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(D) The total mechanical energy of the oscillator is given as $E = 160 \, J$. For a linear harmonic oscillator,the energy associated with the harmonic

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Both $(b)$ and $(c)$

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(D) The total mechanical energy of the oscillator is given as $E = 160 \, J$. For a linear harmonic oscillator,the energy associated with the harmonic oscillation (the kinetic energy part) is given by $E_{osc} = \frac{1}{2} k A^2$. Substituting the given values: $E_{osc} = \frac{1}{2} 	imes (2 	imes 10^6) 	imes (0.01)^2 = 10^6 	imes 10^{-4} = 100 \, J$. This $100 \, J$ represents the maximum kinetic energy $(K_{max})$ of the oscillator. Since the total energy is $160 \, J$ and the oscillatory part is $100 \, J$,there must be an additional constant potential energy $U_0 = 160 - 100 = 60 \, J$. The potential energy varies as $U(x) = U_0 + \frac{1}{2} k x^2$. The maximum potential energy occurs at the extreme positions $(x = \pm A)$,which is $U_{max} = U_0 + \frac{1}{2} k A^2 = 60 + 100 = 160 \, J$. 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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