Small oscillations+normal modes of a horrible spring system

[fluidistic]

Small oscillations+normal modes of a "horrible spring system"

Homework Statement

Let 2 particles of mass m be like in the attached figure. All springs have a constant k and natural length l_0. The particles are only free to move horizontally. Calculate the characteristic frequencies of small oscillations and their corresponding normal modes.

Homework Equations

Lagrangian first... L=T-V.

The Attempt at a Solution

I've called x_1 the position of the upper particle and x_2 the one of the lower one, with respect to the left wall. I've called alpha the angle that makes the middle spring with respect to the horizontal.
The Lagrangian of the 2 masses are only worth T because they don't have any potential energy. The springs on the other hand only carry potential energy.
I reached $L=\underbrace {\frac{m}{2} (\dot x _1 ^2}_{\text{KE of particle 1}}+\dot x_2^2)+\underbrace {k(l_0-x_1)^2}_{\text{PE of the 2 upper springs }}+\underbrace {k(l_0-x_2)^2}_{\text{PE of the 2 lower springs }}+\underbrace {\frac{k}{2}(l^2+l^2 \cot ^2 \alpha -2l_0 \sqrt{l^2+l^2 \cot ^2 \alpha }+l_0 ^2)}_{\text{PE of middle spring}}$.
Is this ok so far? I hope it's right... almost been a torture for me to get till here!

 

[genericusrnme]

If you set $x_1 = x_2 = 0$ as the stationary point on the springs things would look a little neater :3

Your lagrangian looks correct as far as I can see though.
If you chose the nice neat coordinates I suggested you'll get rid of those nasty cots though, I'd suggest you do that first.

 

[fluidistic]

If you set $x_1 = x_2 = 0$ as the stationary point on the springs things would look a little neater :3

Your lagrangian looks correct as far as I can see though.
If you chose the nice neat coordinates I suggested you'll get rid of those nasty cots though, I'd suggest you do that first.

I see. I'd need to get rid of alpha eventually anyway to get only 2 generalized coordinates, right?
I get $L=\frac{m}{2} (\dot x _1 ^2 + \dot x_2 ^2 )-k(x_1 ^2 +x_2^2 )-\frac{k}{2} (|x_1|+|x_2|-l_0)^2$. Does this still looks ok?
By the way I notice I got sign errors for my previous Lagrangian. I wrote L=T+V instead of T-V.

 

[genericusrnme]

I thought it was a bit odd defining your k's as negative quantities :p

The potential energy for the middle spring is wrong though, if x1 and x2 are displaced the same amount from 0 there should be no change in the potential energy. You're going to want something that involves a term like $(x_1 - x_2)^2$.

I drew a picture to help you understand what I mean

 

[fluidistic]

Hmm I see what you mean.
So far I can't find anything better than the potential energy of middle spring $= \frac{k}{2} (|x_1-x_2|+l-l_0)^2$.

 

[genericusrnme]

Yeah, you're probably not going to get anything simpler than that I'm afraid

 

[fluidistic]

So this makes $L=\frac{m}{2} (\dot x _1 ^2 + \dot x_2 ^2 )-k(x_1 ^2 +x_2^2 )-\frac{k}{2} (|x_1-x_2|+l-l_0)^2$.
On the other hand if I take $L=\frac{m}{2} (\dot x _1 ^2 + \dot x_2 ^2 )-k(x_1 ^2 +x_2^2 )-\frac{k}{2}(l^2+l^2 \cot ^2 \alpha -2l_0 \sqrt{l^2+l^2 \cot ^2 \alpha }+l_0 ^2)$. Can I take $\cot \alpha \approx 0$ since I'm asked about small oscillations?
This would reduce the Lagrangian to $L=\frac{m}{2} (\dot x _1 ^2 + \dot x_2 ^2 )-k(x_1 ^2 +x_2^2 )-\frac{k}{2}(l^2-2l_0l+l^2)$.

Anyway would the next step be to approximate the total potential energy by a quadratic, i.e. $U(x)\approx \frac{k}{2}(x-x_0)^2$ which would be worth in my case $U(x)\approx \frac{k}{2}x^2$?
Or with my Lagrangian, I find the Euler-Lagrange equations and then I use the small oscillations conditions (i.e. neglecting second order terms)?
I never solved a 2 dimensional small oscillations problem up till now.

 

[sgd37]

you only need the position of one of the masses and the angle of the spring with it. this is similar to the pendulum attached to a moving mass problem just with added potentials and a variable pendulum length

 

[genericusrnme]

If you're going to set $Cot(\alpha) = 0$ then you can equally set $|x_1 - x_2|=0$ and you'll find that the problem reduces to two one dimensional oscillators with frequency $\sqrt{k}$.

What you want to do is expand your potential as a power series ;)
Then you'll end up with two seperable differential equations which you can solve and find your $\omega$'s

 

[fluidistic]

If you're going to set $Cot(\alpha) = 0$ then you can equally set $|x_1 - x_2|=0$ and you'll find that the problem reduces to two one dimensional oscillators with frequency $\sqrt{k}$.

What you want to do is expand your potential as a power series ;)
Then you'll end up with two seperable differential equations which you can solve and find your $\omega$'s

I see.
So basically I want to approximate $V(x_1)$ and $V(x_2)$ by using a Taylor expansion and keep only terms with order lesser or equal to 2?
Like for example $V(x_1)\approx V(0)+V'(0)x_1+\frac{V''(0)x_1^2}{2}$?
Edit: Probably not. I'm not getting 0 for the linear term... sigh.

 

[fluidistic]

I've Goldstein's book under my eyes. I must approximate $V(x_1, x_2)$ by $\frac{1}{2} \sum _{i,j} \left ( \frac{\partial ^2 V}{\partial q _i \partial q_j} \right )_0 \eta _i \eta _j=\frac{1}{2} \sum V_{ij} \eta _i \eta _j$.
I'm very confused by the notation and the summation. So I don't really understand how to approximate (or expand the potential and keep only the 3rd term since the first 2 terms are either worth 0 or can be taken as 0) a potential function when it contains more than 1 variable.
Can someone help me on this?

Edit: Is it $V(x_1, x_2)\approx \frac{1}{2} [x_1^2 V_{x_1x_1}(0,0)+2x_1x_2V_{x_1x_2}(0,0)+x_2^2V_{x_2x_2}(0,0)]$?
Edit2: If so, then I get $V(x_1, x_2) \approx \frac{k}{2}(x_1^2x_2+2x_1x_2^2+3x^2)$.
Next step is to find Euler-Lagrange equations? Or write T and V as matrices?

Edit3: This cannot be right...
I got Euler-Lagrange equations and they gave me the following system of DE's:
$m \ddot x_1+k(x_1x_2+x_2^2)=0$.
$m\ddot x_2++k(2x_1x_2+3x_2)=0$.
They aren't symmetric for $x_1$ and $x_2$ which isn't normal. Even though they don't ask for small oscillations, if I consider them the system reduces to
$m \ddot x_1 =0$ and $m \ddot x_2 +3kx_2^2=0$. Which is a total non sense.
I hope someone can help me.

 

[fluidistic]

Basically what I've done is just a 2 variables Taylor expansion around (0,0). For some reason this gave me a non sense result. Is the methodology wrong?

 

[genericusrnme]

Oh boy, sorry I haven't replied for so long, I've not been checking my thread subscriptions :p

What was the nonsense result you got?

 

[fluidistic]

Oh boy, sorry I haven't replied for so long, I've not been checking my thread subscriptions :p

What was the nonsense result you got?

No problem, I'm very lucky to have you (and others) to help me!
The non sense I'm getting is detailed in post #11.

 

[genericusrnme]

Ah..
I don't know then, I'm completely lost for ideas atm
Sorry buddy :\

 

[fluidistic]

Ah..
I don't know then, I'm completely lost for ideas atm
Sorry buddy :\

No problem :)
If someone else has an idea, feel free to post. Or if someone knows if the method is right (doing a 2 variables Taylor expansion to approximate $V(x_1,x_2)$) please say it. It would mean I likely made an algebra error somewhere.

 

[fluidistic]

I asked a professor, he said the proceedure is ok (Taylor expansion of order 2 for 2 variables in this case). So basically I made an algebra mistake somewhere.
I've redone everything and I think I solved the problem now.
I don't have much time thesedays to write all in latex :(
Anyway, once I have the approximated result for the potential ($V(x_1,x_2) \approx \frac{k}{2} \left ( 3- \frac{l_0}{l} \right ) (x_1^2+x_2^2)+k\left ( \frac{l_0}{l}-1 \right ) x_1x_2$) the next step is to write it under matricial form. Do the same for the kinetic energy expression, i.e. write it under matricial form. Then solve the characteristic equation $|V-\omega ^2 T|=0$.
I reached $\omega _1 = \sqrt {\frac{2k}{m}}$ and $\omega _2 = \sqrt {\frac{2k}{m} \left ( 1-\frac{l_0}{l} \right ) }$. I notice that if $l=l_0$, there's only one characteristic frequency. I am not sure this is possible (so I could have made another error(s)) but it is worth noticing it.