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

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$\frac{1}{\sqrt{2}}$

$\frac{1}{2}$

$\sqrt{2}$

$\frac{\sqrt{3}}{2}$

answer is A.

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Detailed Solution

Given KE = PE $\Rightarrow \frac{1}{2} {mv}^{2} = \frac{1}{2} {kx}^{2}$

$\Rightarrow \frac{1}{2} {mω}^{2} (a^{2} - x^{2}) = \frac{1}{2} {mω}^{2} x^{2}$

$\Rightarrow a^{2} - x^{2} = x^{2} \Rightarrow x^{2} = \frac{a^{2}}{2} \Rightarrow \frac{x}{a} = \frac{1}{\sqrt{2}}$

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