C.1 Simple harmonic motion

Start with the restoring force

Simple harmonic motion occurs when the restoring force is proportional to displacement from equilibrium and always points back toward equilibrium: $F\propto -x$. For constant mass this gives $a\propto -x$.

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.

Common trap

“Periodic” motion alone is not enough. The defining condition is the restoring acceleration $a=-\omega^2x$, including the opposite direction and proportional dependence.

Write the definition

Simple harmonic motion is defined by $a=-\omega^2x$, where $x$ is displacement from equilibrium and $\omega$ is angular frequency. The acceleration is proportional to displacement and points in the opposite direction.

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.

Common trap

Do not write $a=+\omega^2x$ or measure $x$ from an arbitrary origin. The displacement must be relative to equilibrium, and the sign must restore the particle toward it.

Name each quantity

The equilibrium position is the central position where the resultant restoring force is zero. Displacement $x$ is the signed distance from equilibrium. Amplitude $x_0$ is the maximum magnitude of 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.

Common trap

Do not call the distance travelled in one cycle the amplitude. Amplitude is measured from equilibrium to an extreme position, not from one extreme to the other.

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 $f=196\,\mathrm{Hz}$.

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

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

Common trap

Do not mix this general conversion objective with the separate spring and pendulum period models. Also, $\omega$ is $2\pi$ times $f$, not $f/(2\pi)$.

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 $k=84\,\mathrm{N\,m^{-1}}$,

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

Boundary

This model assumes a linear spring and that the stated oscillating mass includes any effective mass the question requires. Do not substitute amplitude for $m$.

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

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

Boundary

The equation is the small-angle approximation, where $\sin\theta\approx\theta$ with $\theta$ in radians. Large amplitudes do not follow this period exactly.

Describe one complete cycle

In ideal SHM, total mechanical energy is constant. As the particle moves from an extreme position to equilibrium, stored potential energy changes into kinetic energy; from equilibrium to the opposite extreme, kinetic energy changes back into potential energy.

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.

Common trap

Do not claim that the energy transfer stops at equilibrium. The particle has maximum speed there, so the transfer reverses direction as it continues toward the next extreme.