A particle is vibrating in a simple harmonic motion with an amplitude of $4\, cm.$ At what displacement from the equilibrium position, is its energy half potential and half kinetic
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(d) Let $x$ be the point where $K.E. = P.E.$
Hence $\frac{1}{2}m{\omega ^2}({a^2} - {x^2}) = \frac{1}{2}m{\omega ^2}{x^2}$
==> $2{x^2} = {a^2}$
Hence $\frac{1}{2}m{\omega ^2}({a^2} - {x^2}) = \frac{1}{2}m{\omega ^2}{x^2}$
==> $2{x^2} = {a^2}$
==> $x = \frac{a}{\sqrt{2}} = \frac{a}{\sqrt{2}} = 2\sqrt{2}\,cm$
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