The potential energy function for a particle executing linear simple harmonic motion is given by $U{(}_x)=\frac{1}{2}k{x}^2$, where $k$ is the force constant. For $k=0.5N{m}^{-1}$,the graph of $U(x)$  versus $x$ is shown in figure. Show that a particle of total energy $1J$ moving under this potential ${}^{\prime }turnsbac{k}^{\prime }$ when it reaches $x=\pm 2m$.