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How to find if equilibrium points of a force is un/stable?

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    U = Ax2 - Bx3

    2. Relevant equations
    du/dx = 2Ax - 3Bx2

    3. The attempt at a solution

    If I was given a potential energy function U = Ax2 - Bx3 and am asked to find:

    1) The expression for the force as a function of x.

    2) The equilibrium points and determine if are they stable or unstable?

    So, for 1):
    Would I differential the function giving like so?

    U' = f'(x) = 2Ax - 3Bx2

    Now for 2):
    Would I set f'(x) = 0 to find the equilibrium points?

    f'(x) = 2Ax - 3Bx2 = 0

    In return I get the points of x through the quadratic equation:
    x = 0 and X = A/B

    If this is all correct how can I determine if a equilibrium point is stable or unstable?
  2. jcsd
  3. Apr 7, 2016 #2


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    Homework Helper

    Recall the force is negative derivative of the potential energy.
    X=0 is correct, but you have a mistake in the other equilibrium point.

    The equilibrium is stable if the potential energy is minimum in that point and unstable if it is maximum.
  4. Apr 7, 2016 #3
    Oh sorry, x = 2A/(3B).
    How I find the max and min of potential energy?
  5. Apr 7, 2016 #4


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    Homework Helper

    You found the positions of the extremes, at x=0 and at x=2A/3B.
    Have you learnt what should be the second derivative at a maximum and at a minimum?
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