A particle moving on x - axis has potential energy `U = 2 - 20x + 5x^(2)` joule along x - axis. The particle is relesed at `x = -3`. The maximum value of `x` will be (`x` is in metre)

Options

A. 5 m
B. 3 m
C. 7 m
D. 8 m

Answer

C

Explanation

`U = 2 -20x +5x_(2)`
`F = (dU)/(dx) = 20-10x`
At equilibrium position `F = 0`
`20 - 10 x =0`
`rArr x = 2`
Since particle is released
at `x = -3`, therefore amplitude of particle is 5.

It will oscillate about x = 2 with an amplitude of 5
`:.` maximum value of x will be 7.