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(C) The total energy $(E)$ of a simple harmonic oscillator is given by the sum of its potential energy $(U)$ and kinetic energy $(K)$.At the mean posi

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$10$

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(C) The total energy $(E)$ of a simple harmonic oscillator is given by the sum of its potential energy $(U)$ and kinetic energy $(K)$.At the mean position,the potential energy is given as $U_{0} = 2\,J$.The total energy is $E = U_{0} + K_{max} = U_{0} + \frac{1}{2}kA^2$.The mean kinetic energy over a cycle is given by $\langle K angle = \frac{1}{4}kA^2 = 4\,J$.Therefore,the maximum kinetic energy is $K_{max} = \frac{1}{2}kA^2 = 2 	imes \langle K angle = 2 	imes 4\,J = 8\,J$.The total energy is $E = U_{0} + K_{max} = 2\,J + 8\,J = 10\,J$.

The potential energy of a simple harmonic oscillator at the mean position is $2\,J$. If its mean kinetic energy $(K.E.)$ is $4\,J$,its total energy will be .... $J$

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