Calculating speed of an oscillator
- The speed of an object in simple harmonic motion varies as it oscillates back and forth
- Its speed is the magnitude of its velocity
SHM solution equation for speed
- The speed of an oscillator released from the equilibrium position in SHM is defined by:
v = v0 cos(⍵t)
- Where:
- v = speed (m s-1)
- v0 = maximum speed (m s-1)
- ⍵ = angular frequency (rad s-1)
- t = time (s)
Interpreting the graph v = v0 cos(⍵t)
The variation of the speed of a mass on a spring in SHM over one complete cycle
- This is a cosine function because the object starts oscillating from the equilibrium position (x = 0 when t = 0)
- This is shown on the SHM graphs revision note
- 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
- The maximum speed of an oscillator is the amplitude, v0 of the velocity-time graph
SHM equation for speed
- The speed v changes with the oscillator’s displacement x according to the equation:
v = ±ω
- Where:
- v = speed (m s-1)
- x0 = amplitude (m)
- ± = ‘plus or minus’, this value can be positive or negative
- ⍵ = angular frequency (rad s-1)
- x = displacement (m)
- This equation shows that when an oscillator has a greater amplitude x0, it has to travel a greater distance in the same amount of time and hence has greater speed v
- Both speed equations will be given on your data sheet in the exam
Maximum speed
- An oscillator reaches its greatest speed at the equilibrium position i.e. when its displacement is 0 (x = 0)
- At maximum speed the SHM speed equation becomes:
v0 = ⍵x0
Worked example
A simple pendulum oscillates with simple harmonic motion with an amplitude of 15 cm. The frequency of the oscillations is 6.7 Hz.
Calculate the speed of the pendulum at a position of 12 cm from the equilibrium position.
Answer:
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
v = ±ω
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
The 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)
Examiner Tip
You often have to convert between time period T, frequency f and angular frequency ⍵ for many exam questions – so make sure you revise the equations relating to these.