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(A) The total mechanical energy $(TE)$ of a linear harmonic oscillator is given by $TE = PE_{max} = \frac{1}{2} k A^2 + C_0$,where $C_0$ is the consta

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Maximum potential energy is $160\,J$

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(A) The total mechanical energy $(TE)$ of a linear harmonic oscillator is given by $TE = PE_{max} = \frac{1}{2} k A^2 + C_0$,where $C_0$ is the constant potential energy offset.Given $k = 2 	imes 10^6\,N/m$,$A = 0.01\,m$,and $TE = 160\,J$.The potential energy at the extreme position $(x = A)$ is the maximum potential energy,which is equal to the total mechanical energy.$PE_{max} = \frac{1}{2} k A^2 + C_0 = 160\,J$.Calculating the elastic potential energy component: $\frac{1}{2} k A^2 = \frac{1}{2} 	imes (2 	imes 10^6) 	imes (0.01)^2 = 10^6 	imes 10^{-4} = 100\,J$.Since $TE = 160\,J$,we have $100\,J + C_0 = 160\,J$,which gives $C_0 = 60\,J$.The minimum potential energy occurs at the equilibrium position $(x = 0)$,where $PE_{min} = \frac{1}{2} k(0)^2 + C_0 = C_0 = 60\,J$.The maximum potential energy occurs at the extreme positions $(x = \pm A)$,where $PE_{max} = \frac{1}{2} k A^2 + C_0 = 100\,J + 60\,J = 160\,J$.Thus,the maximum potential energy is $160\,J$.

$A$ linear harmonic oscillator of force constant $2 \times 10^6\,N/m$ and amplitude $0.01\,m$ has a total mechanical energy of $160\,J$. Which of the following statements is correct?

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