Conditions for Simple Harmonic Motion (AQA A Level Physics): Revision Note

Exam code: 7408

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Conditions for simple harmonic motion

  • Simple harmonic motion (SHM) is a specific type of oscillation where:

    • there is repetitive movement back and forth through an equilibrium, or central, position

    • the maximum horizontal or vertical displacement on one side the equilibrium position is equal to the maximum horizontal displacement on the other

    • the time interval of each complete vibration is the same (periodic)

    • the force responsible for the motion (restoring force) is always directed horizontally or vertically towards the equilibrium position and is directly proportional to the distance from it

Examples of SHM

  • Examples of oscillators that undergo SHM are:

    • the pendulum of a clock

    • a child on a swing

    • the vibrations of a bowl

    • a bungee jumper reaching the bottom of his fall

    • a mass on a spring

    • guitar strings vibrating

    • a ruler vibrating off the end of a table

    • the electrons in alternating current flowing through a wire

    • the movement of a swing bridge when someone crosses

    • a marble dropped into a bowl

Illustration titled “Examples of SHM” with the following examples: a clock pendulum, child on a swing, vibrating bowls, vertical mass on a spring, marble in a bowl, bungee jumper and a suspension bridge.
Everyday oscillations, from clocks to guitar strings, can all be modelled as simple harmonic motion

Modelling SHM

  • Not all oscillations are as simple as SHM

    • This is a particularly simple kind

    • It is relatively easy to analyse mathematically

    • Many other types of oscillatory motion can be broken down into a combination of SHMs

  • An oscillation is defined to be SHM when:

    • the acceleration is proportional to the horizontal or vertical displacement

    • the acceleration is in the opposite direction to the displacement (directed towards the equilibrium position)

  • The time period of oscillation is independent of the amplitude of the oscillation, for small angles of oscillation

  • So, for acceleration a and horizontal displacement x

a ∝ −x

  • Calculations on and explanations of two models of simple harmonic motion are required:

    • A simple pendulum oscillating from side to side attached to a fixed point above

    • Small angle approximations are applied to the simple pendulum swing, meaning that the conditions for SHM described above are valid for horizontal displacement, acceleration, and restoring force

      A pendulum mass displaced to the right of its equilibrium position. The displacement x is rightward, while the restoring force F and acceleration a act leftwards towards equilibrium and in the same direction.
      Force, acceleration and displacement of a simple pendulum in SHM
    • A mass-spring system oscillating vertically up and down or horizontally back and forth

A vertically oscillating mass–spring system where the mass is displaced downward by x from its equilibrium position, with an applied force downward and a restoring force F and acceleration a both acting upward in the same direction.
Force, acceleration and displacement of a mass-spring system

An example of not SHM

  • A person jumping on a trampoline is not an example of simple harmonic motion because:

    • the restoring force on the person is not proportional to their displacement from the equilibrium position and always acts down

    • when the person is not in contact with the trampoline, the restoring force is equal to their weight, which is constant

    • this does not change, even if they jump higher

Two views of a person bouncing on a trampoline, standing on the trampoline and airborne above it. Red upward arrows indicate the person’s upward motion in both positions.
The restoring force of the person bouncing is equal to their weight and always acts downwards

The defining equation of SHM

  • The acceleration of an object oscillating in simple harmonic motion is given by the equation:

a = −ω2x

  • Where:

    • a = acceleration (m s-2)

    • ω = angular frequency (rad s-1)

    • x = displacement (m)

  • The equation demonstrates:

    • acceleration reaches its maximum value when the displacement is at a maximum ie. x = x0 at its amplitude

    • the minus sign shows that when the object is displaced to the right, the direction of the acceleration is to the left and vice versa (a and x are always in opposite directions to each other)

  • Consider the oscillation of the bob pendulum again:

    • The bob speeds up as it heads towards the midpoint

    • Velocity is at a maximum when it passes through the equilibrium position 

    • The pendulum slows down as it continues towards the other extreme of oscillation

    • v = 0 at A as it changes direction

    • The pendulum then reverses and starts to accelerate again towards the midpoint

The graph representing a = −⍵2x

  • The graph of acceleration against displacement is a straight line through the origin sloping downwards (similar to y = −x)

Graph of acceleration against displacement for a = −ω²x. A straight line through the origin is drawn, with x from −A to +A and gradient a/x = −ω². ±A mark the oscillation amplitudes.
The acceleration of an object in SHM is directly proportional to the negative displacement
  • The key features of the graph are:

    • the gradient is equal to −⍵2

    • the maximum and minimum displacement x values are the amplitudes −A and +A

Displacement equation

  • The SHM displacement equation is:

x = Acos(ωt)

  • Where:

    • A = amplitude (m)

    • ω = angular frequency (rad s-1)

    • t = time (s)

  • Because:

    • the graph of x = Acos(ωt) starts from amplitude A when t = 0

    • the displacement is at its maximum when cos(⍵t) equals 1 or −1, when x = A

  • The A Level revision notes on the graphs of trigonometric functions may help with understanding trigonometric graphs

Graph of y equals sin x and y equals cos x, with x from −180 to 360 degrees and y from −1 to 1. The sine curve passes through the origin, while the cosine curve passes through 1 at zero degrees.
The graph of y = cos (x) has maximum displacement when x = 0
  • If an object is oscillating from its equilibrium position (x = 0 at t = 0) then the displacement equation will be:

x = Asin(ωt)

  • The displacement will be at its maximum when sin(⍵t) equals 1 or −1, when x = A

  • This is because the sine graph starts at 0, whereas the cosine graph starts at a maximum

Two displacement–time graphs for a mass on a spring. Starting at equilibrium gives x = A sin ωt, while starting at maximum displacement gives x = A cos ωt. Both show repeated oscillations about x = 0.
These two graphs represent the same SHM. The difference is the starting position

Speed equation

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

    • Where its speed is the magnitude of its velocity given by the equation:

v=±ωA2 − x2 

  • Where:

    • v = speed (m s-1)

    • A = amplitude (m)

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

    • ω = angular frequency (rad s-1)

    • x = displacement (m)

  • This comes from the fact that acceleration is the rate of change of velocity

    • When the standard equation of simple harmonic motion is differentiated using a differential equation, the above equation for velocity is obtained

  • This equation shows that when an oscillator has a greater amplitude A, it has to travel a greater distance in the same time and hence has greater speed v

Worked Example

A mass of 55 g is suspended from a fixed point by means of a spring. The stationary mass is pulled vertically downwards through a distance of 4.3 cm and then released at t = 0. The mass is observed to perform simple harmonic motion with a period of 0.8 s.

Calculate the displacement x, in cm, of the mass at time t = 0.3 s.

[3]

Answer:

Step 1: Write down the SHM displacement equation

Since the mass is released at t = 0 at its maximum displacement, the displacement equation will be with the cosine function:

x = Acos(ωt)

Step 2: Calculate angular frequency

ω = 2πT = 2π0.8 = 7.85 rad s−1 [1 mark]

Remember to use the value of the time period given, not the time where you are calculating the displacement from

Step 3: Substitute values into the displacement equation

x = 4.3cos (7.85 × 0.3) [1 mark]

x = –3.0369… = –3.0 cm (2 s.f.) [1 mark]

Make sure the calculator is in radians mode

The negative value means the mass is 3.0 cm on the opposite side of the equilibrium position from where it started (3.0 cm above it). The downward direction is taken as positive. 

Read the question carefully. It asks for the displacement in cm and not m, so you do not need to convert any measurement units.

Worked Example

A mass–spring system oscillates horizontally with simple harmonic motion with an amplitude of 15 cm. The frequency of the oscillations is 6.7 Hz.

Calculate the speed of the mass at a position of 12 cm from the equilibrium position.

[3]

Answer:

Step 1: Write out the known quantities

  • Amplitude of oscillations, A = 15 cm = 0.15 m

  • Displacement from equilibrium, x = 12 cm = 0.12 m

  • Frequency, f = 6.7 Hz

Step 2: Oscillator speed with displacement equation

v=±ωA2 − 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 = 42.097 rad s−1 [1 mark]

Step 4: Substitute in values and calculate

v = (2π × 6.7) × (0.15)2 − (0.12)2 [1 mark]

v = 3.8 m s−1 (2 s.f.) [1 mark]

Examiner Tips and Tricks

Since displacement is a vector quantity, remember to keep the minus sign in your solutions if they are negative, you could lose a mark if not! Also, remember that your calculator must be in radians mode when using the cosine and sine functions. This is because the angular frequency ⍵ is calculated in rad s-1, not degrees. 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.

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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.

Tim

Reviewer: Tim

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.