Edexcel International A Level Physics

Revision Notes

5.19 Conditions for Simple Harmonic Motion

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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

13-1-examples-of-shm_edexcel-al-physics-rn

  • 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: 
    • 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  

SHM pendulum, downloadable AS & A Level Physics revision notes

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

Worked example

A 200g toy robot is attached to a pole by a spring, with a spring constant of 90 N m-1, and made to oscillate horizontally.

(a) What force will act on the robot when it is at its amplitude position of 5 cm from equilibrium? 

(b) How fast will the robot accelerate whilst at this amplitude position? 

Part (a) 

Step 1: Convert amplitude into m

5 cm = 0.05 m

Step 2: Substitute values into the restoring force equation

-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. 

13-1-worked-example_edexcel-al-physics-rn

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, a space equals space F over mfraction numerator negative 4.5 over denominator 0.2 end fraction = -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

Exam Tip

Even with this topic you must make sure you convert all quantities into standard SI units

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