# 5.19 Conditions for Simple Harmonic Motion

## 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:
• 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

#### 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

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

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 ### Get unlimited access

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