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# 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?**