The ratio between kinetic and potential energ | vedclass

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(D) The kinetic energy $KE$ at a displacement $x = \frac{a}{N}$ is given by:$KE = \frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 \left[ a

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$N^2-1$

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(D) The kinetic energy $KE$ at a displacement $x = \frac{a}{N}$ is given by:$KE = \frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 \left[ a^2 - \left( \frac{a}{N} \right)^2 \right]$The potential energy $PE$ at a displacement $x = \frac{a}{N}$ is given by:$PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left( \frac{a}{N} \right)^2$Taking the ratio of $KE$ to $PE$:$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 \left[ a^2 - \frac{a^2}{N^2} \right]}{\frac{1}{2} m \omega^2 \frac{a^2}{N^2}}$Simplifying the expression:$\frac{KE}{PE} = \frac{a^2 (1 - \frac{1}{N^2})}{\frac{a^2}{N^2}} = \frac{\frac{N^2 - 1}{N^2}}{\frac{1}{N^2}} = N^2 - 1$

The ratio between kinetic and potential energies of a body executing simple harmonic motion,when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is

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