Simple harmonic motionEdexcel International A Level Physics: Revision notes

An object moves with simple harmonic motion (SHM) when the resultant force on it is proportional to its displacement from the equilibrium position and is always directed towards that position:

$F = -kx$

The minus sign shows that the force is opposite to the displacement. Since $F = ma$, the acceleration is also proportional to displacement and in the opposite direction.

SHM occurs for a mass on a spring that obeys Hooke's law, and for a simple pendulum swinging through small angles. It does not occur if the force is constant, depends on the square of the displacement, or is not directed towards the equilibrium position.

Combining $F = -kx$ with $F = ma$ gives:

$a = -\omega^2 x$

where $\omega$ is the angular frequency in rad s⁻¹, with $\omega = 2\pi f$ and $T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}$.

For an oscillation starting at maximum displacement $A$ (the amplitude) at $t = 0$:

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

The maximum speed is $v_{max} = A\omega$ (at equilibrium) and the maximum acceleration is $a_{max} = A\omega^2$ (at the ends). Keep the calculator in radians when using these equations.

A particle has $A = 4.0$ cm and $f = 2.5$ Hz. Then $\omega = 2\pi \times 2.5 = 15.7$ rad s⁻¹, so $v_{max} = 0.040 \times 15.7 = 0.63$ m s⁻¹ and $a_{max} = 0.040 \times 15.7^2 = 9.9$ m s⁻².

A buoy with $A = 0.60$ m and $T = 5.0$ s has $\omega = 2\pi/5.0 = 1.26$ rad s⁻¹. At $t = 1.0$ s: $x = 0.60\cos(1.26) = 0.19$ m, $v = -0.60 \times 1.26 \times \sin(1.26) = -0.72$ m s⁻¹ and $a = -\omega^2 x = -0.29$ m s⁻². The negative signs show that the velocity and acceleration are downwards.

For a mass on a spring, $F = -kx$ gives $a = -(k/m)x$, so $\omega^2 = k/m$ and

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

The period increases with mass and decreases with a stiffer spring. Quadrupling the mass doubles the period.

For a simple pendulum of length $l$, the restoring force along the arc is $mg\sin\theta$. For small angles, $\sin\theta \approx \theta = x/l$, so $F \approx -(mg/l)x$, and

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

The mass cancels, so the period does not depend on the mass of the bob. A mass-spring system of 0.250 kg and 40 N m⁻¹ has $T = 2\pi\sqrt{0.250/40} = 0.50$ s.