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

File:Mode11.png

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.

$M=\left[\begin{array}{cc}{m}_1& 0\\ 0& m_2\end{array}\right]$

$d=\left[\begin{array}{c}{d}_1\\ d_2\end{array}\right]$

$C=\left[\begin{array}{cc}{C}_1+C_2& -C_2\\ -C_2& C_2+C_3\end{array}\right]$

$K=\left[\begin{array}{cc}(k_1+k_2)& -k_2\\ -k_2& (k_2+k_3)\end{array}\right]$

$k_1=1,k_2=3,k_3=3$

$K=\left[\begin{array}{cc}3& -2\\ -2& 5\end{array}\right]$

$[K-$γ$I]x=$ $\left[\begin{array}{cc}3& -2\\ -2& 5\end{array}\right]$ $-$γ $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ $)$ $\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}$ $=$ $\left\{\begin{array}{c}0\\ 0\end{array}\right\}$


$det\left\{\begin{array}{cc}3-\gamma & -2\\ -2& 5-\gamma \end{array}\right\}={\gamma }^2-8\gamma +11=0$

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 ${\gamma }_2$


${\gamma }_2=4+\sqrt{5}$

$[K-{\gamma }_{2}I]x=$ $\left[\begin{array}{cc}-1-\sqrt{5}& -2\\ -2& 1-\sqrt{5}\end{array}\right]$ $\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}$ $=$ $\left\{\begin{array}{c}0\\ 0\end{array}\right\}$


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

File:Mode11.png

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 ${\gamma }_1$


${\gamma }_1=4-\sqrt{5}$

$[K-{\gamma }_{1}I]x=$ $\left[\begin{array}{cc}-1+\sqrt{5}& -2\\ -2& 1+\sqrt{5}\end{array}\right]$ $\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}$ $=$ $\left\{\begin{array}{c}0\\ 0\end{array}\right\}$


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 ${\gamma }_2$


${\gamma }_2=4+\sqrt{5}$

$[K-{\gamma }_{2}I]x=$ $\left[\begin{array}{cc}-1-\sqrt{5}& -2\\ -2& 1-\sqrt{5}\end{array}\right]$ $\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}$ $=$ $\left\{\begin{array}{c}0\\ 0\end{array}\right\}$


Set $x_1=1$


$(-1-\sqrt{5})x_1-(2)x_2=0$


$x_2=\frac{-1-\sqrt{5}}{2}$