# 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.

The potential energy of the 4gm mass

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

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 $$d^{2}x/dt^{2}$$ = -1600x

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

Hence frequency of oscillation

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