# Normal modes for small oscillations

1. Jul 20, 2011

### fluidistic

1. The problem statement, all variables and given/known data
I'm stuck at understanding how to find the kinetic and potential energy matrices such that the determinant $|V- \omega ^2 T|=0$ when solved for $\omega$, gives the normal modes (characteristic frequencies?) of the considered system.
For example in Goldstein's book for a molecule of the type m---M---m where the masses are linked via springs of constant k, $V=\frac{k}{2} (\nu ^2 _1+2 \nu ^2 _2 + \nu ^2 _3 -2 \nu _1 \nu _2 - 2 \nu _2 \nu _3)$ where $\nu _i =x_i-x_{0i}$.
He then says "hence the V matrix has the form $\begin{bmatrix} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{bmatrix}$". I notice it's symmetric as it should, since I guess it has been diagonalized.
He continues on by saying that the kinetic energy $T=\frac{m}{2}( \dot \nu ^2 _1+ \dot \nu ^2 _3)+\frac{M}{2} \dot \nu ^2 _2$. I've no problem understanding this, but I do have a problem when he says that thus the T matrix is diagonal and worth $\begin{bmatrix} m & 0 & 0 \\ 0 & M & 0 \\ 0 & 0 & m \end{bmatrix}$. How did he get this matrix?!
When I look for the definitions of both the T and V matrix, it's even more confusing to me.
$T=\frac{1}{2} \dot \Xi ^t \dot \Xi$. And it's not even clear to me what is the definition of the $\Xi$ matrix. It seems it has to see with coordinates but I don't understand anything else than this. While $V=\frac{1}{2} \Xi ^t \lambda \Xi$. Where the ^t mean transpose (yes, they are matrices).

So I'm given the following problem:
Find the normal modes of longitudinal vibrations of the chain of masses like this: Wall-k-m-k-m-k-m-k-Wall. Where the "-" are springs.

2. Relevant equations
Not sure.
3. The attempt at a solution
I feel I can't do my problem if I don't understand how to calculate the matrices V and T from the expression of V and T. I do understand how to get V and T (scalars), but not their link to the T and V matrices.

2. Jul 20, 2011

### mathfeel

Just a fancy way of writing in terms of matrices and vector:
$$T=\frac{m}{2}( \dot{\nu_1}^2+\dot{\nu_3}^2)+\frac{M}{2} \dot{\nu_2}^2 = \frac12 \begin{pmatrix}\dot \nu_1 \\ \dot \nu_2 \\ \dot \nu_3\end{pmatrix}^{T} \begin{pmatrix}m & 0 & 0 \\ 0 & M & 0 \\ 0 & 0 & m\end{pmatrix} \begin{pmatrix}\dot \nu_1 \\ \dot \nu_2 \\ \dot \nu_3\end{pmatrix}$$
More generally, we can say the kinetic energy has this form: $T = \frac12 \dot{\boldsymbol{v}}^{T} \cdot \boldsymbol{T} \cdot \dot{\boldsymbol{v}}$, where $\boldsymbol{T}$ is some matrix and $v$ is some vector.

The same goes for V:$$V = \frac{k}{2} (\nu ^2 _1+2 \nu ^2 _2 + \nu ^2 _3 -2 \nu _1 \nu _2 - 2 \nu _2 \nu _3) = \frac12 \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3\end{pmatrix}^{T} \begin{pmatrix} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3\end{pmatrix}$$
Generally, it can be written as: $V = \frac12 {\boldsymbol{v}}^{T} \cdot \boldsymbol{V} \cdot {\boldsymbol{v}}$ for some matrix $\boldsymbol{V}$

Write down the Lagrangian $L = T - V$, then do the usual Eurler-Lagrange, you get the equation of motion:
$$\boldsymbol{T} \cdot \ddot{\boldsymbol{v}} + \boldsymbol{V} \cdot \boldsymbol{v}= 0$$
Now you are looking for harmonic motion, solution of the type $\boldsymbol{v} = \bar{\boldsymbol{v}}_{\omega} e^{i \omega t}$, Just substitute into EOM, you get:
$$\left[ -\boldsymbol{T} \omega^2 + \boldsymbol{V} \right] \cdot \bar{\boldsymbol{v}}_{\omega}= 0$$

This is an eigenvalue problem, if you can't see it, just multiply by $\boldsymbol{T}^{-1}$ on both sides:
$$\boldsymbol{T}^{-1} \boldsymbol{V} \cdot \bar{\boldsymbol{v}}_{\omega} =\omega^2 \bar{\boldsymbol{v}}_{\omega}$$
So to find $\omega$ and the corresponding $\bar{\boldsymbol{v}}_{\omega}$ is just solving for eigenvalues and eigenvectors.

3. Jul 20, 2011

### fluidistic

Thanks a lot for your reply. Now it's a bit clearer to me. But I still don't get how it's trivial to obtain $V = \frac{k}{2} (\nu ^2 _1+2 \nu ^2 _2 + \nu ^2 _3 -2 \nu _1 \nu _2 - 2 \nu _2 \nu _3) = \frac12 \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3\end{pmatrix}^{T} \begin{pmatrix} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3\end{pmatrix}$. How do you find this matrix?
In other words, how do you generally find the matrix V?

4. Jul 20, 2011

### mathfeel

Easy, instead of writing $v_1^2$, write it as $v_1 v_1$.
For $v_1 v_2$ etc, write it as $v_i v_j = \frac{1}{2} (v_i v_j + v_j v_i)$ etc.
$$V = \frac{k}{2} (\nu ^2 _1+2 \nu ^2 _2 + \nu ^2 _3 -2 \nu _1 \nu _2 - 2 \nu _2 \nu _3) = \frac{k}{2} \left[ v_1 v_1 + 2 v_2 v_2 + v_3 v_3 - v_1 v_2 - v_2 v_1 - v_2 v_3 - v_3 v_2 \right]$$

Then the ij then element of the matrix is the coefficient of $v_i v_j$.

Last edited: Jul 20, 2011
5. Jul 20, 2011

### fluidistic

Thank you very very much for all. I'm going to refer to your replies for the future. Now I think it's clear. If I have any doubt I'll ask.

6. Jul 21, 2011

### fluidistic

I get $V=\begin{bmatrix} k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{bmatrix}$ and $\begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m \end{bmatrix}$. I am unsure of my V matrix. I get exactly the same as Goldstein for his problem of 3 masses. But in my case the masses are linked via 2 walls and I didn't know how to include this information in the equations.
$|V- \omega ^2 T |=0$ gives $(k- \omega ^2 m)[(k-\omega ^2m)(2k-\omega ^2m)+k^2]+k[k(k-\omega ^2m)]=0$. With the obvious solutions $\omega = \pm \sqrt {\frac{k}{m}}$. I know there are more solutions, I'll look into it. But so far, did I make any error?

7. Jul 22, 2011

### ehild

All masses are connected to two springs. There are four springs instead of two. Take it care in the potential.

ehild

8. Jul 22, 2011

### mathfeel

Imagine the problem without one or two of the walls and ask would matrix T be different? Would matrix V be different? So in fact, by writing the the V like you had, you already implicitly included the fact that two ends are connected to unmovable walls.

The dimension of the matrix equal to the number of degrees of freedom. There are 3 degrees of freedom in this problem, so you should get 3x3 matrices.

I don't know about the answer, but the procedure seems legit. I can make an educated guess that the other two of the modes frequencies are $\omega=\pm\sqrt{\frac{2k}{m}}$ and $\omega=\pm\sqrt{\frac{2k}{3m}}$. But you should really try to solve the polynomial.

9. Jul 22, 2011

### ehild

No. The V matrix shown by Fluidistic corresponds to free ends:

OγγγγγOγγγγγO.

He needs the potential energy for the arrangement

|γγγγγOγγγγγOγγγγγOγγγγγ|,

with four springs instead of two. The potential energy obviously differs from the first one.
with a sign error.

ehild

10. Jul 22, 2011

### ehild

Yes, apart from the fact that it is the equation for the free-ends system, there should be a - sign in front of the terms with k^2. Correct the signs, factorize the equation, you get the roots at once.

ehild

11. Jul 22, 2011

### mathfeel

You are right. All the diagonal elements should be $2k$.

12. Jul 22, 2011

### ehild

It is right this way

ehild

13. Jul 22, 2011

### fluidistic

Thanks for pointing the sign error for k² (good eye!), now I see it (I made it when calculating the determinant of the 3x3 matrix).
I'm having a hard time to include the walls in the equations.
4 springs but only 3 masses, this is really confusing me.
Furthermore when I compress 1 spring, the following spring is stretched. Hmm.
I'll have a try: Potential energy stored in the first spring: $V_1=(k/2)(x_{01}-x_1-b)^2$ where b is the length of any non stretched spring, $x_1$ is the position of the mass 1 and $x_{01}$ is the natural position of the mass (when the spring is unstretched).
V_2 would be the potential energy stored in the second spring. So $V_2=(k/2)(x_2-x_1-b)^2$.
In the same fashion, $V_3=(k/2)(x_3-x_2-b)^2$.
$V_4=(k/2)(x_3)^2$.
So that the potential term would be $V=V_1+V_2+V_3+V_4$. Is this right? I realize I don't get a 4x4 matrix because there are only 3 masses (variable coordinates here), despite having 4 springs.
I'm so tired, I would be surprised if this is right so far.

Edit: If what I did is good, I indeed reach the potential matrix with 2k's on the diagonal. I reach for the frequencies... $\omega = \pm \sqrt {\frac{2k}{m}}$ and $\omega = \pm \sqrt {\frac{2k(1-k)}{m}}$. It seems like I made a stupid algebra mistake somewhere, since I don't think I'm supposed to get a condition on the value of k.
Here's the factorized equation I got: $(2k - \omega ^2 m)[(2k- \omega ^2 m)-2k^2]=0$.

Last edited: Jul 22, 2011
14. Jul 22, 2011

### ehild

You need the potential energies of the objects, not that of the springs in the matrix. So the dimension is still 3x3.
I suggest to start with the Lagrangian and find the Euler-Lagrange equations from it, instead of trying to find out the matrix form at once. And use the small displacements from equilibrium instead of the coordinates.

You made an error again in the determinant. It is $(2k - \omega ^2 m)[(2k- \omega ^2 m)^2-2k^2]=0$ . There are three frequencies.

ehild

15. Jul 23, 2011

### fluidistic

Isn't it the same? I mean the potential energy of the 3 masses and the one stored in all springs? I do get a 3x3 matrix using the one stored in the 4 springs.
Fine, but I still have a doubt on how to calculate the potential energy of the system; I need to pass by the same methodology than by trying to get the V matrix at once... or I'm also wrong on this?

Oops I see my error, thanks once again for pointing it out. It wasn't in the determinant itself, it was when I factorized (2k- omega ²m), I simply dropped the ² by error.