the displacement of particle along x axis is given by

Dear Student
The displacement of a particle along the x-axis is given by x = a sin² ωt. 
For a particle executing SHM
$$\begin{gathered} acceleration \alpha -x .. \left(1\right) \\ that is acceleration = -\omega x \\ Here, \\ x = a{\sin }^2\omega t \\ v =\frac{dx}{dt} = a \frac{d}{dt}\left({\sin }^2\omega t\right) \\ v =2a\omega \sin \omega t \cos \omega t \\ v = a\omega \left(\sin 2 \omega t\right) \\ Acceleration = \frac{dv}{dt} =a\omega . \cos 2 \omega t. 2\omega =2a{\omega }^2\cos 2 \omega t \\ So this does not satisfies condition 1 \\ Hence motion is not simple harmonic motion \end{gathered}$$
So correct option is C

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