Obtain instantaneous acceleration of a $SHM$ particle with the help of reference circle.

(N/A) $1$. Consider a particle $P$ moving in a circle of radius $A$ with constant angular velocity $\omega$. The projection of this motion on the diameter (say $x$-axis) represents $SHM$.
$2$. The position of the particle at any time $t$ is given by $x = A \cos(\omega t + \phi)$.
$3$. The velocity of the particle is $v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$.
$4$. The instantaneous acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-A\omega \sin(\omega t + \phi))$.
$5$. Differentiating,we get $a = -A\omega^2 \cos(\omega t + \phi)$.
$6$. Since $x = A \cos(\omega t + \phi)$,we can substitute this into the acceleration equation to get $a = -\omega^2 x$.