A particle is vibrating in S.H.M. with an amp | vedclass

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(D) The kinetic energy $(K.E.)$ of a particle in $S.H.M.$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(P.E.)$ is giv

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$2 \sqrt{2} \,cm$

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(D) The kinetic energy $(K.E.)$ of a particle in $S.H.M.$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(P.E.)$ is given by $P.E. = \frac{1}{2} m \omega^2 x^2$.Given that the energy is half potential and half kinetic, we have $K.E. = P.E.$Substituting the expressions: $\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2$.Simplifying this, we get $A^2 - x^2 = x^2$, which implies $A^2 = 2x^2$.Therefore, $x = \frac{A}{\sqrt{2}}$.Given the amplitude $A = 4 \,cm$, we find $x = \frac{4}{\sqrt{2}} = 2\sqrt{2} \,cm$.

$A$ particle is vibrating in $S.H.M.$ 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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