User:EML4507.s13.team4ever.Tobin/R.5.3

Honor Pledge:
On my honor I have neither given nor received unauthorized aid in doing this problem.

Problem Description:
We are tasked with finding the eigenvector $$x_1$$ corresponding to the eigenvalue $$\lambda_1$$ for the same system as described in R4.1. We must also plot the mode shapes and compare with those from R4.1. Lastly we are tasked with creating an animation for each mode shape.

Given:
from Fead.s13.sec53b(4).djvu:

$$ K= \begin{bmatrix} 3 & -2\\ -2 & 5\\ \end{bmatrix} $$

$$ \gamma_1 = 4 - \sqrt{5} $$

$$ \gamma_2 = 4 + \sqrt{5} $$

and, from R4.1

$$ X_2 = \begin{bmatrix} -0.618 \\ 1 \\ \end{bmatrix} $$

Solution:
$$ \begin{bmatrix} K - \gamma_1*I\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

So,

$$ \begin{bmatrix} -1 + \sqrt{5} & -2\\ -2 & 1 + \sqrt{5}\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

Combining and reducing to row echelon form. $$ \begin{bmatrix} 1 & -0.618 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$

we may set $$ x_1 = 1 $$ and obtain:

$$ X_1 = \begin{bmatrix} 1 \\ 1.618 \\ \end{bmatrix} $$

Proving that they are orthogonal by ensuring the dot product between the two is zero: $$ X_1 $$ $$ \dot{} $$ $$ X_2 = (1)(-0.618)+ (1)(1.618) = 0$$ The eigenvalue corresponding to this eigenvector is the smallest possible and therefore this must be a fundamental mode unlike the one from R.4.1. Mode shape plotted in picture below as well as Matlab graph showing the mode 1 in blue and mode 2 in green. The oscillation moves from zero out to the displacement for each respective mass. It can be seen that the oscillation for mode 2 is indeed fundamental as both masses move in the same direction. Picture is adapted from Fead.s13.sec53b(4).djvu p.53-16.