• Support PF! Buy your school textbooks, materials and every day products Here!

Equation of motion for isotropic harmonic oscillator

  • Thread starter JulienB
  • Start date
  • #1
408
12

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.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
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##?
 
  • #3
408
12
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
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
You are asked to find the EoM, not to solve it.
 
  • #5
408
12
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.
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
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).
 
  • #7
408
12
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.
 
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
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}
 
  • #9
408
12
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.
 
  • #10
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
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.
 
  • #11
408
12
Mm can I then present it that way?

##r = \pm \int \frac{L}{\sqrt{2m(E - V_{eff})}} d\varphi##
 
  • #12
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
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.
 
  • #13
408
12
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.
 
  • #14
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
Is it a differential equation describing how r varies with phi?
 
  • #15
408
12
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##?
 
  • #16
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
6,454
  • #17
408
12
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:

Related Threads on Equation of motion for isotropic harmonic oscillator

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
5K
Replies
9
Views
9K
Replies
1
Views
2K
Replies
5
Views
8K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
7K
Replies
3
Views
1K
  • Last Post
3
Replies
57
Views
7K
Top