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(D) Given:Amplitude $A = 0.01 \, m = 10^{-2} \, m$Force constant $k = 2 	imes 10^6 \, N/m$Total mechanical energy $E_T = 160 \, J$The maximum kinetic

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$(B)$ and $(C)$ both

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(D) Given:Amplitude $A = 0.01 \, m = 10^{-2} \, m$Force constant $k = 2 	imes 10^6 \, N/m$Total mechanical energy $E_T = 160 \, J$The maximum kinetic energy $(K.E.)_{\max}$ of a linear harmonic oscillator is given by:$(K.E.)_{\max} = \frac{1}{2} k A^2$$(K.E.)_{\max} = \frac{1}{2} 	imes (2 	imes 10^6) 	imes (10^{-2})^2$$(K.E.)_{\max} = 10^6 	imes 10^{-4} = 100 \, J$In a linear harmonic oscillator,the total mechanical energy $E_T$ is equal to the maximum potential energy $(P.E.)_{\max}$ (when kinetic energy is zero at extreme positions) and also equal to the maximum kinetic energy $(K.E.)_{\max}$ (when potential energy is zero at the mean position,assuming $P.E. = 0$ at the mean position).However,here $E_T = 160 \, J$ and $(K.E.)_{\max} = 100 \, J$. This implies that the potential energy at the mean position is not zero. Since $E_T = (K.E.)_{\max} + (P.E.)_{\min}$,we have $160 = 100 + (P.E.)_{\min}$,so $(P.E.)_{\min} = 60 \, J$.The maximum potential energy $(P.E.)_{\max}$ occurs at the extreme positions where kinetic energy is zero,so $(P.E.)_{\max} = E_T = 160 \, J$.Thus,both statements $(B)$ and $(C)$ are correct.

$A$ linear harmonic oscillator of force constant $2 \times 10^6 \, N/m$ and amplitude $0.01 \, m$ has a total mechanical energy of $160 \, J$. Its

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