# Equation of motion for isotropic harmonic oscillator

• JulienB
In summary, the problem provides a mass point under the influence of a central force and asks for the equation of motion in terms of angular momentum. The potential and equations of motion are determined, and the total energy equation is manipulated to eliminate time derivatives. It is then shown that for a circle, the radial acceleration must be zero, leading to the conclusion that the total energy must be equal to the effective potential.

## 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?

Julien.

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##?

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

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

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.

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).

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.

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

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.

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.

Mm can I then present it that way?

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

JulienB said:
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.

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.

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

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##?

JulienB said:
Nope! :) Then it is probably:
Yes it is...

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.

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