#### Question

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

#### Solution 1

Here, force constant k = 0.5 Nm^{-1} and total energy of particle E = 1J. The particle can go up to a maximum distance x_{m}, where its total energy is transformed into elastic potential energy.

`1/2 kx_m^2 =E`

`=> x_m =sqrt((2E)/K) = sqrt((2xx1)/(0.5)) = sqrt4 = +- 2m`

#### Solution 2

Total energy of the particle, *E* = 1 J

Force constant, *k* = 0.5 N m^{–1}

Kinetic energy of the particle, K = `1/2mv^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 = 1/2kx^2`

`1/2xx0.5x^2 = 1`

`x^2 = 4`

`x =+-2`

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