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(D) The total energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $(K.E.)$ and potential energy $(P.E.)$.Given that at

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$2\sqrt{2}\, cm$

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(D) The total energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $(K.E.)$ and potential energy $(P.E.)$.Given that at displacement $x$,the kinetic energy is equal to the potential energy,i.e.,$K.E. = P.E.$We know that $P.E. = \frac{1}{2}m\omega^2x^2$ and $K.E. = \frac{1}{2}m\omega^2(a^2 - x^2)$,where $a$ is the amplitude.Setting $K.E. = P.E.$:$\frac{1}{2}m\omega^2(a^2 - x^2) = \frac{1}{2}m\omega^2x^2$$a^2 - x^2 = x^2$$2x^2 = a^2$$x^2 = \frac{a^2}{2}$$x = \frac{a}{\sqrt{2}}$Given $a = 4\, cm$,we have:$x = \frac{4}{\sqrt{2}} = 2\sqrt{2}\, cm$.Thus,the correct option is $D$.

$A$ 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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