The kinetic energy and the potential energy of a particle executing $S.H.M.$ are equal. The ratio of its displacement and amplitude will be
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(a) Given $K.E. = P.E.$$ \Rightarrow \frac{1}{2}m{v^2} = \frac{1}{2}k{x^2}$
$ \Rightarrow \frac{1}{2}m{\omega ^2}({a^2} - {x^2}) = \frac{1}{2}m{\omega ^2}{x^2}$
$ \Rightarrow {a^2} - {x^2} = {x^2}$$ \Rightarrow {x^2} = \frac{{{a^2}}}{2}$$ \Rightarrow \frac{x}{a} = \frac{1}{{\sqrt 2 }}.$
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