Show that for a particle executing $SHM$,velocity and displacement have a phase difference of $\frac{\pi}{2}$.

(N/A) The displacement of a particle executing $SHM$ is given by:
$x = A \cos(\omega t)$,where $A$ is the amplitude.
The phase of displacement is $\theta_{1} = \omega t$.
To find the velocity,we differentiate the displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}[A \cos(\omega t)]$
$v = -A\omega \sin(\omega t)$
Using the trigonometric identity $\sin(\theta) = \cos(\frac{\pi}{2} + \theta)$,we can rewrite the velocity as:
$v = A\omega \cos(\omega t + \frac{\pi}{2})$
The phase of velocity is $\theta_{2} = \omega t + \frac{\pi}{2}$.
The phase difference between velocity and displacement is:
$\Delta\theta = \theta_{2} - \theta_{1}$
$\Delta\theta = (\omega t + \frac{\pi}{2}) - \omega t$
$\Delta\theta = \frac{\pi}{2}$
Thus,the velocity leads the displacement by a phase of $\frac{\pi}{2}$.