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(C) The total energy $E$ of a particle executing $SHM$ is given by the formula $E = \frac{1}{2}m{\omega ^2}{A^2}$,where $m$ is the mass,$\omega$ is th

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$\frac{9}{16}E$

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(C) The total energy $E$ of a particle executing $SHM$ is given by the formula $E = \frac{1}{2}m{\omega ^2}{A^2}$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.Since the time period $T$ is kept constant,the angular frequency $\omega = \frac{2\pi}{T}$ also remains constant.Therefore,the total energy is directly proportional to the square of the amplitude: $E \propto A^2$.Let the initial amplitude be $A$ and the initial energy be $E$. The new amplitude is $A' = \frac{3}{4}A$.The new energy $E'$ is given by $\frac{E'}{E} = \left( \frac{A'}{A} \right)^2$.Substituting the value of $A'$,we get $\frac{E'}{E} = \left( \frac{\frac{3}{4}A}{A} \right)^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.Thus,the new energy is $E' = \frac{9}{16}E$.

The amplitude of a particle executing $SHM$ is made three-fourth keeping its time period constant. Its total energy will be

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