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(D) The potential energy $(P.E.)$ of a particle in $SHM$ at displacement $x$ is given by $P.E. = \frac{1}{2} kx^2$.The kinetic energy $(K.E.)$ of the

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(D) The potential energy $(P.E.)$ of a particle in $SHM$ at displacement $x$ is given by $P.E. = \frac{1}{2} kx^2$.The kinetic energy $(K.E.)$ of the particle at displacement $x$ is given by $K.E. = \frac{1}{2} k(A^2 - x^2)$,where $A$ is the amplitude.Given that the displacement $x = \frac{A}{2}$.Substituting $x$ into the potential energy formula: $P.E. = \frac{1}{2} k(\frac{A}{2})^2 = \frac{1}{2} k(\frac{A^2}{4}) = \frac{1}{8} kA^2$.Substituting $x$ into the kinetic energy formula: $K.E. = \frac{1}{2} k(A^2 - (\frac{A}{2})^2) = \frac{1}{2} k(A^2 - \frac{A^2}{4}) = \frac{1}{2} k(\frac{3A^2}{4}) = \frac{3}{8} kA^2$.The ratio of potential energy to kinetic energy is $\frac{P.E.}{K.E.} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$.

$A$ particle is executing Simple Harmonic Motion $(SHM)$. The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

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