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Potential energy and equilibrium

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]


    2. Relevant equations

    dU/dx = 0

    d^2U/dx^2 > 0 = minimum (stable equilibrium)

    d^2U/dx^2 < 0 = maximum (unstable equilibrium)


    3. The attempt at a solution

    I think i've found the solution. I just want someone to see if i'm correct or not:

    dU/dx = Uo [(3/Xo^3)*X^2 + (2a/Xo^2)*X + (4/Xo)] = 0 --->

    I then used the quadratic formula to find the roots using --->
    a = (3/Xo^3)
    b = (2a/Xo^2)
    c = (4/Xo) ----> and ended up with two solutions for when X=0 --->

    X1 = Xo(-2a + [tex]\sqrt{4a^2 - 48}[/tex] / 6) ---> and --->

    X2 = Xo(-2a - [tex]\sqrt{4a^2 - 48}[/tex] / 6)

    I then found d^2U/dx^2 = --->

    Uo [(6/Xo^3)*X + (2a/Xo^2)] ---> I then plugged in the different X values -->

    For X1 =
    Uo * [(6/Xo^3) * Xo(-2a + [tex]\sqrt{4a^2 - 48}[/tex] / 6] + 2a/Xo^2 --->

    Uo * [(-2a + [tex]\sqrt{4a^2 - 48}[/tex]) / Xo^2 + 2a/Xo^2] --->

    Uo * ([tex]\sqrt{4a^2 - 48}[/tex]/Xo^2)

    For X2 =

    The opposite sign which =

    Uo * (-[tex]\sqrt{4a^2 - 48}[/tex]/Xo^2)

    Therefore -->

    -[tex]\sqrt{12}[/tex] > a > [tex]\sqrt{12}[/tex]

    where X1 computed in d^2U/dx^2 > 0

    and X2 computer in d^2U/dx^2 < 0

    Therefore, for the two equilibria, X1 is in stable equilibrium

    and X2 is in unstable equilibrium
     
  2. jcsd
  3. Oct 28, 2007 #2
    hmmm, not sure if this is correct
     
  4. Oct 28, 2007 #3
    nothing but crickets
     
  5. Oct 31, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Oops. I thought I had responded, but obviously only in my mind.

    Your solution is perfect. The only thing I would change is this:

    I'd write it as:
    [tex]a^2 > 12[/tex]

    Or:
    [tex]a \in (-\infty, -\sqrt{12}) \cup (\sqrt{12}, +\infty) [/tex]
     
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