$k=2 \times 10^{6} N / m$
$E_{T}=160 J$
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 $160 \mathrm{J}$


| $A (mm \,\,s^{-2}$) |
$16$ |
$8$ |
$0$ |
$- 8$ |
$- 16$ |
|
$x\;(mm)$ |
$- 4$ |
$- 2$ |
$0$ |
$2$ |
$4$ |