Question

What is the displacement of an object in S. H. M when K. E=P. E??

Answer 1

$\frac{amplitude}{\sqrt{2}}$

We know that in SHM, KE = $\frac{1}{2} m\omega^2 (A^2 - x^2)$

A stands for amplitude.

And, PE = $\frac{1}{2} m\omega^2 x^2$

So, performing PE = KE:

$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m\omega^2 x^2$

So, $A^2 - x^2 = x^2$ (cancelling the useless stuff)

And $A^2 = 2x^2$

So, $x = \frac{A}{\sqrt{2}}$