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(D) The total energy $(TE)$ of a particle executing $SHM$ is constant,given by $TE = \frac{1}{2} m \omega^2 A^2$.The potential energy $(PE)$ at a disp

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(D) The total energy $(TE)$ of a particle executing $SHM$ is constant,given by $TE = \frac{1}{2} m \omega^2 A^2$.The potential energy $(PE)$ at a displacement $x$ from the mean position is given by $PE = \frac{1}{2} m \omega^2 x^2$.The difference between total energy and potential energy is the kinetic energy $(KE)$:$KE = TE - PE$$KE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2$$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$This equation represents a downward-opening parabola with its vertex at $x = 0$ and roots at $x = \pm A$. This matches the shape shown in graph $D$.

Which graph represents the difference between total energy and potential energy of a particle executing $SHM$ versus its distance from the mean position?

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