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(C) For a simple harmonic oscillator,the displacement $d$ is given by $d = A \sin(\omega t + \phi).$The potential energy $(PE)$ is given by $PE = \fra

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(C) For a simple harmonic oscillator,the displacement $d$ is given by $d = A \sin(\omega t + \phi).$The potential energy $(PE)$ is given by $PE = \frac{1}{2} k d^2,$ which represents a parabola opening upwards with its vertex at the origin $(0,0).$The kinetic energy $(KE)$ is given by $KE = E - PE = \frac{1}{2} k A^2 - \frac{1}{2} k d^2,$ where $E$ is the total energy. This represents an inverted parabola (a parabola opening downwards) with its maximum value at $d = 0$ and zero value at $d = \pm A.$Comparing these with the given options,the graph in option $C$ correctly shows $KE$ as an inverted parabola and $PE$ as a parabola opening upwards.

For a simple pendulum,a graph is plotted between its kinetic energy $(KE)$ and potential energy $(PE)$ against its displacement $d.$ Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

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