The ratio between kinetic and potential energies of a body exccuting simple harmonic motion, when it is at a distance of 1/N of its amplitude from the mean position is

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N²

N² – 1

N² + 1

$\frac{1}{N^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} y = \frac{A}{N}; KE = \frac{1}{2} {mω}^{2} (A^{2} - y^{2}) \\ = \frac{1}{2} {mω}^{2} (A^{2} - \frac{A^{2}}{N^{2}}) \dots \dots \dots (1) \\ PE = \frac{1}{2} {mω}^{2} y^{2} = \frac{1}{2} {mω}^{2} \frac{A^{2}}{N^{2}} \\ \frac{eq (1)}{eq (2)} \frac{KE}{PE} = \frac{1 - \frac{1}{N^{2}}}{\frac{1}{N^{2}}} = N^{2} - 1 \end{array}$

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