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# 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.

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#### The potential energy of the 4gm massU = mV =$$4\times(200x^{2} + 150)$$ = $$800x^{2} + 600 \ ergs$$The force F acting on the particle is given byF = $$-dU/dx$$ = $$d/dx (800x^{2} + 600) dyne = -1600x$$Then the equation of motion of the particle is given bym $$d^{2}x/dt^{2}$$ = -1600x$$d^{2}x/dt^{2}$$ = -1600/4 x = -400xHence frequency of oscillation$$n = 1/2\pi$$ $$\sqrt{400}$$ = 10/$$\pi$$ = $$3.2\ sec^{-1}$$

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