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(B) The potential energy $(U)$ of a body in simple harmonic motion at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $

Vedclass · Accepted Answer

$A/2$

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(B) The potential energy $(U)$ of a body in simple harmonic motion at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $(E)$ of the body is $E = \frac{1}{2} m \omega^2 A^2$.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 $E$:$\frac{1}{2} m \omega^2 y^2 = \frac{1}{4} \left( \frac{1}{2} m \omega^2 A^2 \right)$Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides:$y^2 = \frac{A^2}{4}$Taking the square root of both sides:$y = \pm \frac{A}{2}$Thus,at a displacement of $A/2$ from the mean position,the potential energy is one-fourth of the total energy.

$A$ body executes simple harmonic motion with an 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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