Simple Harmonic Motion

An object undergoes simple harmonic motion (SHM) when its acceleration is directly proportional to its displacement from a fixed equilibrium position and is always directed back towards that position. Both parts of the condition matter: proportionality alone is not enough, and the direction requirement is what makes the motion repeat.

The defining equation is:

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

Here the negative sign encodes the restoring direction and $\omega$ is the angular frequency.

Physically, SHM requires a restoring force that grows linearly with displacement. A trolley tethered between two stretched springs is a good example: displace it $2\ \text{cm}$ to the right and the net spring force pulls it back to the left; displace it $4\ \text{cm}$ and that force doubles. A bouncing ball is periodic but not SHM, because the forces acting on it are not proportional to its displacement.

You must be able to:

• state both conditions precisely
• recognise SHM from descriptions or graphs (an $a$–$x$ graph for SHM is a straight line through the origin with negative gradient)
• use $a = -\omega^2 x$ to show that the maximum acceleration $a_{\text{max}} = \omega^2 x_0$ occurs at the extremes of the motion

Three linked quantities measure how rapidly an oscillation repeats:

• The period $T$ is the time for one complete cycle, in seconds.
• The frequency $f = \frac{1}{T}$ is the number of cycles completed each second, in hertz ($\text{Hz}$).
• The angular frequency $\omega$, in $\text{rad s}^{-1}$, expresses the same rate in radians, given by:

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

One full cycle corresponds to a phase change of $2\pi\ \text{rad}$, so multiplying the frequency by $2\pi$ converts cycles per second into radians per second.

For example, a loudspeaker cone vibrating at $f = 50\ \text{Hz}$ has $T = \frac{1}{50} = 0.020\ \text{s}$ and $\omega = 2\pi \times 50 \approx 314\ \text{rad s}^{-1}$. Ideal SHM is isochronous: the period stays the same whatever the amplitude, which is why pendulums could regulate early clocks even as their swings died away.

You should be able to:

• convert fluently between $T$, $f$ and $\omega$
• read the period directly from a displacement–time graph (peak to peak along the time axis)
• substitute $\omega$, never $f$, into SHM equations such as $a = -\omega^2 x$

Mixing up $f$ and $\omega$ is one of the quickest ways to lose marks in this topic.

Two standard oscillators appear throughout this topic. A mass–spring system has period given by:

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

A larger mass oscillates more slowly, while a stiffer spring (larger spring constant $k$) shortens the period. Because $g$ does not appear, a horizontal mass–spring oscillator would keep the same period on the Moon.

A simple pendulum has period:

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

This is set only by its length and the local gravitational field strength; the mass of the bob and the amplitude of the swing do not appear at all.

The pendulum result needs care, because a pendulum does not strictly execute SHM. The restoring force is the tangential component of gravity, $F = -mg \sin\theta$, and since displacement along the arc is $s = l\theta$, this yields $F = -mg \sin(s/l)$ — proportional to $\sin(s/l)$, not to $s$.

However, for small angles (where $\theta \le 10^\circ$, or $\approx 0.17\ \text{rad}$), $\sin\theta \approx \theta$, and substituting this approximation gives $F \approx -\frac{mg}{l} s$, restoring the linear relationship required for SHM with angular frequency $\omega = \sqrt{\frac{g}{l}}$. The period is therefore independent of mass and amplitude only under this small-angle constraint.

You must be able to:

• calculate periods
• rearrange both formulae for $m$, $k$, $l$ or $g$
• describe the classic experiment that determines $g$ by plotting $T^2$ against $l$ (the gradient is $\frac{4\pi^2}{g}$)

As an undamped oscillator moves through one cycle, its total mechanical energy stays constant while continuously converting between kinetic and potential forms. Kinetic energy is maximum at the equilibrium position, where the speed is greatest, and zero at the extremes $x = \pm x_0$, where the object is momentarily at rest; potential energy does the opposite, peaking at maximum displacement.

A complete exchange between the two forms happens every quarter of a cycle.

Graphs make this concrete. Plotted against displacement:

• potential energy is an upward-opening parabola
• kinetic energy is a downward-opening one
• their sum is a horizontal line at $E_T$

Plotted against time, each energy varies as $\sin^2$ or $\cos^2$, always positive and repeating twice per oscillation.

The same physics can be drawn in a velocity–displacement plot. For a mass–spring oscillator, conservation of energy gives:

$$\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E_T$$

Dividing both sides by $E_T$ yields $\frac{v^2}{2E_T/m} + \frac{x^2}{2E_T/k} = 1$, the equation of an ellipse. As time progresses, the system's state $(x, v)$ traces out this closed curve once per cycle, and any damping force removes energy and makes the trajectory spiral inwards towards the origin (rest at equilibrium).

At SL the energy description is qualitative; calculations with $E_T = \frac{1}{2}m\omega^2 x_0^2$ belong to the HL extension. You should be able to:

• describe the interchange
• sketch both families of graphs
• state where each form of energy is zero or maximum

At HL the motion is written as explicit functions of time. Displacement is given by:

$$x = x_0 \sin(\omega t + \phi)$$

Here the phase angle $\phi$ fixes where in its cycle the oscillator is at $t = 0$: with $\phi = 0$ the object starts at equilibrium moving in the positive direction, while $\phi = \frac{\pi}{2}$ means it starts at $x = +x_0$, since $\sin(\omega t + \frac{\pi}{2}) = \cos(\omega t)$.

The velocity at any time is given directly by $v = \omega x_0 \cos(\omega t + \phi)$, and substituting $x = x_0 \sin(\omega t + \phi)$ into the defining equation $a = -\omega^2 x$ gives the acceleration $a = -\omega^2 x_0 \sin(\omega t + \phi)$. These forms show velocity leading displacement by $\frac{\pi}{2}\ \text{rad}$ and acceleration in antiphase with displacement.

When the time is not known, find the speed at any displacement using:

$$v = \pm\omega\sqrt{x_0^2 - x^2}$$

This gives the maximum speed $v_{\text{max}} = \omega x_0$ at $x = 0$ and zero at the extremes. The energy relations follow directly:

• $E_T = \frac{1}{2}m\omega^2 x_0^2$
• $E_p = \frac{1}{2}m\omega^2 x^2$
• $E_k = \frac{1}{2}m\omega^2(x_0^2 - x^2)$

For instance, a $0.30\ \text{kg}$ mass with $x_0 = 0.10\ \text{m}$ and $\omega = 8.0\ \text{rad s}^{-1}$ has $v_{\text{max}} = 0.80\ \text{m s}^{-1}$ and $E_T = 0.096\ \text{J}$.

You must be able to:

• choose between sine and cosine forms
• extract $\phi$ from a graph or from initial conditions
• keep your calculator in radian mode
• combine these equations to find displacement, velocity, acceleration or energy at any instant