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(C) The potential energy $U$ of a particle executing simple harmonic motion at a displacement $y$ is given by $U = \frac{1}{2}m\omega^2 y^2$.The maxim

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$a/2$

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(C) The potential energy $U$ of a particle executing simple harmonic motion at a displacement $y$ is given by $U = \frac{1}{2}m\omega^2 y^2$.The maximum potential energy $U_{\max}$ occurs at the amplitude $a$,given by $U_{\max} = \frac{1}{2}m\omega^2 a^2$.According to the problem,the potential energy is one-fourth of its maximum value:$U = \frac{1}{4} U_{\max}$Substituting the expressions for $U$ and $U_{\max}$:$\frac{1}{2}m\omega^2 y^2 = \frac{1}{4} (\frac{1}{2}m\omega^2 a^2)$Canceling the common terms $\frac{1}{2}m\omega^2$ from both sides:$y^2 = \frac{1}{4} a^2$Taking the square root of both sides:$y = \frac{a}{2}$

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation,the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is

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