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.

Question

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Answer 1

The potential energy of the 4gm mass

U = mV = $4 \times (200 x^{2} + 150)$ = $800 x^{2} + 600 e r g s$

The force F acting on the particle is given by

F = $- d U / d x$ = $d / d x (800 x^{2} + 600) d y n e = - 1600 x$

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

$n = 1 / 2 π$ $400$ = 10/ $π$ = $3.2 s e c^{- 1}$

Answer 2

The potential energy of the 4gm mass

U = mV =

4 \times (200 x^{2} + 150)

=

800 x^{2} + 600 e r g s

The force F acting on the particle is given by

F =

- d U / d x

=

d / d x (800 x^{2} + 600) d y n e = - 1600 x

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 π

400

= 10/

π

=

3.2 s e c^{- 1}