Conditions for Simple Harmonic Motion (AQA A Level Physics)

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

  • There is repetitive movement back and forth through an equilibrium, or central, position, so the maximum horizontal or vertical displacement on one side of this 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

Examples of objects that undergo SHM

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

• You will be required to perform calculations on and explain two models of simple harmonic motion:

• • 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

Force, acceleration and displacement of a simple pendulum in SHM

• • A mass-spring system oscillating vertically up and down or horizontally back and forth

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

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 = −⍵²x

• Where:
  • a = acceleration (m s^-2)
  • ⍵ = 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)

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 = A cos (⍵t)

• Where:
  • A = amplitude (m)
  • t = time (s)

• Because:
  • The graph of x = cos (t) starts from amplitude A when t = 0
  • The displacement is at its maximum when cos(⍵t) equals 1 or −1, when x = A
• Use the A Level revision notes on the graphs of trigonometric functions to aid your understanding of trigonometric graphs

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 = A sin (⍵t)

• 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

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:

• Where:
  • v = speed (m s^-1)
  • A = amplitude (m)
  • ± = ‘plus or minus’. The value can be negative or positive
  • ⍵ = 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

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(⍵t)

Step 2: Calculate angular frequency

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) = –3.0369… = –3.0 cm (2 s.f)

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 to where it started (3.0 cm above it). The downwards 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.

Step 1: Write out the known quantities

• • Amplitude of oscillations, A = 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

• • 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:

⍵ = 2πf = 2π× 6.7 = 42.097…

Step 4: Substitute in values and calculate

v = 3.789 = 3.8 m s^-1 (2 s.f)