A particle is executing simple harmonic motion with a time period T...

$x = Asin \omegat$

$v = \dfrac{dx}{dt} = a\omega cos \omegat$

$K . E . = \dfrac{1}{2}m{v}^{2}$

$= \dfrac{1}{2}m{A}^{2}{\omega}^{2}\cos^{2} \omegat$

$= \dfrac{1}{2}m{A}^{2}{\omega}^{2}\cos^{2} \dfrac{2\pi}{T}t$

At $t = 0$, $K . E .$ is maximum.

The period of $K . E .$ graph is half of the period of displacement graph.

Comment:- The points on the graph of $x -$axis should be differentiable.

In all option mentioned, the graph has cusps at the points on $x -$axis.