Obtain the velocity of a particle executing simple harmonic motion $(SHM)$ by considering the projection of a particle undergoing uniform circular motion.

(N/A) The linear velocity of a particle moving in a circular path of radius $A$ with angular speed $\omega$ is given by $v = A \omega$.
The direction of the linear velocity of a particle at time $t$ is along the tangent to the circle at that point.
According to the figure,the linear velocity vector $A \omega$ can be resolved into two mutually perpendicular components. The component along the $X$-axis is $A \omega \sin (\omega t + \phi)$,which represents the velocity of the projection of the particle on the $X$-axis at time $t$.
Since the projection is moving towards the origin (opposite to the positive $X$-axis direction),the velocity is given by:
$v(t) = -A \omega \sin (\omega t + \phi)$
This expression represents the instantaneous velocity of a particle executing $SHM$.