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  4. The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude will be

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Q. The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude will be

Oscillations

Solution:

Given $KE = PE \Rightarrow \frac{1}{2} m v^{2}=\frac{1}{2} k x ^{2}$
$\Rightarrow \frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}$
$\Rightarrow a^{2}-x^{2}=x^{2} \Rightarrow x^{2}=\frac{a^{2}}{2} \Rightarrow \frac{x}{a}=\frac{1}{\sqrt{2}}$