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 = ωA

• Where:

  • v_max  = maximum speed (m s^-1)

  • ω = 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

The maximum speed of a mass on a spring is at the equilibrium position. Its speed is 0 at its positive and negative amplitude

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 m s⁻¹ at x = A

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

• The maximum acceleration is given by the equation:

a_max = ω²A

• Where:

  • a_max = maximum acceleration (m s²)

  • ω = angular frequency (rad s^-1)

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

• This comes from the defining equation of SHM:

a = −ω²x

The maximum acceleration of a mass on a spring is at its positive and negative amplitude. Its acceleration is 0 at the equilibrium position