User:EML4507.s13.team4ever.Bonner/Bonnerreport4p1

Problem R4.1
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description
We are to find the eigenvector $$ x_2 $$ corresponding to the eigenvalue $$ \lambda_2 $$ for the spring-mass-damper system on p.53-13. We are then to plot and comment on this mode shape. Finally we are to verify that this eigenvector is orthogonal to the one obtained in the example before.

Given:
We are given from Fead.s13.sec53b.djvu that

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

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

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

and

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

Solution:
So we need to find an eigenvector such that

$$ \begin{bmatrix} K - \gamma_2*I\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

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

Augmenting the right hand matrix onto the left hand matrix and putting this in row reduced echelon form we obtain,

$$ \begin{bmatrix} 1 & -0.618 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$

we may set $$ x_2 = 1 $$ and obtain:

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

We will now show that these two eigenvectors are orthogonal by showing that the dot product between the two is zero.

$$ X_1 $$ $$ \dot{} $$ $$ X_2 = (1.618)(-0.618) + (1)(1) = 0 $$

Therefore, they are orthogonal.

The corresponding eigenvalue for this eigenvector isn't the smallest possible eigenvalue which means this isn't a fundamental mode.

The modal shape is plotted below.