__Question 6.4__:*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.

__Solution__:Total energy of the particle,

*E*= 1 J

*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) X 0.5x

^{2}= 1

x

^{2}= 4

x = ±2

Hence, the particle turns back when it reaches

*x*= ± 2 m.

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