Calculating Maximum Speed & Acceleration (AQA A Level Physics): Revision Note

Maximum speed

• The maximum speed of an oscillator, $v_{max}$, is given by the equation:

$v_{max}=\omega A$

• Where:

  • $v_{max}$  = maximum speed (m s^-1)

  • $\omega$ = angular frequency (rad s^-1)

  • $A$ = amplitude (m)

• This comes from the SHM speed equation

$v=\pm \omega \sqrt{\left(A^2-x^2\right)}$

• Where:

  • $v$ is maximum at the equilibrium position $x=0$

  • So, 

$v_{max}=\pm \omega \sqrt{A^2}=\omega A$

• When an oscillator begins its motion at the equilibrium position then the velocity-time graph is a cosine graph

$v=v_0\cos \left(\omega t\right)$

• The maximum speed of an oscillator is the amplitude, v₀ of the velocity-time graph

• For a mass oscillating on a vertical spring:

  • $v_{max}$ occurs when the spring is in its equilibrium position

  • $v=0$ at the amplitude position

Maximum acceleration

• The maximum acceleration, $a_{max}$, of an oscillator will occur when the gradient of the velocity-time graph is steepest

  • When $v=0\mathrm{m}{\mathrm{s}}^{-1}$ at $x=A$

• Acceleration is zero at the equilibrium position ($x=0$)

• The maximum acceleration is given by the equation:

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

• Where:

  • $a_{max}$ = maximum acceleration (m s^-2)

  • $\omega$ = angular frequency (rad s^-1)

  • $A$ = amplitude (maximum displacement, x) (m)

• This comes from the defining equation of SHM:

$a=-{\omega }^{2}x$