A particle is executing S.H.M. with total mec | vedclass

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(B) The total mechanical energy $E$ of a particle executing $S.H.M.$ is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant a

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$4$

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(B) The total mechanical energy $E$ of a particle executing $S.H.M.$ is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.Since $k = m \omega^2$ is constant for a given system,the energy is directly proportional to the square of the amplitude: $E \propto A^2$.Given $E_1 = 90 \,J$ and $A_1 = 6 \,cm$.When the energy is changed to $E_2 = 40 \,J$,let the new amplitude be $A_2$.Using the ratio: $\frac{E_1}{E_2} = \frac{A_1^2}{A_2^2}$.Substituting the values: $\frac{90}{40} = \frac{6^2}{A_2^2}$.$\frac{9}{4} = \frac{36}{A_2^2}$.$A_2^2 = \frac{36 	imes 4}{9} = 4 	imes 4 = 16$.$A_2 = \sqrt{16} = 4 \,cm$.

$A$ particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$,then its amplitude will become ........ $cm$.

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