Solution:
Max K.E. = $\frac{1}{2} k a^2 = \frac{1}{2} \times 2 \times 10^6 \times (0.01)^2 = 100 \text{ J}$. Total Energy = $160 \text{ J}$. Min P.E. = $160 - 100 = 60 \text{ J}$. Max P.E. = Total Energy = $160 \text{ J}$.
A linear harmonic oscillator of force constant $2 \times 10^6 \text{ N/m}$ and amplitude $0.01 \text{ m}$ has a total mechanical energy of $160 \text{ J}$. Its (1996)
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