Velocity (OCR A Level Physics) : Revision Note

Katie M

Author

Katie M

Last updated

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 open parentheses x space equals space 0 close parentheses

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

v space equals space v subscript 0 space cos space omega t

  • Where:

    • v = velocity (m s-1)

    • v subscript 0 = maximum velocity (m s-1)

    • omega = angular frequency (rad s-1)

    • t = time (s)

  • This is a cosine function if the object starts oscillating from the equilibrium position (x space equals space 0 when t space equals space 0)

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

v space equals space plus-or-minus omega square root of x subscript 0 superscript 2 space minus space x squared end root

  • Where:

    • x = displacement (m)

    • x subscript 0 = amplitude (m)

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

  • This equation shows that

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

  • The maximum velocity v subscript 0 of an oscillator is therefore given by:

v subscript 0 space equals space omega x subscript 0

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, x subscript 0 = 15 cm = 0.15 m

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

  • Frequency,space f = 6.7 Hz

Step 2: Oscillator speed with displacement equation

v space equals space plus-or-minus omega square root of x subscript 0 superscript 2 space minus space x squared end root

  • 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:

omega space equals space 2 straight pi f space equals space 2 straight pi cross times 6.7

Step 4: Substitute in values and calculate

v space equals space open parentheses 2 straight pi cross times 6.7 close parentheses square root of 0.15 squared space minus space 0.12 squared end root

v space equals space 3.789 space equals space 3.8 space straight m space straight s to the power of negative 1 end exponent space open parentheses 2 space straight s. straight f. close parentheses

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

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Expertise: Physics Content Creator

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.