Question

A particle is executing SHM with amplitude A and has maximum velocity $V_0$. Find its speed when it is located at distance of A/2 from mean position.

Answer 1

Let equation of displacement from mean be $x=A\sin \left(\omega t\right)$

Then at $x=\frac{A}{2}$

$\Rightarrow \frac{A}{2}=A\sin \omega t$

$\Rightarrow \sin \omega t=\frac{1}{2}$

$\omega t=\pi /6$

Also,

$v=\frac{dx}{dt}=\left(A\omega \right)\cos \left(\omega t\right)$

As $v_o=A\omega$,

At $\omega t=\pi /6$,

$v=v_o\cos \left(\pi /6\right)=v_o\sqrt{3}/2$

$v=\frac{\sqrt{3}}{2}{v}_o$.