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
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 $$\gamma_2$$ $$\gamma_2=4+\sqrt{5}$$ $$[K-\gamma_2I]x=$$ $$ \begin{bmatrix} -1-\sqrt{5} & -2\\ -2 & 1-\sqrt{5} \end{bmatrix}$$ $$ \begin{Bmatrix} x_1\\ x_2 \end{Bmatrix}$$ $$=$$ $$ \begin{Bmatrix} 0\\ 0 \end{Bmatrix}$$ 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
Find the eigenvectors for $$\gamma_1$$ and $$\gamma_2$$ when setting $$x_1=1$$

Solution
We find the eigenvectors from $$\gamma_1$$ $$\gamma_1=4-\sqrt{5}$$ $$[K-\gamma_1I]x=$$ $$ \begin{bmatrix} -1+\sqrt{5} & -2\\ -2 & 1+\sqrt{5} \end{bmatrix}$$ $$ \begin{Bmatrix} x_1\\ x_2 \end{Bmatrix}$$ $$=$$ $$ \begin{Bmatrix} 0\\ 0 \end{Bmatrix}$$ 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 $$\gamma_2$$ $$\gamma_2=4+\sqrt{5}$$ $$[K-\gamma_2I]x=$$ $$ \begin{bmatrix} -1-\sqrt{5} & -2\\ -2 & 1-\sqrt{5} \end{bmatrix}$$ $$ \begin{Bmatrix} x_1\\ x_2 \end{Bmatrix}$$ $$=$$ $$ \begin{Bmatrix} 0\\ 0 \end{Bmatrix}$$ Set $$x_1=1$$ $$(-1-\sqrt{5})x_1-(2)x_2=0$$ $$x_2=\frac{-1-\sqrt{5}}{2}$$