The potential energy function for a particle executing linear simple harmonic motion is given by V{x) = `kx^2//2`, wherek is the force constant of the oscillator. For `k = 0.5 N m^-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 =overset+- 2 m.

Options

A. V = 0, K = E
B. V = E, K = 0
C. V lt E, K = 0
D. V = 0, K lt E.