Velocity (OCR A Level Physics): Revision Note

Exam code: H556

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

Velocity of an Oscillator

  • The velocity of an object in simple harmonic motion varies as it oscillates back and forth

    • Since velocity is a vector, the velocity of the oscillator is its speed in a certain direction

  • The maximum velocity of an oscillator is at the equilibrium position i.e. when its displacement is zero (x = 0)

  • The velocity of an oscillator in SHM is defined by:

v = v0 cos ωt

  • Where:

    • v = velocity (m s-1)

    • v0 = maximum velocity (m s-1)

    • ω = 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)

  • The velocity v of an oscillator is related to its displacement x by:

v = ±ωx02  x2

  • Where:

    • x = displacement (m)

    • x0 = amplitude (m)

    • ± = ‘plus or minus’. The value can be negative or positive

  • This equation shows that

    • The greater the amplitude x0 of an oscillation, the greater its velocity v when passing through the equilibrium position (at x = 0)

  • The maximum velocity v0 of an oscillator is therefore given by:

v0 = ωx0

Speed SHM graph, downloadable AS & A Level Physics revision notes

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

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 = ±ωx02  x2

  • 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:

ω = 2πf = 2π×6.7

Step 4: Substitute in values and calculate

v = (2π×6.7)0.152  0.122

v = 3.789 = 3.8 m s1 (2 s.f.)

Examiner Tips and Tricks

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:

9-1-3-equations-for-shm-1-ib-hl

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.