# 8.3.3 SHM Energy in the x-domain

From $\displaystyle v=\pm \omega \sqrt{{{{x}_{0}}^{2}-{{x}^{2}}}}$ , we can write $\displaystyle KE=\frac{1}{2}m{{v}^{2}}$ as

$\displaystyle \displaystyle KE=\frac{1}{2}m{{\omega }^{2}}({{x}_{0}}^{2}-{{x}^{2}})$

We can then write PE as

\displaystyle \displaystyle \begin{aligned}PE&=TE-KE\\&=\frac{1}{2}m{{\omega }^{2}}{{x}_{0}}^{2}-\frac{1}{2}m{{\omega }^{2}}({{x}_{0}}^{2}-{{x}^{2}})\\&=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\end{aligned}

A few things to note.

1. Both KE and PE are quadratic graphs and are mirror image of each other
2. The graphs intersect at $\displaystyle E=\frac{1}{2}TE$ , $\displaystyle x=\pm \frac{{{{x}_{0}}}}{{\sqrt{2}}}$.

Video Explanation

Energy-Displacement Graphs of SHM

Concept Test

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