Question

A particle is executing SHM with amplitude A and has maximum velocity $v_s$. Find its speed when at distance of A/2 from mean position.

Answer 1

Given that,

Distance y= A/2

Amplitude = A

We know that,

$v=\omega \sqrt{A^2-y^2}$

Now, the speed is

$v=\omega \sqrt{A^2-\frac{A^2}{4}}$

$v=\omega \sqrt{\frac{3A^2}{4}}$

$v=\omega A\frac{\sqrt{3}}{2}$

Hence, the speed is $\omega A\frac{\sqrt{3}}{2}$