Syllabus Edition

First teaching 2023

First exams 2025

|

Calculating Speed in SHM (CIE A Level Physics)

Revision Note

Test yourself
Ann H

Author

Ann H

Last updated

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)

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

  • This is a cosine function because 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
  • The maximum speed of an oscillator is the amplitudevof the velocity-time graph

SHM equation for speed

  • The speed v changes with the oscillator’s displacement x according to the equation:

v = ±ωsquare root of stretchy left parenthesis x subscript 0 squared space minus space x squared stretchy right parenthesis end root

  • 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 = ±ωsquare root of stretchy left parenthesis x subscript 0 squared space minus space x squared stretchy right parenthesis 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

The equation relating angular frequency and normal frequency:

= 2πf = 2π× 6.7 = 42.097…

Step 4: Substitute in values and calculate

bold italic v bold space bold equals bold space stretchy left parenthesis 2 pi space cross times space 6.7 stretchy right parenthesis bold space bold cross times bold space square root of bold 0 bold. bold 15 to the power of bold 2 bold space bold minus bold space bold 0 bold. bold 12 to the power of bold 2 end root

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.

You've read 0 of your 10 free revision notes

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?

Ann H

Author: Ann H

Expertise: Physics

Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students no matter their schooling or background.