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Question

(B) The potential energy $U$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $E$ of the particle is given by $

Vedclass · Accepted Answer

$10$

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(B) The potential energy $U$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $E$ of the particle is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.The ratio of potential energy to total energy is $\frac{U}{E} = \frac{y^2}{a^2}$.Given that the displacement $y = \frac{a}{2}$,we substitute this into the ratio:$\frac{2.5}{E} = \frac{(a/2)^2}{a^2} = \frac{a^2/4}{a^2} = \frac{1}{4}$.Therefore,$E = 2.5 	imes 4 = 10 \ J$.

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

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