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(D) The total energy of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.At any displacement $x$,the potential energy is $U = \frac

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$\frac{A}{2}$

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(D) The total energy of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.At any displacement $x$,the potential energy is $U = \frac{1}{2} m \omega^2 x^2$.The kinetic energy is $K = E - U = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Given that $K = \frac{3E}{4}$,the potential energy $U$ must be $E - \frac{3E}{4} = \frac{E}{4}$.Substituting the expressions for $U$ and $E$:$\frac{1}{2} m \omega^2 x^2 = \frac{1}{4} (\frac{1}{2} m \omega^2 A^2)$.Simplifying,we get $x^2 = \frac{A^2}{4}$.Taking the square root,$x = \frac{A}{2}$.

$A$ particle performs $S.H.M.$ from the mean position. Its amplitude is $A$ and total energy is $E$. At a particular instant,its kinetic energy is $\frac{3E}{4}$. The displacement of the particle at that instant is:

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