# University of Florida/Eml4507/s13.team3.GuzyR5

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Problem 5.1

On my honor, I have neither given nor received unauthorized aid in doing this assignment.


## Given

$m_{1}=3$ $m_{2}=2$ $k_{1}=10$ $k_{2}=20$ $k_{3}=15$ ## Find

Solve by hand the gen. eigenvalue problem for the spring-mass-damper system on p.53-13, using the data for the masses in (2) p.53-13b, and the data for the stiffness coefficients in (4) p.53-13b.

## Solution

First plug given values into the stiffness matrix:

$K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$ $K={\begin{bmatrix}(10+20)&-20\\-20&(20+15)\\\end{bmatrix}}$ $K={\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}$ Next plug stiffness matrix in equation below:

$[K-\gamma _{I}]x=0$ Take the determinant:

$det{\begin{Bmatrix}30-\gamma &-20\\-20&35-\gamma \end{Bmatrix}}=[(30-\gamma )(35-\gamma )]-(-20)(-20)=0$ Simplifying:

$1050-30\gamma -35\gamma +\gamma ^{2}-400$ $\gamma ^{2}-65\gamma +650=0$ Using the Quadratic formula to solve:

$\gamma _{1}={\frac {65+{\sqrt {65^{2}-(4)(650)}}}{2}}$ $\gamma _{2}={\frac {65-{\sqrt {65^{2}-(4)(650)}}}{2}}$ Final gamma values:

$\gamma _{1}=52.66$ $\gamma _{2}=12.34$ To find the eigenvectors, first we:

$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-\gamma *{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}{\begin{bmatrix}]x_{1}\\x_{2}\\\end{bmatrix}}=0$ To solve, lets set $x_{1}$ equal to 1:

$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$ $[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}+{\begin{bmatrix}-52.66&0\\0&-52.66\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$ $[K-\gamma _{I}]x=[{\begin{bmatrix}30-52.66&-20\\-20&35-52.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$ $[K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$ Multiplying out the matrices, we obtain:

$-20-17.66x_{2}=0$ $x_{2}=1.1325$ Using the same process we find $x_{1}$ :

$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0$ $[K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0$ Multiplying out the matrices, we obtain:

$-22.66x_{1}-20=0$ $x_{1}=0.8826$ 