The potential energy function for a particle executing linear simple harmonic motion is given by `V(x)= kx^(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. 6. 12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches `x= pm 2m`.