Question

The potential energy function for a particle executing S.H.M. is given by V(x)=0.25x. Show that a particle of total energy 1J moving under this potential must turn back when it reaches x=+ 2m.

Answer 1

Actually the potential energy as a function in the displacement x is,

$\displaystyle\longrightarrow\sf{V(x)=0.25x^2}$

At x = ±2 m,

$\displaystyle\longrightarrow\sf{V(x)=0.25(\pm2)^2}$

$\displaystyle\longrightarrow\sf{V(x)=0.25\times 4}$

$\displaystyle\longrightarrow\sf{V(x)=1\ J}$

This means the particle utilised its whole energy to maintain its potential energy at these points, since the total energy of the particle is 1 J.

By total mechanical energy conservation we get that the kinetic energy of the particle is zero at this moment.

By total mechanical energy conservation, we have,

$\displaystyle\longrightarrow\sf{E=V(x)+K(x)}$

At $\displaystyle\sf {x=\pm2,}$

$\displaystyle\longrightarrow\sf{E=V(\pm2)+K(\pm2)}$

$\displaystyle\longrightarrow\sf{1=1+K(\pm2)}$

$\displaystyle\longrightarrow\sf{K(\pm2)=0\ J}$

This means the velocity of the particle becomes zero here, and hence the particle will not further move forward. Thus it must turn backward since it is under SHM.

QED