The potential energy function for a particle executing linear simple harmonic motion is given by
V(x)=kx2/2, where k is the force constant of the oscillator.
For k=0.5Nm−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=±2m.
![419813_18c97bdaadbd4e62ba2318385a06f40f.png](https://search-static.byjusweb.com/question-images/toppr_ext/questions/419813_18c97bdaadbd4e62ba2318385a06f40f.png)