A particle which is constant to move along the `x- axis` , is subjected to a force in the same direction which varies with the distance `x` of the particle from the origin as `F(x) = -Kx + ax^(3)` . Hero `K` and `a` are positive constant . For `x ge 0`, the fanctional from of the patential every `U(x) of the particle is

Options

A.
B.
C.
D.

Answer

D

Explanation

`F(x) = -kx + ax^2 implies F = - (dU)/(dx)`
`int dU = -int Fdx " " implies int dU = - int (-kx + ax^2)dx`
`U = -(-(kx^2)/2 + (ax^3)/(3)) implies u = (kx^2)/2`
`-(ax^3)/3`
Since `(dU)/(dx) = kx - ax^2` So For U to be maximum or minimum `(dU)/(dx) = 0,`
So `x (k - ax) = 0`
`implies x = 0 , x = k//a`
`(d^2U)/(dx^2) = k - 2ax`
(I) at `x = 0, (d^2 U)/(dx^2) lt 0`, So, `U_("min") = 0`
(ii) at `x = k//a`
`implies (d^2 U)/(dx^2) = k - 2a k/a implies (d^2 U)/(dx^2) = - k`
`U_("max") = k/2 (k/a)^2 - a/3 (k/a)^3 implies U_("max")`
`= (k^3)/(2a^2) - (k^3)/(3a^2) implies U_("max") = (k^3)/(6a^2)`
again for `u = 0, (kx^2)/(2) - (ax^3)/3 = 0`
`x^2 (k/2 - (ax)/3) = 0 implies x = 0, x = (3k)/(2a)`