Simple harmonic motionAQA A-Level Physics: Revision notes

An object oscillating about an equilibrium position performs simple harmonic motion (SHM) if its acceleration is directly proportional to its displacement from equilibrium and always directed towards equilibrium. In symbols, $a \propto -x$: the minus sign shows that $a$ and $x$ have opposite directions.

The amplitude $A$ is the maximum displacement from equilibrium. The time period $T$ is the time for one complete oscillation, and the frequency is $f = 1/T$.

The condition for SHM is written as the defining equation

$a = -\omega^2 x$

where $\omega$ is the angular frequency in rad s⁻¹, with $\omega = 2\pi f = 2\pi/T$. A graph of $a$ against $x$ is a straight line through the origin with negative gradient $-\omega^2$, so this is the test for SHM from data.

The period depends on $\omega$ only, so it does not depend on the amplitude.

If the oscillation starts at maximum displacement ($x = A$ at $t = 0$) the displacement is

$x = A\cos(\omega t)$

The velocity at displacement $x$ has magnitude

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

The $\pm$ shows the body can pass the same position in either direction. The speed is zero at $x = \pm A$ and a maximum at $x = 0$.

Use radians on your calculator when evaluating $\cos(\omega t)$.

Starting from maximum displacement, the displacement–time graph is a cosine curve. The velocity is the gradient of the x–t graph and the acceleration is the gradient of the v–t graph, so:

• $v$–$t$ is a negative sine curve: zero at the extremes, $-\omega A$ at $t = T/4$ and $+\omega A$ at $3T/4$
• $a$–$t$ is a negative cosine curve: $-\omega^2 A$ at $t = 0$, zero at $T/4$ and $+\omega^2 A$ at $T/2$

$v$ leads $x$ by a quarter of a cycle, and $a$ is in antiphase with $x$ (half a cycle out of phase), as $a = -\omega^2 x$.

The speed is greatest at equilibrium and the acceleration is greatest at the extremes:

$v_{max} = \omega A \qquad a_{max} = \omega^2 A$

The time period is $T = 1/f$.

Worked example: a body oscillates with $A = 0.10$ m at $f = 2.0$ Hz. Then $\omega = 2\pi \times 2.0 = 12.6$ rad s⁻¹, $v_{max} = 12.6 \times 0.10 = 1.3$ m s⁻¹ and $a_{max} = 12.6^2 \times 0.10 = 16$ m s⁻². At $x = 0.060$ m, $v = 12.6\sqrt{0.10^2 - 0.060^2} = 1.0$ m s⁻¹.