Calculations and Examples with SHM

Calculations and Examples with SHM

The definition for simple harmonic motion tells us that:

a α -s

We can get rid of the proportionality sign by putting in a constant. In this case, the constant is (2πf)2, so:

a = - (2 πf)2 s

Example:

A road drill vibrates up and down with SHM at a frequency of 20Hz.

What's the maximum acceleration of the pick head if the amplitude of the oscillation is 5cm?

Calculations and Examples with SHM

Answer:

The maximum acceleration occurs at the point of maximum displacement - for instance, the amplitude. So,

a = -(40π)2 x 5 x 10-2

a = 790ms-2

Once you've found the acceleration, you can calculate the forces involved - so long as you are told the mass of the pick head.

So if the mass of the pick head is 3kg...

F = ma = 3 x 790 = 2370N (3sf)

As shm oscillations follow a sine or cosine wave, we can find the displacement at any point using:

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Where:

A = amplitude - not acceleration!

Velocity can be found using:

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Example:

A person uses a rope swing to get across a stream. They run and grab the rope at A and swing to the other side at B, 6m from the start.

They hang on to the rope until it stops at the end of the swing at C, 6.5 m from the start.

The centre of the swing (and of the stream), D, is now 3.5 m away, so the amplitude of the swing is 3.5 m.

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a) If the frequency of the swing is 0.2 Hz, what was the person's speed as they passed the opposite bank of the stream at B, 3 m from the centre of the swing?

b) What was the maximum speed during the swing and where exactly was the person then?

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