Obtain the instantaneous velocity of a particle executing $SHM$.

(N/A) The instantaneous velocity of an $SHM$ particle is the rate of change of displacement with respect to time.
Suppose the displacement of an $SHM$ particle at time $t$ with amplitude $A$ and angular frequency $\omega$ is given by:
$x(t) = A \cos(\omega t + \phi) \quad \dots (1)$
where $\phi$ is the initial phase.
Differentiating equation $(1)$ with respect to $t$ gives the instantaneous velocity:
$v(t) = \frac{d[x(t)]}{dt} = \frac{d}{dt}[A \cos(\omega t + \phi)]$
$v(t) = -A \omega \sin(\omega t + \phi)$
Since $\sin(\omega t + \phi) = \pm \sqrt{1 - \cos^2(\omega t + \phi)}$,we substitute this into the velocity expression:
$v(t) = -A \omega (\pm \sqrt{1 - \cos^2(\omega t + \phi)})$
$v(t) = \pm \omega \sqrt{A^2 - A^2 \cos^2(\omega t + \phi)}$
Since $x^2 = A^2 \cos^2(\omega t + \phi)$,we get:
$v = \pm \omega \sqrt{A^2 - x^2}$
Special Cases:
$(1)$ At the mean position,$x = 0$:
$v = \pm \omega \sqrt{A^2 - 0} = \pm \omega A$
Thus,the maximum velocity is $v_{\max} = A \omega$.
$(2)$ At the extreme points,$x = \pm A$:
$v = \pm \omega \sqrt{A^2 - A^2} = 0$
Thus,the velocity at the extreme points is zero.