3.6.1.2
Maximum speed and maximum acceleration have simple expressions
Simple Harmonic Motion — AQA A-Level Physics
$$v_{\max} = \omega A$$
- $v_{\max}$: maximum 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)
$$a_{\max} = \omega^2 A$$
- $a_{\max}$: maximum accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². (m \(s^{-2}\))
- $\omega$: 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)
- $v_max occurs at \times = 0 (equilibriumAn object is in equilibrium when the resultant force on it is zero. The object is either stationary or moving at constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. position).$
- $a_max occurs at \times =\;\text{A} (maximum displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m).).$
- Both come directly from substituting extremes into the general equations.
Worked Example
An oscillator has time periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). 0.4 s and amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m). 2.8 m. Calculate the maximum accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻²..
Show Solution
1
Find omega from T
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.4} = 15.7 \text{ rad s}^{-1}$$
2
Apply a_max = omega^2 A
$$a_{\max} = (15.7)^2 \times 2.8 = 690 \text{ m s}^{-2} \text{ (2 s.f.)}$$
Answer
$a_{\max} = 690$ m s$^{-2}$
Common Mistake
MEDIUM
Students often: Confusing lowercase a (accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻².) with uppercase A (amplitude).
Instead: $a = acceleration (m s^{-2}). A = amplitude$, the maximum displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m). Be careful with handwriting in exams.
Instead: $a = acceleration (m s^{-2}). A = amplitude$, the maximum displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m). Be careful with handwriting in exams.