AQA A Level Physics复习笔记6.2.3 Calculating Maximum Speed & Acceleration

• The speed v of an oscillator will vary in SHM. It is:
  • Maximum at the equilibrium position (x = 0)
  • Zero at the amplitude (x = A)

   

• The maximum speed, 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)

   

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 acceleration a of an oscillator will also vary in SHM. It is:
  • Maximum at the amplitude (x = A)
  • Zero at the equilibrium position (x = 0)

   

• This is because the accelerationis directly proportional to the displacement of an oscillator
• 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 (m)

   

• Although at the amplitude, the speed is zero, the oscillator has changed direction
• This means that it has a non–zero velocity, and since acceleration is the rate of change of velocity, the oscillator has an acceleration at the amplitude too

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