(N/A) For a particle executing $SHM$ along the $X$-axis,the displacement $x$ at any time $t$ is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
The instantaneous acceleration $a$ is the second derivative of displacement with respect to time:
$a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = -A\omega^2 \sin(\omega t + \phi)$.
Since $x = A \sin(\omega t + \phi)$,we can substitute this into the expression for acceleration:
$a = -\omega^2 x$.
Thus,the formula for instantaneous acceleration is $a = -\omega^2 x$.