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)#

1 Answer
Sep 7, 2017

(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]}