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Equation of motion for isotropic harmonic oscillator

  1. Jun 4, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm a bit stuck in this problem, hopefully someone can help me to make progress there:

    A mass point ##m## is under the influence of a central force ##\vec{F} = - k \cdot \vec{x}## with ##x > 0##.
    a) Determine the equation of motion ##r = r(\varphi)## for the angular momentum ##|\vec{L}| \neq 0##.
    b) For which value of ##E## do we have a circle?

    2. Relevant equations

    Lagrange equations, ##E = T+V##, ##V_{eff} = V(\vec{r}) + \frac{L^2}{2mr^2}##

    3. The attempt at a solution

    Okay so first I determined the potential:

    ##V(\vec{r}) = \int_{|\vec{r}|}^{0} -kr dr = \frac{1}{2} k r^2##

    Then I found the equations of motion

    ##\ddot{r} = r\dot{\varphi}^2 - \frac{k}{m}r## and
    ##\ddot{\varphi} = \frac{2}{r} \dot{r} \dot{\varphi}##

    and I can deduce

    ##\frac{\partial V}{\partial t} = 0 \implies E = const.## and
    ##\frac{\partial L}{\partial \varphi} = 0 \implies \frac{d}{dt} \frac{\partial L}{\partial \dot{\varphi}} = 0 \implies \frac{\partial L}{\partial \dot{\varphi}} = mr^2 \dot{\varphi} = L = const.##

    so I can rewrite ##\dot{r}## as

    ##\dot{r} = \frac{dr}{dt} = \frac{dr}{d\varphi} \dot{\varphi} = \frac{dr}{d\varphi} \frac{L}{mr^2}##

    which I can substitute in my ##E##:

    ##E = \frac{1}{2} \frac{L^2}{mr^4} \bigg(\frac{dr}{d\varphi}\bigg)^2 + V_{eff}##

    and after a few manipulations I get:

    ##\varphi = \pm \frac{L}{\sqrt{2m}} \int \frac{dr}{r^2 \sqrt{E - V_{eff}}}##

    And that's where I get stuck. ##E## and ##L## are constants but surely ##V_{eff}## is not. In the Kepler problem we substitute with ##u = \frac{1}{r}## but if I am not wrong it is not working in the case of a simple oscillator. Any suggestion?


    Thanks in advance for your answers, I appreciate it!


    Julien.
     
  2. jcsd
  3. Jun 4, 2016 #2

    Orodruin

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    I suggest that you instead look at the equation for the total energy, which as you say has to be constant. What happens when ##\dot r = 0##?
     
  4. Jun 4, 2016 #3
    Thanks a lot for your answer. When ##\dot{r} = 0## then ##E = V_{eff}## I would assume. But isn't that the answer to question b? If I should use that for a, I don't see how for the moment.


    Julien
     
  5. Jun 4, 2016 #4

    Orodruin

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    You are asked to find the EoM, not to solve it.
     
  6. Jun 5, 2016 #5
    Hi @Orodruin and thank you for your answer. Yes but I am asked for the equation of motion in function of ##\varphi##, and I am not sure to understand what you suggest with ##\dot{r} = 0##, because that happens only at the farthest and closest points of the elliptic path to the center right? Well unless it is a circle, then ##\dot{r} = 0## at every point and ##\dot{r} = const##.

    I think I am missing something in my understanding of the problem.


    Julien.
     
  7. Jun 5, 2016 #6

    Orodruin

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    The energy equation you have is the equation of motion for r as a function of phi, there is no time in it. The effective potential depends only on r.

    The ##\dot r=0## applies to (b).
     
  8. Jun 5, 2016 #7
    Wait, really? I don't think I'm supposed to have ##\dot{r}## or ##\dot{\varphi}## in my equation though or? I can rewrite ##mr^2\dot{\varphi}^2## with L, but one ##\dot{\varphi}## remains and that doesn't resolve the issue with ##\dot{r}##.

    EDIT: btw my ##E## is

    ##E = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\varphi}^2) + \frac{1}{2} kr^2##.

    Maybe there is a mistake there?

    Thanks a lot for all your help.


    Julien.
     
  9. Jun 5, 2016 #8

    Orodruin

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    Indeed you should not have time derivatives in your EoM, but you have already gotten rid of them ...
     
  10. Jun 5, 2016 #9
    I have so little self confidence that I thought everything was wrong :D like always when you get the solution, it all makes sense now. ;)

    Okay that's what I get now:

    ##dr^2 = (E - V_{eff}) \frac{2 m r^4}{L^2} d\varphi^2 ##
    ##dr = \pm \frac{r^2}{L} \sqrt{2m(E - V_{eff})} d\varphi##
    ##\frac{1}{r} =\pm \frac{1}{L} \sqrt{2m(E - V_{eff})} \int d\varphi##
    ##r = \pm \frac{L \varphi}{\sqrt{2m(E - V_{eff})}}##

    Is that correct now? The ##\pm## is accessory I guess as ##r## is a distance. Thanks a lot.

    Julien.
     
  11. Jun 5, 2016 #10

    Orodruin

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    You cannot integrate it that easily, r depends on phi! Anyway, you do not need to. Only the EoM was sought, not the solution to the EoM.
     
  12. Jun 5, 2016 #11
    Mm can I then present it that way?

    ##r = \pm \int \frac{L}{\sqrt{2m(E - V_{eff})}} d\varphi##
     
  13. Jun 5, 2016 #12

    Orodruin

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    That is formally the solution to the EoM, not the EoM. I do not understand why you insist on solving the EoM - the problem just asks you to find the EoM.
     
  14. Jun 5, 2016 #13
    Sorry, I'm not trying to solve it, I'm just not sure how to present an equation of motion in that context. May I reveal my ignorance by directly asking which form would be considered the equation of motion there? Simply ##E = \frac{1}{2} \frac{L^2}{mr^4} \bigg( \frac{dr}{d\varphi}\bigg)^2 + V_{eff}## ?

    Thanks a lot for your patience.

    Julien.
     
  15. Jun 5, 2016 #14

    Orodruin

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    Is it a differential equation describing how r varies with phi?
     
  16. Jun 5, 2016 #15
    Nope! :) Then it is probably:

    ##\frac{d}{d\varphi} \frac{1}{r} = \pm \frac{1}{L} \sqrt{2m(E - V_{eff})}##

    Maybe I shouldn't have moved the ##r^2## to the left side? If it is correct, what is the policy about the ##\pm##? Should I just drop it because it is a distance and ##E - V_{eff} > 0##?
     
  17. Jun 5, 2016 #16

    Orodruin

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    Yes it is...
     
  18. Jun 5, 2016 #17
    Okay, thanks a lot and sorry for struggling to get it. For b I'd say that for the path to be a circle, the radial acceleration must be zero hence ##\dot{r} = 0##, which means ##E = V_{eff}##.

    Hopefully this is right, thanks a lot for helping me.


    Julien.
     
    Last edited: Jun 5, 2016
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