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 meter)

Options

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

Answer

C

Explanation

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

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