Equation of motion for isotropic harmonic oscillator

[JulienB]

Homework Statement

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?

Homework Equations

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

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.

 

[Orodruin]

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##?

 

[JulienB]

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

 

[Orodruin]

You are asked to find the EoM, not to solve it.

 

[JulienB]

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.

 

[Orodruin]

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).

 

[JulienB]

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.

 

[Orodruin]

Indeed you should not have time derivatives in your EoM, but you have already gotten rid of them ...

which I can substitute in my EEE:

E=12L2mr4(drdφ)2+VeffE=12L2mr4(drdφ)2+VeffE = \frac{1}{2} \frac{L^2}{mr^4} \bigg(\frac{dr}{d\varphi}\bigg)^2 + V_{eff}

 

[JulienB]

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.

 

[Orodruin]

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.

 

[JulienB]

Mm can I then present it that way?

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

 

[Orodruin]

Mm can I then present it that way?

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

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.

 

[JulienB]

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.

 

[Orodruin]

Is it a differential equation describing how r varies with phi?

 

[JulienB]

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##?

 

[Orodruin]

Nope! :) Then it is probably:

Yes it is...

 

[JulienB]

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.