A linear harmonic oscillator of force constant $2 \times 10^6\,Nm^{-1}$ and amplitude $0.01\, m$ has a total mechanical energy of $160\, J.$ Its
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$A=0.01 m$

$k=2 \times 10^{6} N / m$

$E_{T}=160 J$

The total mechanical energy,

$E_{T}=(K . E)_{\max }+(P . E)_{\min }=160 J \ldots \ldots(1)$

we know that, maximum kinetic energy,

$(K . E)_{\max }=\frac{1}{2} k A^{2} \frac{1}{2} \times 2 \times 10^{6} \times 10^{-6}$

$(K . E)_{\max }=100 J$

from equation $(1),$

$(P . E)_{\min }=160-100=60 J$

The minimum kinetic energy is zero so, the maximum potential energy is $160 \mathrm{J}$

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