# I SHM equation clarification

1. Oct 17, 2016

### ual8658

So in class we were given a problem.
A pendulum is oscillating in SHM with amplitude A. After a while you slow it down so that its amplitude is halved. What happens to its speed at x = ± A/4 ?

The answer is 1/sqrt(5) and I don't get why.

2. Oct 18, 2016

### andrewkirk

There are two things that have changed. One is that, because the amplitude has halved from $A$ to $A/2$, the speed for a given phase point has also halved.

The other is that, because the amplitude has changed, the point $x=A/4$ is at a different phase from what it was when the amplitude was $A$.

You need to compare the velocity when position is at half the amplitude (the new situation) with the velocity when position is one quarter of the amplitude (the old situation).

3. Oct 18, 2016

### sophiecentaur

You need to do the calculation; it isn't an intuitive thing. If you write out the equation for SHM it will show you how the position varies with time. If you differentiate it wrt time, you will get the variation of velocity with time. So you can put your two values of Amplitude into that velocity equation and see what the velocities are for x= A/4. AS the differential of sin is cos, life is pretty easy for you.

4. Oct 18, 2016

### ual8658

Thank you both. I thought there'd be a way to see it intuitively but the calculations show the reason why.