University of Florida/Eml4507/s13.team3.DavidPatrickR4

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

${\displaystyle M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}}$

${\displaystyle d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}}$

${\displaystyle C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}}$

${\displaystyle K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}}$

${\displaystyle k_{1}=1,k_{2}=3,k_{3}=3}$

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

${\displaystyle [K-}$γ${\displaystyle I]x=}$ ${\displaystyle {\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}}$ ${\displaystyle -}$γ ${\displaystyle {\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$ ${\displaystyle )}$ ${\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}$ ${\displaystyle =}$ ${\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}$


${\displaystyle det{\begin{Bmatrix}3-\gamma &-2\\-2&5-\gamma \end{Bmatrix}}=\gamma ^{2}-8\gamma +11=0}$

Find the eigenvector ${\displaystyle x_{2}}$ corresponding to the eigenvalue ${\displaystyle \gamma _{2}}$ 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

Eigenvalues are found
${\displaystyle \gamma _{1}=4+{\sqrt {5}}>0}$
${\displaystyle \gamma _{2}=4-{\sqrt {5}}>0}$

We find the eigenvectors from ${\displaystyle \gamma _{2}}$


${\displaystyle \gamma _{2}=4+{\sqrt {5}}}$

${\displaystyle [K-\gamma _{2}I]x=}$ ${\displaystyle {\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}}$ ${\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}$ ${\displaystyle =}$ ${\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}$


Set ${\displaystyle x_{2}=1}$


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


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

Eigenvectors are orthogonal to each other:

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

ans = 0

Use same given values as in problem 4.1

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

We find the eigenvectors from ${\displaystyle \gamma _{1}}$


${\displaystyle \gamma _{1}=4-{\sqrt {5}}}$

${\displaystyle [K-\gamma _{1}I]x=}$ ${\displaystyle {\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\end{bmatrix}}}$ ${\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}$ ${\displaystyle =}$ ${\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}$


Set ${\displaystyle x_{1}=1}$


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


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





We find the eigenvectors from ${\displaystyle \gamma _{2}}$


${\displaystyle \gamma _{2}=4+{\sqrt {5}}}$

${\displaystyle [K-\gamma _{2}I]x=}$ ${\displaystyle {\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}}$ ${\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}$ ${\displaystyle =}$ ${\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}$


Set ${\displaystyle x_{1}=1}$


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


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