• The speed of an object in?simple harmonic motion?varies as it oscillates back and forth
  • Its speed is the magnitude of its velocity
• The greatest speed of an oscillator is at the equilibrium position ie. when its displacement is 0 (x = 0)
• The speed of an oscillator in SHM is defined by:

v = v_0?cos(?t)

• Where:
  • v = speed (m s^-1)
  • v₀?= maximum speed (m s^-1)
  • ? = angular frequency (rad s^-1)
  • t = time (s)
• This is a cosine function if the object starts oscillating from the equilibrium position (x = 0 when t = 0)
• Although the symbol?v?is commonly used to represent velocity, not speed, exam questions focus more on the magnitude of the velocity than its direction in SHM
• How the speed?v?changes with the oscillator’s displacement?x?is defined by:

• Where:
  • x = displacement (m)
  • x₀?=?amplitude?(m)
  • ± = ‘plus or minus’. The value can be negative or positive
• This equation shows that when an oscillator has a greater amplitude?x₀, it has to travel a greater distance in the same time and hence has greater speed?v
• Both equations for speed will be given on your formulae sheet in the exam
• When the speed is at its maximums (at x = 0), the equation becomes:

v₀?= ?x₀

The variation of the speed of a mass on a spring in SHM over one complete cycle

Step 1: Write out the known quantities

Amplitude of oscillations,?x₀?= 15 cm = 0.15 m

Displacement at which the speed is to be found,?x = 12 cm = 0.12 m

Frequency,?f = 6.7 Hz

Step 2: Oscillator speed with displacement equation

Since the speed is being calculated, the ± sign can be removed as direction does not matter in this case

Step 3: Write an expression for the angular frequency

Equation relating angular frequency and normal frequency:

? = 2πf = 2π× 6.7 = 42.097…

Step 4: Substitute in values and calculate

v = 3.789 = 3.8 m s^-1?(2 s.f)