The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by

We need to identify the correct graph showing the variation of kinetic energy (KE) with displacement (x) for a particle in SHM, from the mean position to the extreme position.

State the formula for KE in SHM.

$$KE = \dfrac{1}{2}m\omega^2(A^2 - x^2)$$

where $$A$$ is the amplitude and $$x$$ is the displacement from the mean position.

Analyze the variation.

At $$x = 0$$ (mean position): $$KE = \dfrac{1}{2}m\omega^2 A^2$$ (maximum).

At $$x = A$$ (extreme position): $$KE = 0$$.

The relationship $$KE = \dfrac{1}{2}m\omega^2 A^2 - \dfrac{1}{2}m\omega^2 x^2$$ shows that KE varies as a downward-opening parabola with respect to $$x$$.

Identify the correct graph.

The graph should be a downward parabola starting from maximum KE at $$x = 0$$ and reaching 0 at $$x = A$$. This is an inverted parabola (concave downward).

The answer is $$\boxed{\text{Option 3}}$$ — KE decreases as a parabola from the mean to extreme position.