The speed $v$ of a particle moving along a straight line, when it is at a distance ($x$) from a fixed point of the line is given by $v^2=108-9x^2$ (all quantities are in $cgs$ units)

$v^2=108-9x^{2}or{v}^2=9\left(12-x^2\right)$

We can compare the above expression with $y$

$v=\omega \sqrt{A^2-x^2},$which is expression of velocity for SHM.

From this, we will get $\omega =3and\mathrm{A}=\sqrt{12}$

SHM is not a uniformly accelerated motion.

Acceleration at a distance $3 \mathrm{cm}$ from the mean position.

$q={\omega }^2(3 \mathrm{cm})=27 \mathrm{cm}/{\mathrm{s}}^2$

Maximum displacement from the mean position =$\mathrm{A}=\sqrt{12}.$