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(B) The total energy $(E)$ of a body in simple harmonic motion is given by $E = \frac{1}{2} kA^2$,where $k$ is the force constant and $A$ is the ampli

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$\frac{A}{2}$

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(B) The total energy $(E)$ of a body in simple harmonic motion is given by $E = \frac{1}{2} kA^2$,where $k$ is the force constant and $A$ is the amplitude.The potential energy $(U)$ at a displacement $x$ from the mean position is given by $U = \frac{1}{2} kx^2$.According to the problem,the potential energy is one-fourth of the total energy:$U = \frac{1}{4} E$Substituting the expressions for $U$ and $E$:$\frac{1}{2} kx^2 = \frac{1}{4} (\frac{1}{2} kA^2)$Canceling $\frac{1}{2} k$ from both sides:$x^2 = \frac{A^2}{4}$Taking the square root of both sides:$x = \pm \frac{A}{2}$Thus,at a displacement of $\frac{A}{2}$ from the mean position,the potential energy is one-fourth of the total energy.

$A$ body performs linear simple harmonic motion of amplitude $A$. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?

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