1. Expression for kinetic energy in SHM

For a particle executing SHM with amplitude $A$ and angular frequency $\omega$,

$v = \omega\sqrt{A^2-x^2}$

So the kinetic energy is

$K = \frac12 mv^2 = \frac12 m\omega^2(A^2-x^2)$

2. Relation between $K$ and $x$

Thus,

$K = \frac12 m\omega^2A^2 - \frac12 m\omega^2x^2$

This is a quadratic function of $x$ opening downward.

So the graph of $K$ versus $x$ is a downward parabola.

3. Behavior from mean position to extreme position

• At mean position, $x=0$: $K_{\max} = \frac12 m\omega^2A^2$
• At extreme position, $x=A$: $K=0$

So as $x$ increases from $0$ to $A$, kinetic energy decreases parabolically from maximum value to zero.

4. Conclusion

Therefore, the correct graph must start from maximum KE at $x=0$ and decrease as a downward parabola to zero at $x=A$.

Hence the correct option is C.