Q.4:- 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 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:-**Total energy of the particle, *E* = 1 J

Force constant, *k* = 0.5 N m^{–1}

Kinetic energy of the particle, K = (1/2)mv^{2}

According to the conservation law:

*E* = *V* + *K*

1 = (1/2)*kx*^{2} + (1/2)*mv*^{2}

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

∴ 1 = (1/2)*kx*^{2}

(1/2) × 0.5*x*^{2} = 1

*x*^{2} = 4

*x* = ±2

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