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(C) The total energy $E$ of a standing wave is given by the formula $E = \frac{1}{4} m \omega^2 A^2$,where $m$ is the mass of the wire,$\omega$ is the

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${E_2} = 4{E_1}$

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(C) The total energy $E$ of a standing wave is given by the formula $E = \frac{1}{4} m \omega^2 A^2$,where $m$ is the mass of the wire,$\omega$ is the angular frequency,and $A$ is the amplitude.In the first experiment,the frequency is $\omega_1 = \omega$ and the amplitude is $A_1 = A$. Thus,$E_1 = \frac{1}{4} m \omega^2 A^2$.In the second experiment,the frequency is $\omega_2 = 2\omega$ and the amplitude is $A_2 = A$. Thus,$E_2 = \frac{1}{4} m (2\omega)^2 A^2 = 4 	imes (\frac{1}{4} m \omega^2 A^2) = 4E_1$.Therefore,the energy in the second experiment is four times the energy in the first experiment,i.e.,${E_2} = 4{E_1}$.

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