Calculations and Examples with SHM
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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?
Answer:
The maximum acceleration occurs at the point of maximum displacement - in other words, 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:
Where:
A = amplitude - not acceleration!
Velocity can be found using:
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.
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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