The variation of kinetic energy (KE) of a par | vedclass

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(D) For a particle executing simple harmonic motion $(SHM)$,the kinetic energy $(KE)$ as a function of displacement $(x)$ is given by the formula:$KE

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(D) For a particle executing simple harmonic motion $(SHM)$,the kinetic energy $(KE)$ as a function of displacement $(x)$ is given by the formula:$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$where $m$ is the mass of the particle,$\omega$ is the angular frequency,and $A$ is the amplitude.$1$. At the mean position $(x = 0)$,the kinetic energy is maximum: $KE_{max} = \frac{1}{2} m \omega^2 A^2$.$2$. At the extreme position ($x = A$ or $x = -A$),the kinetic energy is zero: $KE = 0$.$3$. The equation $KE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2$ represents a downward-opening parabola with respect to $x$.Comparing this with the given options,the graph that shows maximum $KE$ at $x = 0$ and zero $KE$ at $x = A$ with a parabolic shape is represented by option $(D)$.

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

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