v = v0 cos(⍵t)

v0 = ⍵x0

The variation of the speed of a mass on a spring in SHM over one complete cycle
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.
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)
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.
转载自savemyexams
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