The potential energy function for a particle executing linear simple harmonic motion is given by $U(_{x})=21 kx_{2}$, where $k$ is the force constant. For $k=0.5Nm_{−1}$,the graph of $U(x)$ versus $x$ is shown in figure. Show that a particle of total energy $1J$ moving under this potential $_{′}turnsback_{′}$ when it reaches $x=±2m$.