Calculating Maximum Speed & Acceleration (AQA A Level Physics): Revision Note

Exam code: 7408

Ashika

Written by: Ashika

Reviewed by: Caroline Carroll

Updated on

Maximum Speed

 

  • The maximum speed of an oscillator, vmax, is given by the equation:

vmax = ωA

  • Where:

    • vmax  = maximum speed (m s-1)

    • ω = angular frequency (rad s-1)

    • A = amplitude (m)

  • This comes from the SHM speed-equation

v space equals space plus-or-minus omega square root of open parentheses A squared space minus space x squared close parentheses end root

  • Where:

    • v is maximum at the equilibrium position x = 0

    • So, 

v subscript m a x end subscript space equals space plus-or-minus omega square root of A squared space end root space equals space omega A

  • When an oscillator begins its motion at the equilibrium position then the velocity-time graph is a cosine graph

v space equals space v subscript 0 cos open parentheses omega t close parentheses

  • The maximum speed of an oscillator is the amplitudevof the velocity-time graph

  • For a mass oscillating on a vertical spring:

    • vmax occurs when the spring is in its equilibrium position

    • v = 0 at the amplitude position

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

The maximum speed of a mass on a spring is at the equilibrium position. Its speed is 0 at its positive and negative amplitude

Worked Example

Calculate the frequency of an oscillator with a maximum speed of 12 m s-1 and amplitude of 1.4 m.

Answer:

Step 1: State the known values

  • Maximum speed, vmax = 12 m s-1

  • Amplitude, A = 1.4 m

Step 2: Write down the equation 

vmax = ωA

Step 3: Rewrite angular velocity in terms of frequency f

ω = 2πf

vmax = 2πfA

Step 4: Rearrange for frequency, f 

straight f equals fraction numerator straight v subscript max over denominator 2 straight pi straight A end fraction

Step 5: Substitute in the values

straight f equals fraction numerator 12 over denominator 2 straight pi cross times 1.4 end fraction equals 1.364 equals 1.4 space Hz

Maximum Acceleration

  • The maximum accelerationamax of an oscillator will occur when the gradient of the velocity-time graph is steepest

    • When = 0 m s−1 at = A

  • Acceleration is zero at the equilibrium position (x = 0)

  • The maximum acceleration is given by the equation:

amax = ω2A

  • Where:

    • amax = maximum acceleration (m s2)

    • ω = angular frequency (rad s-1)

    • A = amplitude (maximum displacement, x) (m)

  • This comes from the defining equation of SHM:

a = −ω2x

Max & Min Acceleration, downloadable AS & A Level Physics revision notes

The maximum acceleration of a mass on a spring is at its positive and negative amplitude. Its acceleration is 0 at the equilibrium position

Worked Example

Calculate the maximum acceleration of an oscillator with a time period of 0.4 s and amplitude of 2.8 m.

Answer:

Step 1: State the known values

  • Time period, T = 0.4 s

  • Amplitude, A = 2.8 m

Step 2: Write down the equation

amax = ω2A

Step 3: Rewrite maximum acceleration with time period T

straight a subscript max space equals open parentheses fraction numerator 2 straight pi over denominator straight T end fraction close parentheses squared straight A

Step 4: Substitute in the values

straight a subscript max equals open parentheses fraction numerator 2 straight pi over denominator 0.4 end fraction close parentheses squared cross times 2.8 equals 690.97 equals 690 space straight m space straight s to the power of negative 2 end exponent open parentheses 2 space straight s. straight f close parentheses

Examiner Tips and Tricks

Make sure not to get mixed up with lowercase a (acceleration) and uppercase A (amplitude). Make sure you feel confident moving between the equations in their various forms. 

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Ashika

Author: Ashika

Expertise: Physics Content Creator

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

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.