5.19 Conditions for Simple Harmonic Motion

• Simple harmonic motion (SHM) is a specific type of oscillation
• An oscillation is said to be SHM when:
  • The acceleration is proportional to the displacement
  • The acceleration is in the opposite direction to the displacement

• Examples of oscillators that undergo SHM are:
  • The pendulum of a clock
  • A mass on a spring
  • Guitar strings
  • The electrons in alternating current flowing through a wire

• Time period, T:
  • The objects swings are periodic, meaning they are repeated in regular intervals according to their frequency or time period
  • If an object swings freely it always takes the same time to complete one swing

Restoring force

• When an object is moving in SHM a force, called the restoring force, F, is always trying to return the object back to its equilibrium position. 
• The force is proportional to the displacement, x, from that equilibrium position

F = -kx

• Where: 
  • F  is the restoring force
  • x is the displacement of the object from the equilibrium position
  • k is a constant depending on the system
  • the negative sign shows that the acceleration will always be towards the centre of oscillation  

Force, acceleration and displacement of a pendulum in SHM

• This is why a person jumping on a trampoline is not an example of simple harmonic motion:
  • The restoring force on the person is not proportional to their distance from the equilibrium position
  • 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

Part (a) 

Step 1: Convert amplitude into m

5 cm = 0.05 m

Step 2: Substitute values into the restoring force equation

F = -kx = -(90) x (0.05) = - 4.5 N

Step 3: Explain the answer

A force of 4.5 newtons will act on the robot, trying to pull it back towards the equilibrium position. 

Part (b)

Step 1: Convert mass of robot into kg

200 g = 0.2 kg

Step 2: Substitute values into Newton's second law equation: 

F = ma

So,  =  = -22.5 m s^-2

Step 3: Explain the answer

The robot will decelerate at a rate of 22.5 m s^-2 when at this amplitude position