• 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