User:Guy vandegrift/btvs/Effort 6

This effort was before I discovered the generalized eigenvalue problem.

Calculation The dotted line denotes equilibrium. The lowest and highest modes interact the the central "atom"

$E=PE+KE$ {\begin{aligned}E\;&={\tfrac {1}{2}}M_{1}{\dot {x}}_{1}^{2}+{\tfrac {1}{2}}M_{2}{\dot {x}}_{2}^{2}+{\tfrac {1}{2}}M_{3}{\dot {x}}_{3}^{2}\\&+{\tfrac {1}{2}}K_{1}x_{1}^{2}+{\tfrac {1}{2}}K_{2}x_{2}^{2}+{\tfrac {1}{2}}K_{3}x_{3}^{2}\\+&{\tfrac {1}{2}}K_{C}\left(x_{1}-x_{2}\right)^{2}+{\tfrac {1}{2}}K_{C}\left(x_{2}-x_{3}\right)^{2}\end{aligned}} Selecting masses and spring constants

Define $M_{2}=2M$ and $M_{1}=M_{3}=M$ . Also, $K_{2}=K$ and $K_{1}=K_{3}=K$ , and define momentum as $p_{j}=M_{j}{\dot {x}}_{j}^{2}$ for j=1,2,3. The effective coupling effective spring constant is, $K_{C}=T/L$ , where T is a small tension and L is the distance between masses.

{\begin{aligned}E\;&={\frac {1}{2}}{\frac {p_{1}^{2}}{M}}+{\frac {1}{2}}{\frac {p_{3}^{2}}{2M}}+{\frac {1}{2}}{\frac {p_{3}^{2}}{M}}+\\&+{\frac {1}{2}}Kx_{1}^{2}+{\frac {1}{2}}(2K)x_{2}^{2}+{\frac {1}{2}}Kx_{3}^{2}\\+&{\frac {1}{2}}K_{C}\left(x_{1}-x_{2}\right)^{2}+{\frac {1}{2}}K_{C}\left(x_{2}-x_{3}\right)^{2}\end{aligned}} Expand the coupling terms

algebra

{\begin{aligned}E\;&={\frac {1}{2}}{\frac {p_{1}^{2}}{M}}+{\frac {1}{2}}{\frac {p_{3}^{2}}{2M}}+{\frac {1}{2}}{\frac {p_{3}^{2}}{M}}+\\&+{\frac {1}{2}}Kx_{1}^{2}+{\frac {1}{2}}(2K)x_{2}^{2}+{\frac {1}{2}}Kx_{3}^{2}\\&+\quad cross\;terms\end{aligned}} ,

where the $cross\;terms$ are:

=${\frac {1}{2}}K_{C}\cdot \left\{(x_{1}-x_{2})^{2}+(x_{2}-x_{3})^{2}\right\}$ =${\frac {1}{2}}K_{C}\cdot \left\{x_{1}^{2}+2x_{2}^{2}+x_{3}^{2}-2x_{1}x_{2}-2x_{2}x_{3}\right\}$ We can rewrite $-2x_{1}x_{2}-2x_{2}x_{3}$ in matrix form:

${\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}{\begin{pmatrix}0&-1&0\\-1&0&-1\\0&-1&0\end{pmatrix}}{\begin{pmatrix}x_{1}&x_{2}&x_{3}\end{pmatrix}}$ = ${\begin{pmatrix}-x_{2}\\-x_{1}-x_{3}\\-x_{2}\end{pmatrix}}{\begin{pmatrix}x_{1}&x_{2}&x_{3}\end{pmatrix}}$ =$-2x_{1}x_{2}-2x_{2}x_{3}$ Matrix of coupling terms

Define the coupling matrix $\kappa _{ij}$ as:

${\underline {\underline {\kappa }}}={\begin{pmatrix}K_{C}&-K_{C}&0\\-K_{C}&2K_{C}&-K_{C}\\0&-K_{C}&K_{C}\end{pmatrix}}$ Note that the coupling force on the i-th particle is $F_{i}=-\Sigma \kappa _{ij}x_{j}$ were we sum over the j-th index. For the vertical springs:

${\underline {\underline {K}}}={\begin{pmatrix}K_{1}&0&0\\0&K_{2}&0\\0&0&K_{3}\end{pmatrix}}$ = ${\begin{pmatrix}K&0&0\\0&2K&0\\0&0&K\end{pmatrix}}$ Newton's second law: Where I went wrong

The same generalization of the spring constant to a matrix also yields an expression for force that is useful for extracting the normal modes. Assuming sinusoidal motion with angular frequency $\omega$ , we use ${\ddot {x}}_{j}=-\omega ^{2}x_{j}$ to write Newton's second law as:

$-\omega ^{2}{\begin{pmatrix}Mx_{1}\\2Mx_{2}\\Mx_{3}\end{pmatrix}}=-{\begin{pmatrix}K+K_{C}&-K_{C}&0\\-K_{C}&2K+2K_{C}&-K_{C}\\0&-K_{C}&K+K_{C}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}$ Divide the second row by two and divide all rows by M:

Here is my mistake

See http://rotorlab.tamu.edu/Dynamics_and_Vibrations/Lectures%20(pdf)%20and%20homework/Particle%20Kinetics%202/L14-18.pdf

$\omega ^{2}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}K/M+K_{C}/M&-K_{C}/M&0\\-{\frac {1}{2}}K_{C}/M&K/M+K_{C}/M&-{\frac {1}{2}}K_{C}/M\\0&-K_{C}/M&K/M+K_{C}/M\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}$ Eigenvalue problem

Define $\omega _{S}^{2}=K/M$ as the spring/mass natural frequency, and $\omega _{C}^{2}=K_{C}/M$ as a much smaller coupling frequency. Let $K_{C}=\epsilon K$ where $\epsilon$ is a small parameter. Also defined $\lambda =\omega ^{2}/\omega _{S}^{2}$ :

${\begin{pmatrix}1+\epsilon -\lambda &-\epsilon &0\\-{\frac {1}{2}}\epsilon &1+\epsilon -\lambda &-{\frac {1}{2}}\epsilon \\0&-\epsilon &1+\epsilon -\lambda \end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}=0$ Matlab code

Input and output
input
A = [[1+epsilon,   -epsilon,       0;
-epsilon/2,   1+epsilon,  -epsilon/2;
0,           -epsilon,    1+epsilon]]
[V,D] = eig(A)

output
A =
[ epsilon + 1,    -epsilon,           0]
[  -epsilon/2, epsilon + 1,  -epsilon/2]
[           0,    -epsilon, epsilon + 1]
V =
[ 1,  1, -1]
[ 1, -1,  0]
[ 1,  1,  1]
D =
[ 1,             0,           0]
[ 0, 2*epsilon + 1,           0]
[ 0,             0, epsilon + 1]

Output from Matlab

According to MATLAB the columns of V are the right eigenvectors of A: AV=DV

Construct results first using <pre> tag
[ epsilon + 1,    -epsilon,           0]   [ 1,  1, -1]
[  -epsilon/2, epsilon + 1,  -epsilon/2] * [ 1, -1,  0] =
[           0,    -epsilon, epsilon + 1]   [ 1,  1,  1]

[ 1,             0,           0]   [ 1,  1, -1]
[ 0, 2*epsilon + 1,           0] * [ 1, -1,  0]
[ 0,             0, epsilon + 1]   [ 1,  1,  1]

I have interchanged the second and third rows so the 1+2*epsilon  appears las, and I multiplied that column by -1 so that the end mass amplitudes would cancel while the "atom" amplitudes add (initial condition).
|}

MATLAB did not order the eigenvalues by size, be we need to. From lowest to highest eigenvalue we have:

Eigenvalues and eigenvectors

$\lambda _{1}=1:\;{\underline {\xi }}_{1}={\begin{bmatrix}1\\1\\1\end{bmatrix}},$ $\lambda _{2}=1+\epsilon :\;{\underline {\xi }}_{2}={\begin{bmatrix}1\\0\\1\end{bmatrix}},$ $\lambda _{3}=1+2\epsilon :\;{\underline {\xi }}_{3}={\begin{bmatrix}-1\\1\\-1\end{bmatrix}}$ The eigenfrequencs for j=(1,2,3) are given by $\omega _{j}={\sqrt {\lambda _{j}K/M}}={\sqrt {\lambda _{j}}}\;\omega _{S}$ :

$\omega _{1}=\omega _{S},\quad \omega _{2}={\sqrt {1+\epsilon }}\;\omega _{S},\quad \omega _{3}={\sqrt {1+2\epsilon }}\;\omega _{S}.$ Alternative forms

This can be expressed in a number of ways

${\underline {\underline {A}}}\cdot {\underline {\underline {\xi }}}={\begin{pmatrix}\epsilon +1&-\epsilon &0\\-\epsilon /2&\epsilon +1&-\epsilon /2\\0&-\epsilon &\epsilon +\end{pmatrix}}\cdot {\underline {\underline {\xi }}}=A_{ij}\cdot \xi _{jk}={\underline {\underline {A}}}\cdot$ ${\begin{pmatrix}{\underline {\xi }}_{1}&{\underline {\xi }}_{2}&{\underline {\xi }}_{3}\end{pmatrix}}=$ ${\begin{pmatrix}\epsilon +1&-\epsilon &0\\-\epsilon /2&\epsilon +1&-\epsilon /2\\0&-\epsilon &\epsilon +\end{pmatrix}}\cdot {\begin{pmatrix}{\begin{bmatrix}1\\1\\1\end{bmatrix}}&{\begin{bmatrix}1\\0\\1\end{bmatrix}}&{\begin{bmatrix}-1\\1\\-1\end{bmatrix}}\end{pmatrix}}$ $={\begin{pmatrix}\lambda _{1}{\underline {\xi }}_{1}&\lambda _{2}{\underline {\xi }}_{2}&\lambda _{3}{\underline {\xi }}_{3}\end{pmatrix}}$ next

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\widetilde {x_{1}}}{\begin{bmatrix}1\\1\\1\end{bmatrix}}e^{-i\omega _{1}t}+{\widetilde {x_{2}}}{\begin{bmatrix}1\\0\\1\end{bmatrix}}e^{-i\omega _{2}t}+{\widetilde {x_{3}}}{\begin{bmatrix}-1\\1\\-1\end{bmatrix}}e^{-i\omega _{3}t}$ =

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{pmatrix}{\begin{bmatrix}1\\1\\1\end{bmatrix}}&{\begin{bmatrix}1\\0\\1\end{bmatrix}}&{\begin{bmatrix}-1\\1\\-1\end{bmatrix}}&\end{pmatrix}}{\begin{pmatrix}{\widetilde {x_{1}}}e^{-i\omega _{1}t}\\{\widetilde {x_{2}}}e^{-i\omega _{2}t}\\{\widetilde {x_{3}}}e^{-i\omega _{3}t}\\\end{pmatrix}}THISISWRONG$ Here we take only the real part. Don't confuse the indexes (1,2,3) on the RHS marking location of the masses, with LHS indexes (1,2,3) marking the three normal modes. If the RHS is denoted by the column vector ${\underline {x}}$ , this can be written as a sum from over n=1,2,3. This is analogous to Fourier analysis for 3 discrete elements:

${\underline {x}}=\sum _{n}{\widetilde {x_{n}}}{\underline {\xi }}_{n}e^{i\omega _{n}t}$ Aha, now I see the need for the matrix whose columns are the eigenvectors. Following matlab, we call it V: