University of Florida/Eml4507/s13.team3.GuzyR5

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

Given
$$ m_{1}=3 $$

$$ m_{2}=2 $$

$$ k_{1}=10 $$

$$ k_{2}=20 $$

$$ k_{3}=15 $$

Find
Solve by hand the gen. eigenvalue problem for the spring-mass-damper system on p.53-13, using the data for the masses in (2) p.53-13b, and the data for the stiffness coefficients in (4) p.53-13b.

Solution
First plug given values into the stiffness matrix:

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

$$ K= \begin{bmatrix} (10+20) & -20\\ -20 & (20+15)\\ \end{bmatrix}$$

$$ K= \begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix}$$

Next plug stiffness matrix in equation below:

$$[K-\gamma_I]x= 0$$

Take the determinant:

$$det \begin{Bmatrix} 30-\gamma & -20\\ -20 & 35-\gamma \end{Bmatrix} =[(30-\gamma)(35-\gamma)]-(-20)(-20)= 0$$

Simplifying:

$$1050-30\gamma-35\gamma+\gamma^2-400$$

$$\gamma^2-65\gamma+650=0$$

Using the Quadratic formula to solve:

$$\gamma_1=\frac{65+\sqrt{65^2-(4)(650)}}{2}$$

$$\gamma_2=\frac{65-\sqrt{65^2-(4)(650)}}{2}$$

Final gamma values:

$$\gamma_1=52.66$$

$$\gamma_2=12.34$$

To find the eigenvectors, first we:

$$[K-\gamma_I]x= [\begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix} - \gamma* \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix} \begin{bmatrix}] x_{1}\\ x_{2}\\ \end{bmatrix} = 0 $$

To solve, lets set $$ x_{1} $$ equal to 1:

$$[K-\gamma_I]x= [\begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix} - 52.66* \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}] \begin{bmatrix} 1\\ x_{2}\\ \end{bmatrix} = 0 $$

$$[K-\gamma_I]x= [\begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix} + \begin{bmatrix} -52.66 & 0\\ 0 & -52.66\\ \end{bmatrix}] \begin{bmatrix} 1\\ x_{2}\\ \end{bmatrix} = 0 $$

$$[K-\gamma_I]x= [\begin{bmatrix} 30-52.66 & -20\\ -20 & 35-52.66\\ \end{bmatrix} \begin{bmatrix} 1\\ x_{2}\\ \end{bmatrix} = 0 $$

$$[K-\gamma_I]x= [\begin{bmatrix} -22.66 & -20\\ -20 & -17.66\\ \end{bmatrix} \begin{bmatrix} 1\\ x_{2}\\ \end{bmatrix} = 0 $$

Multiplying out the matrices, we obtain:

$$-20-17.66x_{2}=0$$

$$x_{2}= 1.1325$$

Using the same process we find $$ x_{1} $$:

$$[K-\gamma_I]x= [\begin{bmatrix} 30 & -20\\ -20 & 35\\ \end{bmatrix} - 52.66* \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}] \begin{bmatrix} x_{1}\\ 1\\ \end{bmatrix} = 0 $$

$$[K-\gamma_I]x= [\begin{bmatrix} -22.66 & -20\\ -20 & -17.66\\ \end{bmatrix} \begin{bmatrix} x_{1}\\ 1\\ \end{bmatrix} = 0 $$

Multiplying out the matrices, we obtain:

$$-22.66x_{1}-20=0 $$

$$x_{1}=0.8826 $$