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Potential energy graphing

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data

    The potential energy V(R) of a two particle system exhibiting oscillatory behavior near a local minimum at the equilibrium separation Ro. V(R)= -(A/R)+(B/R^2) , where R is the interparticle separation.

    A) Sketch V(R), what happens to V(R) as R→0
    B) At what value of R is there a minima in the potential?
    C) For very small oscillations about this equilibrium point, Ro, write the force on the particle F=-k(R-Ro), define k.

    2. Relevant equations
    V(R)= -(A/R)+(B/R^2)

    3. The attempt at a solution

    I must apologize in advance because I feel that I'm about to ask a stupid question. But how can I plot this function if I don't know what the values of A and B are?
     
  2. jcsd
  3. Feb 7, 2017 #2

    TSny

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    It's a good question.

    What if you plot ##V## in units of ##A^2/B## and ##R## in units of ##B/A##?

    [Edited to correct a mistake.]
     
    Last edited: Feb 7, 2017
  4. Feb 7, 2017 #3

    PeroK

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    I would have just taken ##A = B = 1## for the purpose of a sketch.
     
  5. Feb 7, 2017 #4
    Sorry, I'm not sure how I would plot the function in those terms such as A^2/B and B/A. Do you think I can fix the values like PeroK suggested? My only concern is that the function slightly changes based on what A and B are. But I see that the function runs off to infinity as R approaches 0.
     
  6. Feb 7, 2017 #5

    TSny

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    Yes, you can go with PeroK's suggestion. The graph will be the same as I was suggesting.

    Note that ##A^2/B## has the dimension of energy and ##B/A## has the dimension of length.

    So, you can introduce dimensionless quantities ##\tilde{V} = \frac{V}{A^2/B}## and ##\tilde{R} = \frac{R}{B/A}##.

    If you write the equation in terms of these dimensionless quantities ##\tilde{V}## and ##\tilde{R}##, you should find that ##A## and ##B## disappear.

    Graphing ##\tilde{V}## versus ##\tilde{R}## will give the same graph as graphing ##V## versus ##R## with ##A= B = 1##.
     
  7. Feb 7, 2017 #6
    Thanks for your explanation. I wasn't used to this method so I couldn't see the relationship before. For part B where it asks to find the minimum, taking A=B=1, I found an equilibrium point at R=1. Would it be correct for me to use the first derivative test?
     
  8. Feb 7, 2017 #7

    TSny

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    For A = B = 1, I don't think the equilibrium is at R = 1. But, I'm pretty sure that you are meant to express the equilibrium value of R in terms of A and B.

    Yes, use differentiation to find the minimum.
     
  9. Feb 7, 2017 #8
    So then the equilibrium value of R is at R=(B/A)? I found this by leaving A and B as variables and setting V(R)=0.
     
  10. Feb 7, 2017 #9

    TSny

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    The equilibrium condition is not V(R) = 0. Equilibrium is where the force is zero.
     
  11. Feb 7, 2017 #10
    Since V(R) is given as the potential energy function in the problem, I can find my force function by using F(x)=-du/dx. Is that the right idea?
     
  12. Feb 7, 2017 #11

    TSny

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    Yes. Good.
     
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