# University of Florida/Eml4507/Team 7 Report 5
## Problem 1
### Find

Solve the general eigenvalue problem for the spring-mass-damper system given on p.53-13 of 'Fead.s13.sec.53b', compare results obtained in Pb-53-6 and verify mass orthogonality of eigenvectors.

### Given

$M={\begin{pmatrix}3&0\\0&2\end{pmatrix}}$
$K={\begin{pmatrix}30&-20\\-20&35\end{pmatrix}}$

### Solution

Solving the general eigenvalue problem
$Kx=\lambda Mx$
$\lambda Mx-Kx=0$
$(\lambda M-K)x=0$
$\lambda M-K={\begin{pmatrix}3\lambda -30&20\\20&2\lambda -35\end{pmatrix}}$
$det(\lambda M-K)=0$
Calculating the determinant
$(3\lambda -30)(2\lambda -35)-400=0$
$6\lambda ^{2}-165\lambda +1050=0$
Therefore, the Eigenvalues are: $\lambda _{1}=17.5,\lambda _{2}=10$
The next step is finding the Eigenvectors
$\lambda _{1}=17.5$
$(\lambda _{1}M-K)X_{1}={\begin{pmatrix}22.5&20\\20&0\end{pmatrix}}{\begin{pmatrix}x_{11}\\x_{21}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$
Letting $x_{21}=1$ and calculating $x_{12}$
$\lambda _{1}=10$
$(\lambda _{1}M-K)X_{1}={\begin{pmatrix}0&20\\20&-15\end{pmatrix}}{\begin{pmatrix}x_{12}\\x_{22}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$
Letting $x_{12}=1$ and calculating $x_{22}$
These results compare favorably with those from problem 5.6 using CALFEM.
Mass Orthogonality: $x_{i}^{T}M_{xj}=0$
The eigenvectors are therefore proven to be mass-orthogonal.

## Problem 2
### Find

Do all work with your own matlab FE code, then verify the results with CALFEM; display both results clearly for easy comparison.
Solve the generalized eigenvalue problem of the above truss. Display the results for the lowest 3 eigenpairs. Plot and animate the lowest 3 mode shapes. Verify the mass orthogonality of these 3 eigenvectors.
Now consider the same truss, but with 2 missing braces:
Solve the generalized eigenvalue for this truss, and plot the mode shapes with zero frequency.

### Given

$L_{2}3=L_{4}5=1,L_{1}2=L_{2}4=L_{4}6=1,A=1/2,E=5,\rho =2$

### Solution

The lumped mass matrix, ${\bar {M}}$ , is shown below:
>> M=[1 0 0 0 0 0 0 0 0 0 0 0; 0 1 0 0 0 0 0 0 0 0 0 0; 0 0 1 0 0 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0 0 0 0 0; 0 0 0 0 1 0 0 0 0 0 0 0; 0 0 0 0 0 1 0 0 0 0 0 0; 0 0 0 0 0 0 1 0 0 0 0 0; 0 0 0 0 0 0 0 1 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 0 0; 0 0 0 0 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 0 0 0 0 1];
K=zeros(10,10)
Because they are identical, we can use the CALFEM function “assem” for elements 2 and 6:
>> edof=[2 1 2 5 6; 6 3 4 9 10]; >> l=cos(45); m=sin(45); >> Ke=1.768*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke); For elements 1,4,7, and 10: >> edof=[1 1 2 3 4; 4 5 6 9 10; 7 3 4 7 8; 10 7 8 11 12]; >> l=1; >> m=0; >> Ke=2.5*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke);
For the elements 3 and 8:
>> edof=[3 3 4 5 6; 8 7 8 9 10]; l=0; m=1; Ke=2.5*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke);
For elements 5 and 9:
>> edof=[5 5 6 7 8; 9 9 10 11 12]; >> l=cos(135); >> m=sin(135); >> Ke=1.768*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke)
K =
In order to solve the generalized eigenvalue problem we used the function “eigen”, which returns the eigenvalues, $\lambda$, and the eigenvectors, ‘x’:
Kx=λMx >> [L,X]=eigen(K,M) L =
X =
For the first eigenpair, the eigenvalue and corresponding eigenvector are as follows:
L= 0 X= {-0.406457927237019 0.0381918846465145 -0.406457927237019 0.0381918846465150 -0.406457927237019 0.0381918846465148 -0.406457927237019 0.0381918846465148 -0.406457927237019 0.0381918846465148 -0.406457927237020 0.0381918846465028}
These eigenvectors were used to calculate the displacement of each node on the truss and plotted using the MATLAB code below: % truss original position % start nodes (x/y) / end nodes (x/y)
% the engine
% the engine
For the second eigenpair:
L=0 X={-0.0381918846465119 -0.406457927237036 -0.0381918846465119 -0.406457927237034 -0.0381918846465145 -0.406457927237035 -0.0381918846465120 -0.406457927237031 -0.0381918846465160 -0.406457927237031 -0.0381918846465106 -0.406457927236949}
For the third eigenpair:
L=0.0043 X={0.0124740073031415 -0.216589482612068 0.0122217899111943 -0.192300781266623 -0.0216971729033128 -0.194763453030673 0.0102232821955044 -0.152379327406532 -0.0411885014874805 -0.153869406652153 0.0279665949809443 0.909902450968240}
Analyzing the second truss, we get the lumped mass matrix, ${\bar {M}}$ , shown below: >> M=[1 0 0 0 0 0 0 0 0 0 0 0; 0 1 0 0 0 0 0 0 0 0 0 0; 0 0 1 0 0 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0 0 0 0 0; 0 0 0 0 1 0 0 0 0 0 0 0; 0 0 0 0 0 1 0 0 0 0 0 0; 0 0 0 0 0 0 1 0 0 0 0 0; 0 0 0 0 0 0 0 1 0 0 0 0; 0 0 0 0 0 0 0 0 1 0 0 0; 0 0 0 0 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 0 0 0 0 1];
For element 2 we use the CALFEM function “assem”: >> edof=[2 1 2 5 6]; >> l=cos(45); >>m=sin(45); >> Ke=1.768*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke);
For elements 1,4,7, and 6:
>> edof=[1 1 2 3 4; 4 5 6 9 10; 7 3 4 7 8; 6 7 8 11 12]; >> l=1; >> m=0; >> Ke=2.5*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke);
For the elements 3 and 8:
>> edof=[3 3 4 5 6; 8 7 8 9 10]; l=0; m=1; Ke=2.5*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke); For element 5:
>> edof=[5 5 6 7 8]; >> l=cos(135); >> m=sin(135); >> Ke=1.768*[l^2 l*m -l^2 -l*m; l*m m^2 -l*m -m^2; -l^2 -l*m l^2 l*m; -l*m -m^2 l*m m^2]
Ke =
>> [K]=assem(edof,K,Ke)
>> K
K =
In order to solve the generalized eigenvalue problem we used the function “eigen”, which returns the eigenvalues, $\lambda$, and the eigenvectors, ‘x’:
Kx=λMx >> [L,X]=eigen(K,M)
L =
X =
For the first eigenpair, the eigenvalue and corresponding eigenvector are as follows:
L= 0 X= {0 0 0 0 0 0 0 0 0 0 0 1}
The truss was plotted in MATLAB using the following code:
% truss original position % start nodes (x/y) / end nodes (x/y)
% the engine
% the engine

## Problem 3
### Find

Find, plot and compare the eigenvectors. Compare the mode shapes using two different assumptions. Animate each format in the given sine wave.

### Given

| $\mathbf {|X_{1}X_{2}|} ={\begin{bmatrix}x_{11}&x_{12}\\x_{21}&x_{22}\\\end{bmatrix}}$ | (2.1) |
Solve for the eigenvectors under two differing assumptions:
| $x_{11}=x_{12}=1$ $x_{21}=x_{22}=1$ | (2.2) |
Plot mode shapes according to $y=\mathbf {X_{i}} sin(w_{i}t)$
where $w_{i}t$ is the circular frequency, $w_{i}t={\sqrt {\gamma _{i}}}$,
according to the eigenvalues $\gamma _{1}=4+{\sqrt {5}}$ and $\gamma _{2}=4-{\sqrt {5}}$.

### Solution

Firstly, the eigenvectors must be found. The $x_{21}=x_{22}=1$ assumption is in effect. This value was found with $\gamma _{1}=4+{\sqrt {5}}$
| ${\begin{bmatrix}k-\gamma _{1}I\\\end{bmatrix}}\mathbf {X} ={\begin{bmatrix}0\\\end{bmatrix}}$ | (2.3) |
| ${\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$ | (2.4) |
| $-2x_{1}+1+{\sqrt {5}}=0$ $x_{1}=(1+{\sqrt {5}})/2$ | (2.5) |
and this value was found with $\gamma _{2}=4-{\sqrt {5}}$
| ${\begin{bmatrix}k-\gamma _{2}I\\\end{bmatrix}}\mathbf {X} ={\begin{bmatrix}0\\\end{bmatrix}}$ | (2.6) |
| ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$ | (2.7) |
| $-2x_{1}+1-{\sqrt {5}}=0$ $x_{1}=(1-{\sqrt {5}})/2$ | (2.8) |
With these values solved for, construct the eigenvector:
| ${\begin{bmatrix}{\mathbf {X_{1}} }&{\mathbf {X_{2}} }\end{bmatrix}}={\begin{bmatrix}(1+{\sqrt {5}})/2&(1-{\sqrt {5}})/2\\1&1\\\end{bmatrix}}$ | (2.9) |
Now assuming $x_{11}=x_{12}=1$ the eigenvectors are found using the same eigenvalues $\gamma _{1,2}=4\pm {\sqrt {5}}$
| ${\begin{bmatrix}k-\gamma _{1}I\\\end{bmatrix}}\mathbf {X} ={\begin{bmatrix}0\\\end{bmatrix}}$ | (2.10) |
| ${\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$ | (2.11) |
| $-1+{\sqrt {5}}-2x_{2}=0$ $x_{2}=(-1+{\sqrt {5}})/2$ | (2.12) |
and this value was found with $\gamma _{2}=4-{\sqrt {5}}$
| ${\begin{bmatrix}k-\gamma _{2}I\\\end{bmatrix}}\mathbf {X} ={\begin{bmatrix}0\\\end{bmatrix}}$ | (2.13) |
| ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\\end{bmatrix}}$ | (2.14) |
| $-1-{\sqrt {5}}-2x_{2}=0$ $x_{2}=(-1-{\sqrt {5}})/2$ | (2.15) |
With these values solved for, construct the eigenvector:
| ${\begin{bmatrix}{\mathbf {X_{1}} }&{\mathbf {X_{2}} }\end{bmatrix}}={\begin{bmatrix}1&1\\(-1+{\sqrt {5}})/2&(-1-{\sqrt {5}})/2\\\end{bmatrix}}$ | (2.16) |
For both assumptions, the mode shape are plotted, not according to scale.
Comparing the assumptions: The constant 1 switches to the other point, upon switching the Row = 1 assumption. Going from mode 1 to mode 2 within an assumption leaves the solved variable with a -2.236 reduction in magnitude. Going from one assumption to another, within the same mode, reduces the magnitude of the solved variable by 1.
Modes animated according to $y=\mathbf {X_{i}} sin(w_{i}t)$ where $w_{i}t$ is the circular frequency, $w_{i}t={\sqrt {\gamma _{i}}}$, according to the eigenvalues $\gamma _{1}=4+{\sqrt {5}}$ and $\gamma _{2}=4-{\sqrt {5}}$:

## Problem 4
### Find

The Lowest 3 eigenpairs $(\omega _{j},\phi _{j})forj=1,2,3$ and plot the 3 lowest mode shapes in a gif image.

### Given

$E=100GPa,A=1.0cm^{2},L=0.3m,\rho =5000{\frac {kg}{m^{3}}}$

### Solution

The resulting egienpairs are:
    $\displaystyle {\begin{aligned}&\omega _{1}=1.6890\times 10^{5}\phi _{1}={\begin{bmatrix}0\\-0.0082\\-0.0279\\-0.0757\\-0.0479\\-0.1982\\-0.0605\\-0.3581\\-0.0670\\-0.5384\\-0.0690\\-0.7244\\-0.0690\\-0.9052\\0\\0\\0.0364\\-0.0677\\0.0645\\-0.1906\\0.0847\\-0.3514\\0.0974\\-0.5332\\0.1038\\-0.7213\\0.1058\\-0.9045\end{bmatrix}}\\&\omega _{2}=2.6676\times 10^{6}\phi _{2}={\begin{bmatrix}0\\-0.0745\\0.0227\\-0.4176\\0.0973\\-0.6512\\0.1884\\-0.6252\\0.2603\\-0.3137\\0.2941\\0.1825\\0.2976\\0.6937\\0\\0\\0.0732\\-0.3541\\0.0714\\-0.6149\\0.0153\\-0.6228\\-0.0603\\-0.3368\\-0.1201\\0.1553\\-0.1455\\0.6853\end{bmatrix}}\\&\omega _{3}=7.0219\times 10^{6}\phi _{3}={\begin{bmatrix}0\\-0.0352\\-0.2178\\-0.0416\\-0.3907\\-0.1023\\-0.5191\\-0.1283\\-0.6115\\-0.0751\\-0.6722\\0.0253\\-0.6941\\0.0971\\0\\0\\-0.1204\\-0.0089\\-0.2656\\-0.0780\\-0.4221\\-0.1195\\-0.5663\\-0.0809\\-0.6710\\0.0154\\-0.7178\\0.0940\end{bmatrix}}\\\end{aligned}}$
The mode shapes for the previous eigenpairs are"

## Problem 5
### Find

Do the HW problem on p.21-1 (2-bar truss, unconstrained, no supports, use 6x6 stiffness matrix ) and the HW problem on p.21-3 (4-bar truss, with supports, use reduced stiffness matrix ). For problem on p. 21-1, plot the eigenvectors corresponding to zero evals of 2 bar truss system and interpret results. For problem on p. 21-3, plot eigenvectors corresponding to zero evals for case (a).

### Given

The HW problems on p.21-1 and p.21-3.

### Solution

For problem 21-1:
K x u = 0 x u = 0
Σα x u = W
K x W = K x (Σα x u) = Σα x (K x u) = Σα x 0 = 0
For problem 21-3: K x v = λ x v
where a = b = 1m, E=2, and A=3

## Problem 6
### Find

Using the method of Modal Superposition, find the governing equations of motion for the system shown above.

### Given

$m_{1}=3,m_{2}=2$ $c_{1}=1/2,c_{2}=1/4,c_{3}=1/3$ $k_{1}=10,k_{2}=20,k_{3}=15$ $F_{1}(t)=0,F_{2}(t)=0$ $d_{1}(0)=-1,d_{2}(0)=2$ $d_{1}'(0)=0,d_{2}'(0)=0$

### Solution

First, with the information given we can substitute them into the equation with the form:
* $\mathbf {M} {\ddot {d}}+\mathbf {C} {\dot {d}}+\mathbf {K} d=\mathbf {F} (t)$
Where M, C, and K are the mass, damper and stiffness matrices.
* $\mathbf {M} ={\begin{bmatrix}m_{1}&0\\0&m_{2}\end{bmatrix}}$
* $\mathbf {C} ={\begin{bmatrix}(c_{1}+c_{2})&-c_{2}\\-c_{2}&(c_{2}+c_{3})\end{bmatrix}}$
* $\mathbf {K} ={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\end{bmatrix}}$
Substituting the known values, we now have the equation as matrix form as:
| ${\begin{bmatrix}3&0\\0&2\end{bmatrix}}{\ddot {d}}+{\begin{bmatrix}3/4&-1/4\\-1/4&7/12\end{bmatrix}}{\dot {d}}+{\begin{bmatrix}30&-20\\-20&35\end{bmatrix}}d=\mathbf {F} (t)$ | Eq. 1 |
We can now guess a solution that will solve Equation 1. In this case the following will be used:
* $d(t)={\vec {u}}e^{j\omega t}$
* ${\dot {d(t)}}=j\omega {\vec {u}}e^{j\omega t}$
* ${\ddot {d(t)}}=-j^{2}\omega ^{2}{\vec {u}}e^{j\omega t}$
Where ${\vec {u}}$ represents the eigen vectors and $j={\sqrt {}-1)$ and ${\vec {u}}\neq 0$
The equation can now be written as:
| $\mathbf {M} (-j^{2}\omega ^{2}{\vec {u}}e^{j\omega t})+\mathbf {C} j\omega {\vec {u}}e^{j\omega t}+\mathbf {K} {\vec {u}}e^{j\omega t}=\mathbf {F} (t)$ | Eq. 2 |
The following matrix can be used in order to represent the modes of vibration of the system by its eigenvalues and eigenvectors:
* ${\bigM{\big ]}^{-1}{\bigK{\big ]}={\begin{bmatrix}3&0\\0&2\end{bmatrix}}^{-1}{\begin{bmatrix}30&-20\\-20&35\end{bmatrix}}={\begin{bmatrix}10&-6.667\\-10&17.5\end{bmatrix}}$
Where the inverse matrix of M is determined by:
* ${\bigM{\big ]}^{-1}={\frac {adj(M)}{det(M)}}$
To find the solutions to the system we use the guesses that we made (after plugging them back into the matrix equation, Equation 2) and obtain the following worked out equation:
* ${\big\omega ^{2}-{\bigM{\big ]}^{-1}{\bigK{\big ]}{\big )}d_{0}$
When substituting we obtain:
| $(\omega ^{2}-{\begin{bmatrix}10&-6.667\\-10&17.5\end{bmatrix}})d_{0}$ | Eq. 3 |
And $d_{0}$ are the eigen vectors or mode shapes.
To find $\omega$ we require the determinant for Equation 3:
| $\det(-\omega ^{2}-{\begin{bmatrix}10&-6.667\\-10&17.5\end{bmatrix}})=\det {\begin{bmatrix}\omega ^{2}-10&6.667\\10&\omega ^{2}-17.5\end{bmatrix}}$ | Eq. 4 |
Taking the determinant, we now have the equation:
| $\omega ^{4}-27.5\omega ^{2}+175$ | Eq. 5 |
By using the quadratic formula we can find the respective $\omega$ values.
| $\omega ^{2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ | Eq. 6 |
And now,$\omega _{1}^{2}=17.5$ and $\omega _{2}^{2}=10$ Using these values we can now calculate the mode shapes by first substituting them into the matrix in Equation 4 for each $\omega$ value. For $\omega _{1}^{2}=17.5$ we substitute and obtain:
| ${\begin{bmatrix}296.25&6.667\\10&288.75\end{bmatrix}}{\begin{Bmatrix}d_{1}\\d_{2}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}$ | Eq. 7 |
solving for $d_{1}$:
* $296.25d_{1}+6.667d_{2}=0$
and
* $10d_{1}+288.75d_{2}=0$
Form these equations we obtain:
* $d_{1}={\begin{Bmatrix}-.0225\\-28.875\end{Bmatrix}}$
Now we substitute $\omega _{2}^{2}=10$ into the matrix in equation 4 and repeat the above steps to obtain $d_{2}$ we have:
* ${\begin{bmatrix}90&6.667\\10&82.5\end{bmatrix}}{\begin{Bmatrix}d_{1}\\d_{2}\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}$
Corresponding equations are now:
* $90d_{1}+6.667d_{2}=0$
and
* $10d_{1}+82.5d_{2}=0$
The modal vector is now :
* $d_{2}={\begin{Bmatrix}-13.5\\-.1212\end{Bmatrix}}$
The modal matrix can now be written as:
* $\phi ={\begin{bmatrix}-.0225&-13.5\\-28.875&-.1212\end{bmatrix}}$
Now the method of modal superposition can be implemented by the following equation:
* ${\begin{Bmatrix}d_{1}\\d_{2}\end{Bmatrix}}{\begin{bmatrix}-.0225&-13.5\\-28.875&-.1212\end{bmatrix}}={\begin{Bmatrix}z_{1}\\z_{2}\end{Bmatrix}}$
Where $z_{1}$ and $z_{2}$ are the modal coordinates. For mass matrix:
* ${\big\phi {\big ]}^{T}{\bigM{\big ]}{\big\phi {\big ]}={\begin{bmatrix}-.0225&-28.875\\-13.5&-.1212\end{bmatrix}}{\begin{bmatrix}3&0\\0&2\end{bmatrix}}{\begin{bmatrix}-.0225&-13.5\\-28.875&-.1212\end{bmatrix}}={\begin{bmatrix}1667.5&7.9\\7.9&546.8\end{bmatrix}}$
For damping matrix:
* ${\big\phi {\big ]}^{T}{\bigC{\big ]}{\big\phi {\big ]}={\begin{bmatrix}-.0225&-28.875\\-13.5&-.1212\end{bmatrix}}{\begin{bmatrix}3/4&-1/4\\-1/4&7/12\end{bmatrix}}{\begin{bmatrix}-.0225&-13.5\\-28.875&-.1212\end{bmatrix}}={\begin{bmatrix}486.0388&-95.18\\-95.18&135.878\end{bmatrix}}$
For Stiffness matrix:
* ${\big\phi {\big ]}^{T}{\bigK{\big ]}{\big\phi {\big ]}={\begin{bmatrix}-.0225&-28.875\\-13.5&-.1212\end{bmatrix}}{\begin{bmatrix}30&-20\\-20&35\end{bmatrix}}{\begin{bmatrix}-.0225&-13.5\\-28.875&-.1212\end{bmatrix}}={\begin{bmatrix}29156&-7665\\-7665&5403\end{bmatrix}}$
With the results the equation of motion for the system can be represented by modal coordinates in two 1-degree of freedom equations, with modal superposition we obtain:
| $1667.5{\ddot {z_{1}}}+486.0388{\dot {z_{1}}}+29156z_{1}=0$ | Eq. 8 |
and
| $546.8{\ddot {z_{2}}}+135.878{\dot {z_{2}}}+5403z_{2}=0$ | Eq. 8 |