The total energy of a particle executing S.H. | vedclass

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(D) The total energy $E$ of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 a^2 = 80 \, J$,where $a$ is the amplitude.The potential ene

Vedclass · Accepted Answer

$45$

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(D) The total energy $E$ of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 a^2 = 80 \, J$,where $a$ is the amplitude.The potential energy $U$ at a displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.Taking the ratio of potential energy to total energy:$\frac{U}{E} = \frac{\frac{1}{2} m \omega^2 y^2}{\frac{1}{2} m \omega^2 a^2} = \frac{y^2}{a^2}$.Given that the particle is at a distance $y = \frac{3}{4} a$ from the mean position:$\frac{U}{80} = \frac{(\frac{3}{4} a)^2}{a^2} = \frac{9}{16}$.Solving for $U$:$U = 80 	imes \frac{9}{16} = 5 	imes 9 = 45 \, J$.

The total energy of a particle executing $S.H.M.$ is $80 \, J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position?

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