$x = a{\sin ^2}\omega t = a\left( {{{1 - \cos 2\omega t} \over 2}} \right)$

   $\left(\because {\cos 2\theta = 1 - 2{{\sin }^2}\theta } \right)$

$= {a \over 2} - {{a\cos 2\omega t} \over 2}$

$\therefore$ Velocity, $v = {{dx} \over {dt}} = {{2\omega a\sin 2\omega t} \over 2} = \omega \alpha \sin 2\omega t$

Acceleration, $a = {{dv} \over {dt}} = 2{\omega ^2}a\cos 2\omega t$

For the given displacement x = $\alpha \sin 2\omega t$,

$a \propto - x$ is not satisfied.

Hence, the motion of the particle is non simple harmonic motion.

Note : The given motion is a periodic motion with a time period

$T = {{2\pi } \over {2\omega }} = {\pi \over \omega }$