Simple harmonic motion and its equationsEdexcel A-Level Physics: Revision notes

Simple harmonic motion (SHM) is oscillation in which the resultant force on the object is proportional to its displacement from the equilibrium position and always directed towards that position:

$F = -kx$

The minus sign shows that force and displacement are in opposite directions. By Newton's second law, $F = ma$, so the acceleration obeys

$a = -\omega^2 x$

where $\omega$ is the angular frequency in rad s⁻¹ and $\omega^2 = k/m$. Either equation can be used as the definition of SHM: acceleration proportional to displacement and directed towards equilibrium.

The period $T$ is the time for one complete oscillation, and the frequency $f$ is the number of oscillations per second:

$T = \frac{1}{f} = \frac{2\pi}{\omega}$

so $\omega = 2\pi f$. The amplitude $A$ is the maximum displacement from equilibrium. In SHM the period is independent of the amplitude.

Put your calculator in radian mode for every SHM calculation with $\omega t$.

If the oscillator is released from maximum displacement at $t = 0$:

$x = A\cos\omega t$ $v = -A\omega\sin\omega t$ $a = -A\omega^2\cos\omega t$

The maximum values are:

• maximum speed $v_{max} = A\omega$, at the equilibrium position ($x = 0$)
• maximum acceleration $a_{max} = A\omega^2$, at the ends of the motion ($x = \pm A$)

Worked example: $A = 0.12$ m, $f = 2.5$ Hz. $\omega = 2\pi\times2.5 = 15.7$ rad s⁻¹, so $v_{max} = 0.12\times15.7 = 1.88$ m s⁻¹ and $a_{max} = 0.12\times15.7^2 = 29.6$ m s⁻².

For a mass $m$ on a spring of spring constant $k$ the restoring force is $F = -kx$, so $a = -(k/m)x$. Comparing with $a = -\omega^2x$ gives $\omega = \sqrt{k/m}$ and

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

A stiffer spring (larger $k$) gives a shorter period; a larger mass gives a longer period. For a vertical spring the equilibrium position is where the weight is balanced by the spring force, and the oscillation is again SHM about that point.

Worked example: $m = 0.40$ kg, $k = 25$ N m⁻¹: $T = 2\pi\sqrt{0.40/25} = 0.795$ s.

For a simple pendulum of length $l$ (pivot to centre of the bob) swinging through a small angle (less than about 10°), the restoring force is the component of the weight along the arc, which is approximately proportional to the displacement. The motion is SHM with

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

The period does not depend on the mass of the bob or the amplitude (for small angles). Quadrupling the length doubles the period. Measuring $T$ for a known $l$ gives $g = 4\pi^2 l/T^2$.

To reduce timing errors, time 20 oscillations rather than one, start and stop at the centre of the swing, and take repeat readings.