Question
The potential energy function for a particle executing linear simple harmonic motion is given by $V(x)=k x^2 / 2$, where $k$ is the force constant of the oscillator. For $k=0.5 Nm ^{-1}$, the graph of $V (x)$ versus $x$ is shown in Fig. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches $x= \pm 2 m$.
