Question

A particle of mass 4 gm. lies in a potential field given by V=200$x^2$ + 150 ergs/gm. Deduce the frequency of vibration.

Answer 1

The potential energy of the 4gm mass

U = mV = 4 $(200x^2+150)$ = $800x^2$ + 600 ergs

The force F acting on the particle is given by

F = -dU/dx = d/dx ($800x^2$ + 600) dyne = -1600x

Then the equation of motion of the particle is given by

m $d^{2}x/d{t}^2$ = -1600x

$d^{2}x/d{t}^2$ = -1600/4 x = -400x

Hence frequency of oscillation

n = 1/2$\pi$ $\sqrt{400}$ = 10/$\pi$ = 3.2 $se{c}^{-1}$