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(A) The total energy of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$,where $k = m \omega^2$.Given $E_1 = 90 \,J$ and $A_1 = 6 \,

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$40$

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(A) The total energy of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} k A^2$,where $k = m \omega^2$.Given $E_1 = 90 \,J$ and $A_1 = 6 \,cm = 0.06 \,m$.$90 = \frac{1}{2} k (0.06)^2 \implies k = \frac{180}{0.0036} = 50000 \,N/m$.When the particle is at $x = 4 \,cm = 0.04 \,m$,it is stopped. At this point,its velocity becomes zero,so the new amplitude $A_2$ of the motion becomes the displacement at that instant,i.e.,$A_2 = 4 \,cm = 0.04 \,m$.The new energy of vibration $E_2$ is given by $E_2 = \frac{1}{2} k A_2^2$.$E_2 = \frac{1}{2} 	imes 50000 	imes (0.04)^2$.$E_2 = 25000 	imes 0.0016 = 40 \,J$.

$A$ particle is performing $S.H.M.$ with energy of vibration $90 \,J$ and amplitude $6 \,cm$. When the particle reaches a distance of $4 \,cm$ from the mean position,it is stopped for a moment and then released. The new energy of vibration will be ........... $J$.

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