3.6.1.2
Speed depends on position: fastest at equilibrium, zero at amplitude
Simple Harmonic Motion — AQA A-Level Physics
$$v = \pm\omega\sqrt{A^2 - x^2}$$
- $v$: speed (m \(s^{-1}\))
- $\omega$: angular frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). (rad \(s^{-1}\))
- $A$: amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m). (m)
- $x$: displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m)
- At x = 0 (equilibriumAn object is in equilibrium when the resultant force on it is zero. The object is either stationary or moving at constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹..): $v = \omega A (maximum speed)$.
- At x = A (amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m).): $v = 0 (momentarily stationary)$.
- The plus/minus indicates the object can move in either direction through any given position.
Worked Example
A pendulum oscillates with amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m). 15 cm and frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). 6.7 Hz. Calculate the speed at 12 cm from equilibriumAn object is in equilibrium when the resultant force on it is zero. The object is either stationary or moving at constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹...
Show Solution
1
Convert units and find omega
- $A = 0.15$ m, $x = 0.12$ m
- $\omega = 2\pi f = 2\pi \times 6.7 = 42.1$ rad s$^{-1}$
2
Substitute into speed equation
$$v = \omega\sqrt{A^2 - x^2} = 42.1 \times \sqrt{0.15^2 - 0.12^2}$$
$$= 42.1 \times \sqrt{0.0225 - 0.0144} = 42.1 \times 0.09 = 3.8 \text{ m s}^{-1}$$Answer
$v = 3.8$ m s$^{-1}$ (2 s.f.)