Pendula are not the only oscillating system. Another approximation for many systems is a mass hanging on a spring. Although the forces here are a bit easier to analyse, we do need to make the assumption that the mass and spring system is in space!

This is another example of simple harmonic motion, as the acceleration is proportional to displacement and acting in the opposite direction.

Summary

We know that, for springs obeying Hooke's law, the force is proportional to extension. For a spring in space, the elastic force is the resultant force. The acceleration of a mass hanging from a spring is therefore proportional to its displacement from the equilibrium position.

When a spring is extended it obeys law, this means that the restoring force is proportional to the . According to Newtons law the acceleration of a mass attached to the spring will therefore be proportional to the from the equilibrium position, which complies with the definition of SHM.

A spring extends 2cm when a mass of 100 g is hung from it.

What is the spring constant of the spring?

k = F/x = 1/0.2

A spring extends 2cm when a mass of 100 g is hung from it.

What will the extension be with a mass of 0.3 kg

k is proportional to force.

0.3 kg will exert 3x the force so 3x extension

The graph represents the force against extension for a rubber band

A mass hanging on this sample of rubber:

Between 0 and 1 cm the force is approximately proportional to extension. The graph is approximately linear in the middle region but F = kx +c not kx

The diagram below represents a spring showing the unloaded length, the length when a 100g mass is hanging at rest and the length when the mass is pulled down.

The spring constant is:

K= F/x =1/0.1

The diagram below represents a spring showing the unloaded length, the length when a 100g mass is hanging at rest and the length when the mass is pulled down.

The Extension of the spring in the equilibrium position is:

The middle spring is in the equilibrium position. The extension is from the original length to the new length.

The diagram below represents a spring showing the unloaded length, the length when a 100g mass is hanging at rest and the length when the mass is pulled down.

If the mass is released it will undergo SHM. The ammplitude of the oscillation is:

Amplitude is from the maximum displacement to the equilibrium position

The diagram below represents a spring showing the unloaded length, the length when a 100g mass is hanging at rest and the length when the mass is pulled down.

The mass is replaced by a 50g mass then it is pulled down to the same position. The amplitude of the oscillation will be:

The equilibrium position will be at an extension of 5 cm so when pulled down the displacement will be 15 cm

The diagram below represents a spring showing the unloaded length the length when a 100g mass is hanging at rest and the length when the mass is pulled down.

The mass is replaced by a 50 g mass and it is pulled down to the same position. What will be the effect on the amplitude and frequency?

The equilibrium position will be at an extension of 5 cm so the amplitude will be bigger. The frequency is inversely proportional to √m

The diagram below represents a spring showing the unloaded length, the length when a 100g mass is hanging at rest and the length when the mass is pulled down.

The time period of the oscillation is given by the equation $$T=2\pi \sqrt{\frac{m}{k}}$$ The time period is:

k = 10 Nm^{-1} m = 0.1 kg $$$$T=2π0.110=0.1×2πMathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2da9iaaikdacqaHapaCdaGcaaqaamaalaaabaGaaGimaiaac6cacaaIXaaabaGaaGymaiaaicdaaaaaleqaaOGaeyypa0JaaGimaiaac6cacaaIXaGaey41aqRaaGOmaiabec8aWbaa@3DFF@

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