Question

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 = 0.5 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 = ± 2 m.

Answer 1

Total energy of the particle, E = 1 J
Force constant, k = 0.5 N m^–1
Kinetic energy of the particle, K =1/2mv²

According to the conservation law:​

At the moment of ‘turn back’, velocity (and hence K) becomes zero.

1=1/2kx²

Hence, the particle turns back when it reaches x = ± 2 m.