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.

Question

Philoid · Accepted Answer

Given,

Total energy of the particle, E = 1 J

Force constant, k = 0.5 N m^-1

The total energy is equal to the sum of kinetic energy and potential energy.

So, E = PE + K

⇒ E = kx² + mv²

So, 1 = kx² + mv²

When the velocity of the particle is zero i.e., at the turning point, the kinetic energy is zero.

∴ 1 = kx²

⇒ x² = 2/k

⇒ x² = 2/0.5

⇒ x² = 4

⇒ x = �2

Hence, the particle turns back at x= �2.