# University of Florida/Eml4507/s13.team3.DavidPatrickR4

## Problem 4.1

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


### Given Mode Shape for system. Mode slope increases and crosses into the positive region

Spring-damper-body arrangement as shown. Two separate forces applied to masses.

$M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}$ $d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$ $C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}$ $K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$ $k_{1}=1,k_{2}=3,k_{3}=3$ $K={\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}$ $[K-$ γ$I]x=$ ${\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}$ $-$ γ ${\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ $)$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ $det{\begin{Bmatrix}3-\gamma &-2\\-2&5-\gamma \end{Bmatrix}}=\gamma ^{2}-8\gamma +11=0$ ### Find

Find the eigenvector $x_{2}$ corresponding to the eigenvalue $\gamma _{2}$ for the spring-mass-damper system on p.53-113. Plot and comment on this mode shape. Verify that the eigenvectors are orthogonal to each other

### Solution

Eigenvalues are found
$\gamma _{1}=4+{\sqrt {5}}>0$ $\gamma _{2}=4-{\sqrt {5}}>0$ We find the eigenvectors from $\gamma _{2}$ $\gamma _{2}=4+{\sqrt {5}}$ $[K-\gamma _{2}I]x=$ ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ Set $x_{2}=1$ $(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$ $x_{1}={\frac {2}{-1-{\sqrt {5}}}}$ Eigenvectors are orthogonal to each other:

EDU>> x= [-.8507;-.5257];
EDU>> y= [-.5257;.8507];
EDU>> transpose(y)*x

ans = 0


## Problem 4.2

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


### Given

Use same given values as in problem 4.1

### Find Mode Shape for system. Notice plot is the same even with different initial conditions

Find the eigenvectors for $\gamma _{1}$ and $\gamma _{2}$ when setting $x_{1}=1$ ### Solution

We find the eigenvectors from $\gamma _{1}$ $\gamma _{1}=4-{\sqrt {5}}$ $[K-\gamma _{1}I]x=$ ${\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ Set $x_{1}=1$ $(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$ $x_{2}={\frac {-1+{\sqrt {5}}}{2}}$ We find the eigenvectors from $\gamma _{2}$ $\gamma _{2}=4+{\sqrt {5}}$ $[K-\gamma _{2}I]x=$ ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ Set $x_{1}=1$ $(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$ $x_{2}={\frac {-1-{\sqrt {5}}}{2}}$ 