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Updated on : 2022-09-05

Solution

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The potential energy of the 4gm mass

U = mV =$4×(200x_{2}+150)$ = $800x_{2}+600ergs$

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

$n=1/2π$ $400 $ = 10/$π$ = $3.2sec_{−1}$

Solve any question of Oscillations with:-

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