Obtain the formula for the acceleration of a particle from the formula of displacement of $SHM$.

(N/A) The acceleration of an $SHM$ particle is the second derivative of displacement with respect to time.
The displacement of an $SHM$ particle at time $t$ is given by:
$x(t) = A \cos(\omega t + \phi)$
Differentiating with respect to time $t$ to find velocity $v(t)$:
$v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)$
Differentiating again with respect to time $t$ to find acceleration $a(t)$:
$a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t + \phi)$
Since $x(t) = A \cos(\omega t + \phi)$,we can substitute this into the acceleration equation:
$a(t) = -\omega^2 x(t)$
Generally,$a = -\omega^2 x$.
Special cases:
$(1)$ At the mean position,$x = 0$,so $a = -\omega^2(0) = 0$. Acceleration is zero and velocity is maximum.
$(2)$ At extreme points,$x = \pm A$,so $a = \mp \omega^2 A$. The magnitude of acceleration is maximum,$a_{\max} = A \omega^2$.