Question

A particle located in one dimensional potential field has potential energy function U(x) = `a/x^2 - b/x^3` where a and b are positive constants. The position of equilibrium corresponds to x =?

(1) `(3a)/b`

(2) `(2b)/(3a)`

(3) `(2a)/(3b)`

(4) `(3b)/(2a)`

Answer 1

(4) `x = (3b)/(2a)`

Explanation:

You are looking for a point where the force on the system goes to zero.

`F(x) = -d/dxU(x)`

`-(d)/(dx)(a/x^2 - b/x^3) = (-3 b + 2 a x)/x^4`

By inspection notice that the numerator will be zero when both of those terms are equal. One of the terms includes an `x`. Determine when:

`3b = 2ax`

`(3b)/(2a) = x`

The functional form of `F(x)` will look like this:
graph{(-3 + 2 x)/x^4 [-1, 5, -0.1, 0.2]}
However the particular magnitudes will change depending on the constants `a` and `b`. Here I've set them both to `1`. The equation crosses zero at a point where `x = 3/2`.

The graph of `U(x)` looks like this:
graph{1/x^2 - 1/x^3 [-1, 5, -0.1, 0.2]}