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Classical Mechanics: Finding force, equilibrium points, turning points

  1. Jan 21, 2012 #1
    Classical Mechanics: Finding force, equilibrium points, turning points....

    1. The problem statement, all variables and given/known data
    The potential energy between two atoms in a molecule is
    U(x) = −1/x^6 +1/x^12
    Assume that one of the atoms is very heavy and remains at the origin at rest, and the
    other (m = 1) is much less massive and moves only in the x-direction.
    (a) Find the force F(x).
    (b) Find the equilibrium point x0 and check stability. Give a numerical value for x0
    (c) If the system has a total energy E = −0.2, find the turning points, and the period
    of oscillation.
    (d) If the total energy was E = +0.2, describe the motion of the system.
    (e) Find the period for small oscillations around equilibrium. Can the energy in part
    (c) be considered “small” in this context?


    2. Relevant equations

    F(x)=-dU/dx (not sure though)
    E=K+U


    3. The attempt at a solution

    a)
    dU/dx = -6/x^7 +12/x^13

    b) Set dU/dx to zero and I get:
    6x^6-12=0
    x=6√2 (I said the negative part is ignored since we are only interested in positive x)
    Stability: second derivative of U evaluated at equilibrium point.
    d2U/dx2=-42/x^8+156/x^14 = 12.699, so stable

    c)
    E=K+U=1/2mv^2-1/x^6+1/x^12
    v^2=(2x^6-2-0.4x^12)/x^12
    turning points when v=0,
    5x^6-x^12-5=0

    Now I'm stuck...
     
  2. jcsd
  3. Jan 21, 2012 #2

    ehild

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    Re: Classical Mechanics: Finding force, equilibrium points, turning points....

    You completely ignored the given energy, E=-0.2.

    Edit: The formula is correct, I was wrong.

    ehild
     
    Last edited: Jan 22, 2012
  4. Jan 21, 2012 #3
    Re: Classical Mechanics: Finding force, equilibrium points, turning points....

    Sorry, I skipped a few steps, but I simplified the expression for v with E=-0.2 and I still get the same answer. I believe the answer is complex, I do not know how to deal with this.
     
  5. Jan 22, 2012 #4

    ehild

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    Re: Classical Mechanics: Finding force, equilibrium points, turning points....

    I checked it again, and I see now that your equation 5x^6-x^12-5=0 is correct. Using the quadratic formula, x^6=(5±√ 5)/2. The solutions are real. Take the sixth (real) root of each to get the turning points.

    ehild
     
  6. Jan 22, 2012 #5
    Re: Classical Mechanics: Finding force, equilibrium points, turning points....

    ok thanks! Can you show me how you simplified the expression?
     
  7. Jan 22, 2012 #6

    ehild

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    Re: Classical Mechanics: Finding force, equilibrium points, turning points....

    Denote y=x^6. Your equation transforms to the quadratic -y^2-5y-5=0. Solve with the quadratic formula.

    ehild
     
  8. Jan 23, 2012 #7
    Re: Classical Mechanics: Finding force, equilibrium points, turning points....

    Easy enough, thanks for all the help!
     
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