The ratio between kinetic and potential energ | vedclass

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(D) Let the amplitude of the simple harmonic motion be $a$. The displacement of the body from the mean position is $x = \frac{a}{N}$.The kinetic energ

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$N^2-1$

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(D) Let the amplitude of the simple harmonic motion be $a$. The displacement of the body from the mean position is $x = \frac{a}{N}$.The kinetic energy $(KE)$ of the body is given by:$KE = \frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 \left(a^2 - \frac{a^2}{N^2}\right) = \frac{1}{2} m \omega^2 a^2 \left(1 - \frac{1}{N^2}\right) = \frac{1}{2} m \omega^2 a^2 \left(\frac{N^2 - 1}{N^2}\right) \quad (i)$The potential energy $(PE)$ of the body is given by:$PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left(\frac{a}{N}\right)^2 = \frac{1}{2} m \omega^2 \frac{a^2}{N^2} \quad (ii)$Taking the ratio of $KE$ to $PE$:$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 a^2 \left(\frac{N^2 - 1}{N^2}\right)}{\frac{1}{2} m \omega^2 \frac{a^2}{N^2}} = \frac{N^2 - 1}{1} = N^2 - 1$

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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