The kinetic energy, K of a body performing simple harmonic motion...

Kine tic energy of a body performing simple harmonic motion is given as K=$\frac{1}{2}$m v² where, $$\begin{aligned}& v=\frac{d y}{d t}=\frac{d}{d t} \cdota \sin \omegat \quad[\because y=a \sin \omegat] \\ & v=a \omega \cos \omegat \end{aligned}$$\therefore$From Eq. (i), we get$$\begin{aligned}& K=\frac{1}{2}m(a \omega \cos \omegat)^2=\frac{1}{2}m a² \omega^2 \cos ^2 \omegat \\ & K=\frac{1}{2}m a² \omega^2\left(\frac{1+\cos 2 \omega t}{2}\right) \quad \ldots(\text{ i }) \end{aligned}$$The graph represented in option (1) is correct for expression of kinetic energy represented by Eq. (i).