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(C) The total energy of the particle in $S.H.M.$ is $E = \frac{1}{2}m\omega^2A^2$.At position $x = \frac{\sqrt{3}}{2}A$,the potential energy is $U = \

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$\sqrt{2}A$

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(C) The total energy of the particle in $S.H.M.$ is $E = \frac{1}{2}m\omega^2A^2$.At position $x = \frac{\sqrt{3}}{2}A$,the potential energy is $U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2(\frac{3}{4}A^2) = \frac{3}{8}m\omega^2A^2$.The initial kinetic energy at this position is $K = E - U = \frac{1}{2}m\omega^2A^2 - \frac{3}{8}m\omega^2A^2 = \frac{1}{8}m\omega^2A^2$.An impulsive force increases the kinetic energy by $\Delta K = \frac{1}{2}m\omega^2A^2$.The new kinetic energy at this position is $K' = K + \Delta K = \frac{1}{8}m\omega^2A^2 + \frac{4}{8}m\omega^2A^2 = \frac{5}{8}m\omega^2A^2$.The potential energy $U$ remains unchanged as the position $x$ is the same.The new total energy $E'$ is $E' = K' + U = \frac{5}{8}m\omega^2A^2 + \frac{3}{8}m\omega^2A^2 = m\omega^2A^2$.Since $E' = \frac{1}{2}m\omega^2(A')^2$,we have $\frac{1}{2}m\omega^2(A')^2 = m\omega^2A^2$.Therefore,$(A')^2 = 2A^2$,which gives $A' = \sqrt{2}A$.

$A$ particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}A$ from the mean position,its kinetic energy is increased by an amount $\frac{1}{2}m\omega^2A^2$ due to an impulsive force. What is its new amplitude?

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