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.