Differentiating this with respect to time will give us the velocity of the particle. $\frac{d y}{d t}=A \omega \cos \omega t$

We have, $\sin \omega t =\frac{y}{A}$ and therefore we can calculate $\cos \omega t =\frac{\sqrt{A^2-y^2}}{A}$ (Use Pythagoras theorem to calculate cosine from sine)

Use the value of cosine in equation $(1)$,

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

or, $v=\omega \sqrt{A^2-y^2}$ is our required answer.