Calculating Maximum Speed & Acceleration (AQA A Level Physics)

• 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

• Where:
  • v is maximum at the equilibrium position x = 0
  • So, 

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

• 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

Step 1: State the known values

• • Maximum speed, v_max = 12 m s^-1
  • Amplitude, A = 1.4 m

Step 2: Write down the equation 

v_max = ωA

Step 3: Rewrite angular velocity in terms of frequency f

ω = 2πf

v_max = 2πfA

Step 4: Rearrange for frequency, f 

Step 5: Substitute in the values

• 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

Step 1: State the known values

• • Time period, T = 0.4 s
  • Amplitude, A = 2.8 m

Step 2: Write down the equation

a_max = ω²A

Step 3: Rewrite maximum acceleration with time period T

Step 4: Substitute in the values