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Help with particle in a potential

  1. Jul 20, 2011 #1
    Hello,

    I would like you to check this problem.

    1. The problem statement, all variables and given/known data

    A particle of mass m is moving in a potential V(x) = [itex]\frac{cx}{x^2+a^2}[/itex], where a, c > 0.

    1. Plot V(x)
    2. Find the position of stable equilibrium and the period (small oscillations)
    3. Find the values of velocity for which the particle: (a) oscillates, (b) escapes to [itex]\infty[/itex], (c) escapes to [itex]-\infty[/itex].

    2. Relevant equations

    V(x) = [itex]\frac{cx}{x^2+a^2}[/itex], a, c > 0.

    3. The attempt at a solution

    1. Well, I started by doing x = 0 to find the value of V in the origin.
    Next, x = a --> V(x) = [itex]\frac{c}{2a}[/itex], x = na (n>1) --> V(x) = [itex]\frac{nc}{a(n^2+1)}[/itex]. With c = 1 I can plot V(x) for x = a/2, a, 2a, 3a.
    But if I do x = -na, V(x) turns negative... I think V(x) should be positive along the entire x-axis, shouldn't it? Otherwise, how can the particle oscillate?

    2. I suppose the position of stable equilibrium is x = 0, isn't it?
    Period = ????

    3. Velocity. In order to the particle doesn't go to infinity (i.e. it stays in the region -a[itex] \leq[/itex] x [itex] \leq [/itex] a), its kinetic energy K needs to be [itex] \leq [/itex] V, right? Then, with K = [itex]\frac{mv^2}{2}[/itex] = V(x=a), (the maximum value of) v = [itex]\sqrt{ \frac{2V}{m} }= \frac{1}{\sqrt {ma}} [/itex] . For v > [itex]\frac{1} {\sqrt{ma}} [/itex] the particle escapes to infinity. But, what happens when x is negative?

    Thank you.
     
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 20, 2011 #2
    1) the way i'd think about sketching that potential is:

    i) its an odd function, thus V(-x) = - V(x). Or in words, the function for positive x will look look identical to that for negative x when reflected in x axis, then in V(x) axis.
    ii) At large x, the function can be approximated to c/x, thus is must tend to zero at both positive and negative infinity.
    iii) When x is small (less than 1), the x^2 term can be neglected, thus the function could be approximated to V(x) = (c/a)x about zero.

    Have you actually sketched this function? It would help you with the rest of the answer if you did because it is clearly not positive along the whole x-axis.

    2) A particle is in stable equilibrium when it experiences no *net* force. If you know F = - dV/dx then it should follow how to find the value of x for which it is in equilibrium. (Particles in stable equilibrium reside at the bottom of potential wells) By the way, it is not at x=0.

    To find the period of small oscillations you might want to think of a way to simplify the potential about the point of equilibrium. Heard of taylor or binomial expansion or something similar? Perhaps consider what happens when you displace the mass a small distance from equilibrium.

    Do those first then think about velocity.
     
    Last edited: Jul 21, 2011
  4. Jul 21, 2011 #3
    Not it is not. [itex]V(x) < 0[/itex] when [itex]x<0[/itex]

    Not it is NOT. Equilibrium is when the derivative of the potential is zero. Stable equilibrium is when in addition the second derivative is positive. If you plot V(x) as part 1 asked, it is where V(x) has a minimum and looks locally like a bowl. I can tell without plotting that V(x) has a minimum at some negative x, perhaps x = a.

    Since [itex]\frac{c x}{x^2 + a^2}[/itex] is an odd function, [itex]x=0[/itex] is NOT a stable equilibrium. Actually in this case it is not even an deflection point.

    Again, make a plot of V(x). In order to escape to [itex]-\infty[/itex], the kinet energy has to over come the depth of the bowl. In order to escape to [itex]\infty[/itex], it has too over come the hump on the positive x side.

    As for frequency of small oscillation. Suppose that you found that V(x) has a minimum at [itex]x=x_0[/itex]. Write [itex]x = x_0 + \epsilon[/itex] and do a series expansion of V(x) around [itex]x=x_0[/itex]. Keep up to the second order term, whose coefficient gives you an effective "spring constant" for this potential for small oscillation.
     
  5. Jul 21, 2011 #4
    Thank you very much, JesseC and mathfeel.
    Yes, I did it. But I had the idea of "V(x) positive", so I didn't see immediately what about -x...
    Is it a or a^2?
    OK.

    Ok... This was because thinking about it as a positive potential...
    Yes! I sketched V(x) again and it looks nearly like a "bowl" in x = -a!

    First attempt (wrong...):
    [PLAIN]http://img8.imageshack.us/img8/1402/vx1n.jpg [Broken]

    Second attempt (better?):
    [PLAIN]http://img23.imageshack.us/img23/6619/vx2r.jpg [Broken]

    Now, I will go on trying with the rest of this problem.
     
    Last edited by a moderator: May 5, 2017
  6. Jul 21, 2011 #5
    Sorry, what is [itex]\epsilon[/itex]?


    Oh...
    Is it [itex]\epsilon[/itex]?
     
    Last edited: Jul 21, 2011
  7. Jul 21, 2011 #6
    Yes technically it is a^2, but what is more important than the constants is the general shape of the function. Your second attempt at sketching it is good! :) Let us know if you have any more problems.
     
  8. Jul 21, 2011 #7
    Yes :D your hunch is correct!
     
  9. Jul 21, 2011 #8
    You are right! :)
    Yes, thank you again!!
    Great!!! :D
     
    Last edited: Jul 21, 2011
  10. Jul 21, 2011 #9
    You sketch looks like there is a "kink" at x = -a. This function should be smooth in all derivatives.

    Nearly looking like a "bowl" is the whole point here! Any stable equilibrium looks locally like a bowl. And you can expand the potential in series around any point [itex]x_0[/itex]:[tex]V(x) = V(x_0) + V^{\prime}(x_0) (x-x_0) + \frac{1}{2} V^{\prime\prime}(x_0)(x-x_0)^2 + \cdots[/tex]
    In particular, you can expand around the point of equilibrium so that some of these terms vanishes. Then [itex]V(x)[/itex] is nearly given by a parabola if you ignore all higher order terms. This is justified if [itex]\epsilon \doteq x - x_0[/itex] is small.

    Now, what other potential is given by a parabola? And what kinds of motion does that have? How do you compute the frequency in that case?
     
  11. Jul 21, 2011 #10
    Oops...
    I did it. I got:
    [itex]V(x) = -\frac{c}{2a}+\frac{1}{2} \frac{c}{2a^3}\epsilon^2 [/itex]

    Ok... Harmonic oscillator?

    [itex]T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2ma^3}{c}}[/itex]

    Now I understand why it's necessary to expand the potential in series :smile:

    Next step: velocity!
     
    Last edited: Jul 21, 2011
  12. Jul 22, 2011 #11
    That looks right to me so long as you've got the algebra for V([itex]\epsilon[/itex]) right. Have a stab at the velocity part.
     
  13. Jul 22, 2011 #12
    I hope the algebra and the derivatives are right...

    I have to find the values of v for 3 cases: the particle escapes [itex]\infty[/itex], the particle escapes to [itex]-\infty[/itex], the particle oscillates... I feel a little confused...

    1. [itex]\infty[/itex]
    The hump on the positive x side is on x = a, so kinetic energy K > V(x=a), ok?

    [itex]K = \frac{1}{2}mv^2 > \frac{c}{2a}, [/itex]
    [itex]mv^2 > \frac{c}{a}, [/itex]
    [itex]v > \sqrt{\frac{c}{am}}. [/itex]

    2. [itex]-\infty[/itex]
    The bowl is in x = -a, ---->K > V(x=-a), but I think it only isn't enough if the particle wants to leave the potential well... What else does it need to escape? K > 0, maybe?

    Sorry, I have to go right now... To be continued...
     
  14. Jul 22, 2011 #13
    Suppose the particle is sitting at the bottom of the "bowl", where would you define the "edges"? Are the edges the same height in either direction?

    Don't forget you can define the zero point of potential to be wherever you please, what is important is the change in potential :)
     
  15. Jul 30, 2011 #14
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