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Stable Equilibria

  1. Nov 13, 2005 #1
    Determine the location of the first postivve (x>0) stable equilibrium position for
    V(x) = = Vocos((2pie/lambda)x)

    eventually you will get sin(2pie/lambda)x) = 0

    Is a solution for x pie?
     
  2. jcsd
  3. Nov 13, 2005 #2

    Tide

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    No, that is not a solution.

    Also, when you find a correct equilibrium point, how will you know whether it is stable?
     
  4. Nov 13, 2005 #3
    If it's >0.

    So then what is a soln?
     
  5. Nov 13, 2005 #4

    Tide

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    For what values of x is this true:

    [tex]\sin \frac {2\pi x}{\lambda} = 0[/tex]

    Certainly, it's not just [itex]x = \pi[/itex].

    Also, the sign of the potential alone does not determine stability.
     
  6. Nov 13, 2005 #5
    If the second derivative is > 0 while evaluated at its equilibrium points does not determine stability then what does?

    Also, x = lambda/2 as the first positive eq. pt.

    But when does bounded motion occur?
     
    Last edited: Nov 13, 2005
  7. Nov 14, 2005 #6

    Tide

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    Regarding stability, I never suggested otherwise. That's the first time you have related the second derivative to stability. Go back and read what you wrote previously! :)

    Bounded motion occurs when there is a restoring force about the equilibrium point.
     
  8. Dec 15, 2005 #7
    What is the maximum kinetic energy that the particle can have and still remain bounded for all time?

    If the total energy is T + V and never changes, V is min when T is max. V is min at the stable eq. pts. so Vmin = Vocos(pie) = -Vo

    (V is potential energy)

    Is that right? How can I check my answer?

    If the total energy of the particle E = 0 what is the maximum distance that the particle can travel between its turning points assuming that the eq. pt is at x = lambda/2?

    When E = 0 Does this imply that V = -T?

    Or do I assume that E = 0 = V and thus x is lambda/4 but why?
     
  9. Dec 15, 2005 #8

    Tide

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    The maximum kinetic energy the particle can have without leaving the potential well is [itex]T_{max} = V_{max} - V_{min} = 2V_0[/itex]. This assures that the kinetic energy is 0 at the top of the potential curve (i.e. it cannot go beyond the maximum of potential because v = 0 there).
     
  10. Dec 16, 2005 #9
    Is Vmax = Vo ? and Vmax = Vocos((2pie/lambda)(lambda/2)) ? I thought the max kineitc energy would occur if V =0

    If the total energy of the particle E = 0 what is the maximum distance that the particle can travel between its turning points assuming that the eq. pt is at x = lambda/2?

    so the max distance requires the max kinietic energy and this E = 0 = T + V =>

    1/2 mV^2 = 2Vo + Vocos(2xpie/lambda) = 0

    solve for x?
     
    Last edited: Dec 16, 2005
  11. Dec 16, 2005 #10

    Tide

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    I think you're confusing this with a different problem you may have encountered. You need to realize that the "zero level" of potential or potential energy is arbitrary. What matters in terms of physical interaction is the force acting on a particle which can be expressed as a gradient or derivative of a potential. This means that in a particular situation an arbitrary constant can be added to the potential. Since the gradient of a constant is 0, adding such a constant contributes nothing to the physical forces acting on particles.

    For example, when you consider particles interacting via the Coulomb force or gravity, it is often convenient to set the potential equal to zero out at infinity. That's strictly a matter of convenience but doesn't affect the physical interaction between the particles. You could just as well have set the potential to some nonzero value at infinity with no effect on the actual interaction.

    In your particular problem, someone else has already made the decision on what the constant is in the expression for potential energy. They could just as well have made the potential energy [itex]-V_0 + V_0 \cos (2\pi x/\lambda)[/itex] to make the potential zero at the highest level. In that case, E = 0 would have the interpretation you're suggesting. However, they did not. The quantity you are interested in is the maximum potential and the difference between it and the lowest potential.

    I hope that helps.
     
  12. Dec 16, 2005 #11
    If the total energy of the particle E = 0 what is the maximum distance that the particle can travel between its turning points assuming that the eq. pt is at x = lambda/2?

    do I assume that E = 0 = T since the max imum distance is obtained when T = 0

    E = T + V
    0 = 0 + Vocos(2pie*x*lambda / lambda * 2 )

    thus x is lambda/4
     
  13. Dec 16, 2005 #12
    is it even reasonable to say that the maximum distance lamdba/4 is larger than the equilibirum point? lambda/2
     
  14. Dec 16, 2005 #13

    Tide

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    There are two values of x for which E = 0 when T = 0. You're looking for the separation between those two points.
     
  15. Dec 17, 2005 #14
    When the turning pts occur, E = V but E = 0

    so x is +/- lambda/4

    Now you say its the seperation between those pts so does this imply Vmax - Vmin? I thought Vmax was E.
     
  16. Dec 17, 2005 #15

    Tide

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    You're not paying attention! Your question was what is the distance between the turning points when E = 0? I told you there are two such turning points (about the first stable equilibrium, again as you specified in your original post). The separation between those two points is the separation between them!

    The first turning point occurs at [itex]x = \lambda/4[/itex] and the second occurs at [itex]3\lambda/4[/itex] if the potential you gave in your original post is correct.
     
  17. Dec 17, 2005 #16
    I see now, thanks
     
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