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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