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(Small oscillations) Finding Normal modes procedure.

by Lavabug
Tags: modes, normal, oscillations, procedure
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Lavabug
#1
May15-11, 09:30 AM
P: 899
1. The problem statement, all variables and given/known data
The first part of the problem is just finding the Lagrangian for a system with 2 d.o.f. and using small angle approximations to get the Lagrangian in canonical/quadratic form, not a problem. I am given numerical values for mass, spring constants, etc. and am told to find the normal modes of oscillation and the normal coordinates.
2. Relevant equations



3. The attempt at a solution

I find the eigenvalues (eigenfreq, [tex]\lambda = \omega^{2}[/tex]) by diagonalizing: |V-[tex]\lambda[/tex]T| = 0. (V and T are matrices, also called M and K, they're the matrix of coefficients of the velocities and coordinates from the canonical Lagrangian respectively)

I get 2 ugly numbers for my eigenfrequencies, approximately 0.3 and 15.7.

Now I sub the first one into V-[tex]\lambda[/tex]T * (a11, a21) = 0
to solve for the eigenvector (a11, a21) which are the amplitudes of oscillation of both of my 2 d.o.f.(1 and 2) for this normal mode (1).

Question: plugging them in gives me a system of 2 equations of 2 variables. I am told that necessarily, one of the equations is a multiple of the other, but it doesn't seem to be true (using approximate values might have to do with this but I'm not sure). Is this correct?

If I believe what I am told, I pick one of the equations (the end result depends on which one I pick!) and solve for a11 as a function of a21, then use the orthonormalization condition: (a11,a21)(transposed)*T*(a11,a21) = 1 which allows me to find one of the amplitudes, which I then use to go back and solve for the other.

I am also very worried about approximating everything, at this stage of the problem I've already approximated decimal points on at least 3 occasions.

Can someone PLEASE help clarify this to me? Am I on the right track?
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grey_earl
#2
May15-11, 09:34 AM
P: 170
You are on the right track, but don't plug in numbers until the very end. For a 2x2 matrix, you can do everything analytically. If you are stuck, show some more of your calculations :)
Lavabug
#3
May15-11, 10:05 AM
P: 899
Thanks. The problem is my eigenvalues are generally irrational numbers so in order to not drag on mistakes from finding the roots, I just approximate the eigenvalues which is what my prof does and suggests doing even for 2 d.o.f. problems.

I have a problem with the orthonormalization process, I'm never quite sure if I'm multiplying the A(transposed)TA vectors properly. Do I first calculate A(transposed)*T, giving me a column vector, which I then multiply like a scalar product with the last term A(not transposed)?

grey_earl
#4
May15-11, 11:42 AM
P: 170
(Small oscillations) Finding Normal modes procedure.

You must have a prof with a technical background :)
The order in which you do the multiplications doesn't matter. I prefer to first calculate T*a, which gives another vector and then take the scalar product with a transposed, but that's just because I find multiplying with a matrix from the left easier than from the right.
Lavabug
#5
May15-11, 11:51 AM
P: 899
Quote Quote by grey_earl View Post
You must have a prof with a technical background :)
The order in which you do the multiplications doesn't matter. I prefer to first calculate T*a, which gives another vector and then take the scalar product with a transposed, but that's just because I find multiplying with a matrix from the left easier than from the right.
Aha so its valid both ways, I can just use the "scissor fingers" rule. :)

The prof is actually a theoretician. :P

To get the normal coordinates, once I have the general solution both coordinates q (linear combo of normal modes), I multiply each by the inverse of the A matrix? Is this the whole matrix or just the corresponding eigenvector?
grey_earl
#6
May15-11, 12:00 PM
P: 170
That's the whole matrix. You can use your normalization equation to simplify things:
[tex]\vec{N} = A^{-1} \vec{q} = A^{\mathrm{T}} T \vec{q}[/tex] since [tex]A^{\mathrm{T}} T A = 1[/tex].
Lavabug
#7
May15-11, 12:11 PM
P: 899
That 3rd expression you've got there... Didn't think of that. :P I'd much rather transpose and multiply 2 square matrices with a vector than invert a matrix any day! Thanks, I've learned how to do 2 things more efficiently today thanks to you. :)


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