SHM Graphs (AQA A Level Physics): Revision Note

Exam code: 7408

Katie M

Written by: Katie M

Reviewed by: Timothy Sit

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SHM graphs

  • The displacement, velocity and acceleration of an object in simple harmonic motion can be represented by graphs against time

  • All undamped SHM graphs are represented by periodic functions

    • This means they can all be described by sine and cosine curves

  • Each graph needs to be recognised, along with how it relates to the other graphs

  • Remember that:

    • velocity is the rate of change of displacement v = st

    • acceleration is the rate of change of velocity a = ∆vt

Graphs that start at the equilibrium position

  • When oscillations start from the equilibrium position then:

    • the displacement-time graph is a sine curve

    • the velocity-time graph is the gradient of the displacement-time graph, so a cosine graph and 90° out of phase with the displacement-time graph

    • the acceleration-time graph is the gradient of the velocity-time graph, so a negative sine graph and 90° out of phase with the velocity-time graph

  • More information on this can be found in the A-level Maths revision notes on trigonometric differentiation

Graphs of displacement against time and velocity against time for SHM starting at equilibrium. x reaches ±0.02 metres every 0.4 seconds, while v reaches ±0.30 metres per second every 0.4 seconds and is zero at maximum displacement.
Graph of acceleration against time for SHM starting at equilibrium. Acceleration a reaches ±5 metres per second squared every 0.4 seconds, and is reflected from the displacement graph.
The displacement, velocity and acceleration graphs in SHM are all 90° out of phase with each other

Graphs that start at the amplitude position

  • When oscillations start from the amplitude position then:

    • the displacement-time graph is a cosine curve

    • the velocity-time graph is the gradient of the displacement-time graph, so a negative sine graph and 90° out of phase with the displacement-time graph

    • the acceleration-time graph is the gradient of the velocity-time graph, so a negative cosine graph and 90° out of phase with the velocity-time graph

Graphs of displacement against time, velocity against time and acceleration against time for SHM starting at maximum amplitude. Displacement starts at positive amplitude, velocity at zero and acceleration at negative amplitude. The curves are respectively cosine, negative sine and reflected cosine, with period T. Velocity-time graph is 90 degrees out of phase with the displacement-time graph.
When an oscillation starts at maximum displacement, the displacement-time graph is a cosine curve and the velocity and acceleration graphs follow from its gradient

Relationship between graphs

  • Key features of the displacement-time graphs:

    • The amplitude of oscillations A is the maximum value of x

    • The time period of oscillations T is the time taken for one full wavelength cycle

  • Key features of the velocity-time graphs:

    • The velocity of an oscillator at any time can be determined from the gradient of the displacement-time graph:

      v = ∆x∆t

    • An oscillator moves the fastest at its equilibrium position

    • Therefore, the velocity is at its maximum when the displacement is zero (the amplitude of the graph is vmax)

  • Key features of the acceleration-time graph:

    • The acceleration graph is a reflection of the displacement graph in the x-axis

    • This means when a mass has positive displacement (to the right) the acceleration is in the opposite direction (to the left) and vice versa (from a = −ω2x)

    • The acceleration of an oscillator at any time can be determined from the gradient of the velocity-time graph:

      a = ∆v∆t

    • The maximum value of the acceleration is when the oscillator is at its maximum displacement

Worked Example

A swing is pulled 5 cm and then released. The variation of the horizontal displacement x of the swing with time t is shown on the graph below.

A displacement-time graph for a swing, starting at 5 cm when t = 0 and falling through zero, with time marked in intervals along the horizontal axis. The curve starts at +5 cm, crosses x = 0 at 0.2 s, reaches −5 cm at 0.4 s, then crosses zero again at 0.6 s.

The swing exhibits simple harmonic motion. Use data from the graph to determine at what time the velocity of the swing is first at its maximum.

[1]

Answer:

Step 1: The velocity is at its maximum when the displacement x = 0

Step 2: Reading value of time when x = 0

A displacement-time graph for a swing, starting at 5 cm when t = 0 and falling through zero, with time marked in intervals along the horizontal axis. The curve starts at +5 cm, crosses x = 0 at 0.2 s, reaches −5 cm at 0.4 s, then crosses zero again at 0.6 s. A green circle and arrow highlight where the curve crosses x = 0 at t = 0.2 seconds.
  • From the graph, this is equal to 0.2 s [1 mark]

Examiner Tips and Tricks

These graphs might not look identical to what is in your textbook, because they depend on the starting position of the oscillation of the object when t = 0. If there is no damping, they will be a sine or cosine curve.

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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.

Timothy Sit

Reviewer: Timothy Sit

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.