitude while Hooke's law remains valid.

Worked example from local textbook question 6

For $T=1.00\,\mathrm s$ and $k=84\,\mathrm{N\,m^{-1}}$,

$$m=k\left(\frac{T}{2\pi}\right)^2=84\left(\frac{1.00}{2\pi}\right)^2=2.13\,\mathrm{kg}$$

Simple-pendulum period

For a pendulum of length $l$ undergoing small-angle oscillations,

$$T=2\pi\sqrt{\frac{l}{g}}$$

Measure $l$ from the pivot to the bob's centre of mass and use $g$ in $\mathrm{m\,s^{-2}}$.

Read the dependence

$T\propto\sqrt l$ and $T\propto1/\sqrt g$. Bob mass does not appear, so changing mass alone does not change the ideal period.

Worked example from local practice question 9

Changing $M$ to $4M$ has no effect. Changing $l$ to $0.25l$ gives

$$T'=2\pi\sqrt{\frac{0.25l}{g}}=0.5T$$

Use quarter-cycle checkpoints

At an extreme, speed and kinetic energy are zero while the relevant potential energy is maximum. At equilibrium, speed and kinetic energy are maximum while that potential energy is minimum. The energy pattern repeats every cycle.

Connect the model

A circular-motion picture can help visualize phase: the projected coordinate oscillates between two extremes and passes equilibrium twice per cycle. Use it as a representation of the oscillation, while the energy explanation remains based on the SHM position and speed.