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 $x_{1}=1$


$(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$


$x_{2}={\frac {-1-{\sqrt {5}}}{2}}$