User:EGM4313.f14.Team1.Linehan/Report 3

=Report 3=

Problem Statement
Compare the standard eigenvalue problem and generalized eigenvalue problem. Discuss in detail the similarities and the differences between

the eigenpairs (eigenvalues and eigenvectors) obtained from problem R2.5 and those obtained from problem R2.6

Solution
The following eigenvalues were obtained for problem Problem 2.5.


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$$ \beta_1 = 5 $$
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$$ \beta_2 = 15 $$
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The values above yielded the following eigenvectors.


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$$ \boldsymbol \phi_1 = \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$
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$$ \boldsymbol \phi_2 = \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} $$
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The following eigenvalues were obtained for problem Problem 2.6.


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$$ \beta_1 = \frac{50}{13} $$
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$$ \beta_2 = \frac{150}{13} $$
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The values above yielded the following eigenvectors.


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$$ \boldsymbol \phi_1 = \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$
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$$ \boldsymbol \phi_2 = \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} $$
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As displayed above, while problems 2.5 and 2.6 produced different eigenvalues, the eigenvectors ended up being the same values.

While problems 2.5 and 2.6 were both given the same problem to solve, problem 2.5 was solved in standard eigenvalue form while problem 2.6 was solved in general eigenvalue form.

The difference in the methods of solving the problems (standard vs general) explains why different eigenvalues were obtained but the same eigenvectors were produced at the end of the problems.

General eigenvalue problems take the form


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$$ \mathbf A \mathbf x = \mathbf \lambda \mathbf M \mathbf x $$
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while standard eigenvalue problems take the form.


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$$ \mathbf A \mathbf x = \mathbf \lambda \mathbf x $$
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Solving a standard eigenvalue problem starts by assuming a trial solution that takes the form of the displacement.


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$$ \mathbf d (t) = e^{\alpha t} \boldsymbol \phi $$
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taking the second derivative of the trial solution recovers,


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$$ \mathbf d'' (t) = \alpha^2 e^{\alpha t} \boldsymbol \phi $$
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Substituting this back into the undamped free vibration equation yields,


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$$ \mathbf M ( \alpha^2 e^{\alpha t} \boldsymbol \phi ) + \mathbf K (e^{\alpha t} \boldsymbol \phi) = \mathbf 0 $$
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After combining like terms


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$$ [ \alpha^2 \mathbf M + \mathbf K] e^{\alpha t} \boldsymbol \phi = \mathbf 0 $$
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Noting that $$ e^{\alpha t} $$ is never zero thus permits


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$$ [ \mathbf M \alpha^2 + \mathbf K ] \boldsymbol \phi = \mathbf 0 \Rightarrow \mathbf K \boldsymbol \phi = - \alpha^2 \mathbf M \boldsymbol \phi $$
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where we then restrict $$ \alpha $$ to be imaginary and let,


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$$ \beta := -\alpha^2 > 0 $$
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This is the basis for all standard eigenvalue problems.

To recap, while both problems produced the same eigenvectors, each problem had a different set of eigenvalues.

The difference in eigenvalues was due to the different methods in which the problems were solved.

These methods were the standard and general methods for solving eigenvalue problems. Both forms were highlighted above.

Problem Statement
Generalized eigenvalue problem, eigenvalues, eigenvectors, normalized eigenvectors, orthogonality with respect to mass and stiffness matrices (continuation of R1.6, R2.5, R2.6).

Similar to sec.53d, Pb.53-7, but set the 2nd coefficient in each eigenvector to -5 (instead of to 1), i.e., set

$$ \boldsymbol \phi_i = \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} = \begin{Bmatrix} \phi_{i1} \\ -5 \end{Bmatrix}, \ i = 1,2 $$

Then, find the 1st coefficient of each eigenvector.

Normalize the resulting eigenvectors with respect to the mass matrix, and animate each eigenvector using desmos.

Compare the animated eigenvectors in this problem to those in R2.6, in which the 1st coefficient in each eigenvector was set to 1.

Provide screenshots and web address of desmos animations.

Solution
Eigenvalues

Referring back to the work from R1.6, we found that


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$$ \mathbf M \mathbf d'' + \mathbf C \mathbf d' + \mathbf K \mathbf d = F(t) $$
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$$ \mathbf M = \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} $$
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$$ \mathbf C = \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} $$
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$$ \mathbf K = \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} $$
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$$ \mathbf F = \begin{Bmatrix} f_1(t) \\ f_2(t) \end{Bmatrix} $$
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$$ \mathbf d = \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} $$
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The combined equation


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$$ \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} + \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} \begin{Bmatrix} d'_1 \\ d'_2 \end{Bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} = \begin{Bmatrix} f_1(t) \\ f_2(t) \end{Bmatrix} $$
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Generalized Eigenvalue Problem (GEP)


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$$ \mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
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$$ [ \mathbf K - \beta \mathbf M ] \boldsymbol \phi = 0 $$
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To have zero solutions for phi would be trivial. Therefore, we must have:


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$$ \det [ \mathbf K - \beta \mathbf M ] = 0 $$
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Evaluate what we are finding the determinant of.


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$$ \mathbf K - \beta \mathbf M = \begin{bmatrix} K_{11} & K_{12} \\ K_21 & K_22 \end{bmatrix} - \beta \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} = \begin{bmatrix} K_11 - m_1 \beta & K_12 \\ K_21 & K_22 - m_2 \beta \end{bmatrix} $$
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Plugging in the given values gives the following equation.


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$$ \mathbf K - \beta \mathbf M = \begin{bmatrix} 10 - 1.3 \beta & -5 \\ -5 & 10 - 1.3 \beta \end{bmatrix} $$
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The determinant of this can then be calculated.


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$$ \det \begin{bmatrix} 10 - \beta 1.3 & -5 \\ -5 & 10 - \beta 1.3 \end{bmatrix} = 0 $$
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The following quadratic is produced.


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$$ 1.69 \beta^2 -26\beta + 75 = 0 $$
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Two roots or eigenvalues are found from the quadratic.


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$$ \beta_1 = \frac{50}{13} $$
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$$ \beta_2 = \frac{150}{13} $$
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Eigenvectors

From the GEP and the two eigenvaules we just found,


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$$ \mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
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$$ \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} = \beta_i \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} $$
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where i = 1, 2

When i = 1 and when setting the second coefficient in each eigenvector equal to negative five as advised in the problem


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$$ ( \mathbf K - \beta \mathbf M ) \boldsymbol \phi_1 = \begin{bmatrix} 10 - 1.3 \beta_1 & -5 \\ -5 & 10 - 1.3 \beta_1 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\ -5 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
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Plugging in the value of $$ \beta_1 $$ and simplifying yields the following


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$$ ( \mathbf K - \beta \mathbf M ) \boldsymbol \phi_1 = \begin{bmatrix} 5 & -5 \\ -5 & 5 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\ -5 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
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$$ ( \mathbf K - \beta \mathbf M ) \boldsymbol \phi_1 = \begin{Bmatrix} 5 \phi_{11} + 25 \\ -5 \phi_{11} - 25 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
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Now $$ \phi_{11} $$ can be solved for to get


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$$ \phi_{11} = -5 $$
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Therefore, the first eigenvector is


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$$ \boldsymbol \phi_1 = \begin{Bmatrix} -5 \\ -5 \end{Bmatrix} $$
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The second eigenvector is found the same way as the first, except $$ \beta_2 $$ is used. This yields


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$$ \boldsymbol \phi_2 = \begin{Bmatrix} 5 \\ -5 \end{Bmatrix} $$
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Normalized Eigenvectors

Modal mass of mode 1:


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$$ \boldsymbol \phi_1^{T} \mathbf M \boldsymbol \phi_1 = \begin{Bmatrix} -5 & -5 \end{Bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} -5 \\ -5 \end{Bmatrix} = 65 $$
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1st normalized eigenvector:


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$$ \bar{\boldsymbol \phi_1} = \frac{\boldsymbol \phi_1}{\sqrt{\bar{m_1}}} = \frac{1}{\sqrt{65}} \begin{Bmatrix} -5 \\ -5 \end{Bmatrix} $$
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Doing the same process to find the 2nd normalized eigenvector yields,


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$$ \boldsymbol \phi_2^{T} \mathbf M \boldsymbol \phi_2 = \begin{Bmatrix} 5 & -5 \end{Bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 5 \\ -5 \end{Bmatrix} = 65 $$
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$$ \bar{\boldsymbol \phi_2} = \frac{1}{\sqrt{65}} \begin{Bmatrix} 5 \\ -5 \end{Bmatrix} $$
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Orthogonality

Use the dot product to find if the eigenvectors are orthogonal to eachother.


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$$ \boldsymbol \phi_1 \cdot \boldsymbol \phi_2 = \begin{Bmatrix} -5 \\ -5 \end{Bmatrix} \cdot \begin{Bmatrix} 5 \\ -5 \end{Bmatrix} = 0 $$
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Check if the eigenvectors are orthogonal to the mass matrix.


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$$ \boldsymbol \phi_1^{T} \mathbf M \boldsymbol \phi_2 = \begin{Bmatrix} -5 & -5 \end{Bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 5 \\ -5 \end{Bmatrix} = 0 $$
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Therefore, the eigenvectors are orthogonal to the mass matrix.

Verify whether the Damping Matrix is Diagonalizable

The damping matrix, as found in R1.6, is


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\mathbf C = \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} $$
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The transpose of this matrix is as follows.


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\mathbf C^T = \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} $$
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It can be seen that the damping matrix and it's transpose are equivalent.

Now it must be checked that $$ \mathbf C \mathbf C^T = \mathbf C^T \mathbf C $$.

Since $$ \mathbf C = \mathbf C^T $$ the multiplication of these two matrices in either order will yield the same result.


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\begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 12 & -16 \\ -16 & 12 \end{bmatrix} $$
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Therefore, the damping matrix is diagonalizable.

Animations

video Mode 1 link: https://www.desmos.com/calculator/0fxmi5y06k

video Mode 2 link: https://www.desmos.com/calculator/g4jjfkirkd

Problem Statement
Generalized eigenvalue problems, comparing the eigenpairs from the following slightly different models.

Case 1: use the example in my lecture notes, with the non-uniform stiffness / mass / damping coefficients defined on p.53c-25 (i.e., sec.53c, p.53-21).

i showed how to obtain the eigenpairs for this example, and the corresponding animations in class. you just use the results i obtained to do the comparison / discussion.

Case 2: use the example in R2.6, with the uniform stiffness / mass / damping coefficients defined in sec.53d, Pb.53-7, for which you have

solved for the eigenpairs in your report R2 (by setting the 1st coefficient in each eigenvector to 1). you just use the solution for this problem to do the comparison / discussion.

Case 3: similar to case 2, but with the non-uniform mass coefficients as in case 1; the uniform stiffness / damping coefficients

are the same as in case 2. you need to solve for the eigenpairs (set the 1st coefficient in each eigenvector to 1), and create the animation for these vibration modes.

Case 4: similar to case 2, but with the non-uniform stiffness coefficients as in case 1; the uniform mass / damping coefficients are

the same as in case 2. you need to solve for the eigenpairs (set the 1st coefficient in each eigenvector to 1), and create the animation for these vibration modes.

For mode 1, create a single desmos animation page showing the motion of the 2 masses for these 4 different cases, with each case on a different

ordinate in the graph area; plot also displacement versus time for these 4 cases.

For mode 2, ditto (do the same as for mode 1).

In your solution, provide screenshots and links to your animations on the desmos site.

Provide a detailed discussion (similarities and differences) of the results. Provide screenshots and web address of desmos animations.

Solution
From deriving the equation of motion of a two degree of freedom spring-mass-damper system, we know the following:


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$$ \mathbf M = \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix}$$
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$$\mathbf K = \begin{bmatrix} K_{11} & K_{12} \\ K_{21} & K_{22} \end{bmatrix}$$, where $$K_{11} = k_1 + k_2$$, $$K_{12} = K_{21} = -k_2$$, and $$K_{22} = k_2 + k_3$$
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In order to solve for the eigenpairs for given values of $$\mathbf K$$ and $$\mathbf M$$, we must first get the Generalized Eigenvalue Problem (GEP).

Because doing so requires looking at the undamped free vibration equation of motion, the damping coefficients have no effect on the values of the eigenpairs.


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$$ \mathbf M \mathbf d'' + \mathbf K \mathbf d = \mathbf 0 $$     (1)
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The trial solution of (1) is assumed to be in the form


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$$ \mathbf d(t) = e^{\alpha t} \boldsymbol \phi $$
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Substituting the trial solution into (1) gives


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$$ \mathbf M \alpha^2 e^{\alpha t} \boldsymbol \phi + \mathbf K e^{\alpha t} \boldsymbol \phi =\mathbf 0 $$
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We can divide out $$e^{\alpha t}$$, because it is never equal to 0. We can also factor out matrix $$\boldsymbol \phi$$. If we make the substitution $$\alpha^2 = -\beta$$, the equation becomes


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$$ (\mathbf K - \beta \mathbf M) \boldsymbol \phi = \mathbf 0 $$     (2)
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(2) is known as the Generalized Eigenvalue Problem, where $$\beta$$ represents one eigenvalue and $$\boldsymbol \phi$$ represents the corresponding eigenvector. Because we want there to be infinite solutions to the GEP, we want


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$$ \begin{vmatrix}\mathbf K - \beta \mathbf M\end{vmatrix} = \mathbf 0 $$     (3)
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In a two degree of freedom system, solving (3) will give both eigenvalues. A higher order system will give the corresponding number

of eigenvalues (three eigenvalues for three degrees of freedom, four for four, etc.). The eigenvalues will come from substituting each

eigenvalue back into (2), setting either coefficient of the eigenvector to any number, and solving for the second coefficient. Each eigenvalue

with its corresponding eigenvector is called an eigenpair. Once both eigenvalues are reached, (2) can be written as the following two equations:


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$$ \mathbf K \boldsymbol \phi_1 = \beta_1 \mathbf M \boldsymbol \phi_1 $$     (4)
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$$ \mathbf K \boldsymbol \phi_2 = \beta_2 \mathbf M \boldsymbol \phi_2 $$     (5)
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Where $$\beta_1$$ and $$\beta_2$$ are both eigenvalues, and $$\boldsymbol \phi_1$$ and

$$\boldsymbol \phi_2$$ are the eigenvectors corresponding to the respective eigenvalues.

To get the modal equations, the eigenvectors need to be normalized with respect to the mass matrix.

To do so, we need to calculate the modal mass of each eigenvector for each case. The modal mass is calculated by the equation


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$$ \overline {m_i} = \boldsymbol \phi_i^T \mathbf M \boldsymbol \phi_i $$     (6)
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Where $$i$$ is $$1$$ or $$2$$, depending on which eigenvector is being normalized.

Once the modal mass is calculated, the eigenvector is normalized by the equation


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$$ \overline {\boldsymbol \phi_i} = \frac{\boldsymbol \phi_i}{\sqrt{\overline{m_i}}} $$     (7)
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Case 1: Non-uniform mass and stiffness coefficients

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$$k_1=14-\sqrt{6}$$
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$$k_2=\sqrt{6}$$

$$k_3=3-\sqrt{6}$$
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$$m_1=2$$
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$$m_2=\frac{1}{2}$$
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Substituting the values into the matrix equations gives


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$$ \mathbf K = \begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & 3 \end{bmatrix} $$
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$$ \mathbf M = \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} $$
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Substituting $$\mathbf K$$ and $$\mathbf M$$ into (2) gives


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$$ \begin{bmatrix} 14-2\beta & -\sqrt{6} \\ -\sqrt{6} & 3-\frac{1}{2}\beta \end{bmatrix} \boldsymbol \phi = \mathbf 0 $$
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(3) is now


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$$ \begin{vmatrix} 14-2\beta & -\sqrt{6} \\ -\sqrt{6} & 3-\frac{1}{2}\beta \end{vmatrix} = 0 $$
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Which is equivalent to the equation


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$$ (14-2\beta)(3-\frac{1}{2}\beta)-(-\sqrt{6})(-\sqrt{6}) = 0 $$
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Which simplifies to the quadratic equation


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$$ \beta^2-13\beta+36= 0 $$     (8)
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The two roots of (8) are 4 and 9, which are the two eigenvalues of Case 1. We can now rewrite (4) as


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$$ \begin{bmatrix} 6 & -\sqrt{6} \\ -\sqrt{6} & 1\end{bmatrix} \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} = \mathbf 0 $$
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For simplicity, we set $$\phi_{11}$$ equal to 1, although we could set either coefficient of the eigenvector

to any number, and we solve for $$\phi_{12}$$.


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$$ 6(1)-\sqrt{6}\phi_{12} = 0 $$
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So, $$\phi_{12}=\sqrt{6}$$. Therefore, the first eigenpair of Case 1 is $$\beta_1=4, \boldsymbol \phi_1=\begin{Bmatrix} 1 \\ \sqrt{6} \end{Bmatrix}$$.

Now, we do the same for $$\beta_2$$ to get the second eigenpair.


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$$ \begin{bmatrix} -4 & -\sqrt{6} \\ -\sqrt{6} & \frac{-3}{2}\end{bmatrix} \begin{Bmatrix} \phi_{21} \\ \phi_{22} \end{Bmatrix} = \mathbf 0 $$
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As before, we set $$\phi_{21}$$ equal to 1 for simplicity and solve for $$\phi_{22}$$.


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$$ -4(1)-\sqrt{6}\phi_{22} = 0 $$
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 * }
 * }

So, $$\phi_{22}=\frac{-4}{\sqrt{6}}$$. Therefore, the second eigenpair of Case 1 is $$\beta_2=9, \boldsymbol \phi_2=\begin{Bmatrix} 1 \\ \frac{-4}{\sqrt{6}} \end{Bmatrix}$$

Using (6) to calculate the modal mass of the first eigenpair,


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & \sqrt{6} \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ -\sqrt{6} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & \sqrt{6} \end{bmatrix} \begin{Bmatrix} 2 \\ -\frac{\sqrt{6}}{2} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = 2 + \frac{6}{2} = 5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the first modal mass is $$5$$. We calculate the second modal mass the same way.


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & \frac{-4}{\sqrt{6}} \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ \frac{-4}{\sqrt{6}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & \frac{-4}{\sqrt{6}} \end{bmatrix} \begin{Bmatrix} 2 \\ \frac{-2}{\sqrt{6}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = 2 + \frac{8}{6} = \frac{10}{3} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the second modal mass is $$\frac{10}{3}$$

We can now normalize each eigenvector by (7).


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ \sqrt{6} \end{Bmatrix}}{\sqrt{5}} = \begin{Bmatrix} \frac{1}{\sqrt{5}} \\ \sqrt{\frac{6}{5}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ \frac{-4}{\sqrt{6}} \end{Bmatrix}}{\sqrt{\frac{10}{3}}} = \begin{Bmatrix} \sqrt{\frac{10}{3}} \\ \frac{-2}{\sqrt{5}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 2: Uniform mass and stiffness coefficients

 * {| style="width:100%" border="0"

$$k_1=k_2=k_3=5$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$m_1=m_2=1.3$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting the values into the matrix equations gives


 * {| style="width:100%" border="0"

$$ \mathbf K = \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M = \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting $$\mathbf K$$ and $$\mathbf M$$ into (2) gives


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 10-1.3\beta & -5 \\ -5 & 10-1.3\beta \end{bmatrix} \boldsymbol \phi = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

(3) is now


 * {| style="width:100%" border="0"

$$ \begin{vmatrix} 10-1.3\beta & -5 \\ -5 & 10-1.3\beta \end{vmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which is equivalent to the equation


 * {| style="width:100%" border="0"

$$ (10-1.3\beta)^2-(-5)^2 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which simplifies to the quadratic equation


 * {| style="width:100%" border="0"

$$ 1.69\beta^2-26\beta+75= 0 $$     (9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The two roots of (9) are $$\frac{50}{13}$$ and $$\frac{150}{13}$$, which are the two eigenvalues of Case 2.

To get the first eigenvector, we now rewrite (4) as


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 5 & -5 \\ -5 & 5\end{bmatrix} \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For simplicity, we set $$\phi_{11}$$ equal to 1, although we could set either coefficient of the eigenvector

to any number, and we solve for $$\phi_{12}$$.


 * {| style="width:100%" border="0"

$$ 5(1)-5\phi_{12} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\phi_{12}=1$$. Therefore, the first eigenpair of Case 2 is $$\beta_1=\frac{50}{13}, \boldsymbol \phi_1=\begin{Bmatrix} 1 \\ 1 \end{Bmatrix}$$.

Now, we do the same for $$\beta_2$$ to get the second eigenpair.


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} -5 & -5 \\ -5 & -5\end{bmatrix} \begin{Bmatrix} \phi_{21} \\ \phi_{22} \end{Bmatrix} = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

As before, we set $$\phi_{21}$$ equal to 1 for simplicity and solve for $$\phi_{22}$$.


 * {| style="width:100%" border="0"

$$ -5(1)-5\phi_{22} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\phi_{22}=-1$$. Therefore, the second eigenpair of Case 2 is $$\beta_2=\frac{150}{13}, \boldsymbol \phi_2=\begin{Bmatrix} 1 \\ -1 \end{Bmatrix}$$

Using (6) to calculate the modal mass of the first eigenpair,


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{Bmatrix} 1.3 \\ 1.3 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = 1.3 + 1.3 = 2.6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the first modal mass is $$2.6$$. We calculate the second modal mass the same way.


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & -1 \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ -1 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & -1 \end{bmatrix} \begin{Bmatrix} 1.3 \\ -1.3 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = 1.3 + 1.3 = 2.6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the second modal mass is $$2.6$$

We can now normalize each eigenvector by (7).


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ 1 \end{Bmatrix}}{\sqrt{2.6}} = \begin{Bmatrix} \frac{1}{\sqrt{2.6}} \\ \frac{1}{\sqrt{2.6}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ -1 \end{Bmatrix}}{\sqrt{2.6}} = \begin{Bmatrix} \frac{1}{\sqrt{2.6}} \\ \frac{-1}{\sqrt{2.6}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 3: Non-uniform mass coefficients, uniform stiffness coefficients

 * {| style="width:100%" border="0"

$$k_1=k_2=k_3=5$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$m_1=2$$
 * style="width:95%" |
 * style="width:95%" |

$$m_2=\frac{1}{2}$$
 * }
 * }

Substituting the values into the matrix equations gives


 * {| style="width:100%" border="0"

$$ \mathbf K = \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M = \begin{bmatrix} 2 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting $$\mathbf K$$ and $$\mathbf M$$ into (2) gives


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 10-2\beta & -5 \\ -5 & 10-\frac{1}{2}\beta \end{bmatrix} \boldsymbol \phi = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

(3) is now


 * {| style="width:100%" border="0"

$$ \begin{vmatrix} 10-2\beta & -5 \\ -5 & 10-\frac{1}{2}\beta \end{vmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which is equivalent to the equation


 * {| style="width:100%" border="0"

$$ (10-2\beta)(10-\frac{1}{2}\beta)-(-5)^2 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which simplifies to the quadratic equation


 * {| style="width:100%" border="0"

$$ \beta^2-25\beta+75= 0 $$     (10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The two roots of (10) are $$\frac{-5(-5+\sqrt{13})}{2}$$ and $$\frac{5(5+\sqrt{13})}{2}$$, which are the two eigenvalues of Case 3. To get the first eigenvector, we now rewrite (4) as


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} -15+5\sqrt{13} & -5 \\ -5 & \frac{15}{4}+\frac{5\sqrt{13}}{4}\end{bmatrix} \begin{Bmatrix} \phi_{21} \\ \phi_{22} \end{Bmatrix} = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For simplicity, we set $$\phi_{11}$$ equal to 1, although we could set either coefficient of the eigenvector to any number, and we solve for $$\phi_{12}$$.


 * {| style="width:100%" border="0"

$$ (-15+5\sqrt{13})(1)-5\phi_{22} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\phi_{12}=-3+\sqrt{13}$$. Therefore, the first eigenpair of Case 3 is $$\beta_1=\frac{-5(-5+5\sqrt{13})}{2}, \boldsymbol \phi_1=\begin{Bmatrix} 1 \\ -3+\sqrt{13} \end{Bmatrix}$$

Now, we do the same for $$\beta_2$$ to get the second eigenpair.


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} -15-5\sqrt{13} & -5 \\ -5 & \frac{15}{4}-\frac{5\sqrt{13}}{4}\end{bmatrix} \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

As before, we set $$\phi_{21}$$ equal to 1 for simplicity and solve for $$\phi_{22}$$.


 * {| style="width:100%" border="0"

$$ (-15-5\sqrt{13})(1)-5\phi_{22} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\phi_{22}=-3-\sqrt{13}$$. Therefore, the second eigenpair of Case 3 is

$$\beta_2=\frac{5(5+\sqrt{13})}{2}, \boldsymbol \phi_2=\begin{Bmatrix} 1 \\ -3-\sqrt{13} \end{Bmatrix}$$.

Using (6) to calculate the modal mass of the first eigenpair,


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & -3+\sqrt{13} \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ -3+\sqrt{13} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & -3+\sqrt{13} \end{bmatrix} \begin{Bmatrix} 2 \\ \frac{-3}{2}+\frac{\sqrt{13}}{2} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = 2 + \frac{1}{2}(-3+\sqrt{13})^2 = -3\sqrt{13}+13 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the first modal mass is $$-3\sqrt{13}+13$$. We calculate the second modal mass the same way.


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & -3-\sqrt{13} \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ -3-\sqrt{13} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & -3-\sqrt{13} \end{bmatrix} \begin{Bmatrix} 2 \\ \frac{-3}{2}-\frac{\sqrt{13}}{2} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = 2 + \frac{1}{2}(-3-\sqrt{13})^2 = 3\sqrt{13}+13 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the second modal mass is $$3\sqrt{13}+13$$

We can now normalize each eigenvector by (7).


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ -3+\sqrt{13} \end{Bmatrix}}{\sqrt{-3\sqrt{13}+13}} = \begin{Bmatrix} \frac{1}{\sqrt{-3\sqrt{13}+13}} \\ \frac{-3+\sqrt{13}}{\sqrt{-3\sqrt{13}+13}} \end{Bmatrix} \approx \begin{Bmatrix} 0.6768 \\ 0.4098 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ -3-\sqrt{13} \end{Bmatrix}}{\sqrt{3\sqrt{13}+13}} = \begin{Bmatrix} \frac{1}{\sqrt{3\sqrt{13}+13}} \\ \frac{-3-\sqrt{13}}{\sqrt{3\sqrt{13}+13}} \end{Bmatrix} \approx \begin{Bmatrix} 0.2049 \\ -1.3535 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 4: Uniform mass coefficients, non-uniform stiffness coefficients

 * {| style="width:100%" border="0"

$$k_1=14-\sqrt{6}$$
 * style="width:95%" |
 * style="width:95%" |

$$k_2=\sqrt{6}$$

$$k_3=3-\sqrt{6}$$
 * }
 * }


 * {| style="width:100%" border="0"

$$m_1=m_2=1.3$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting the values into the matrix equations gives


 * {| style="width:100%" border="0"

$$ \mathbf K = \begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & 3 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M = \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Substituting $$\mathbf K$$ and $$\mathbf M$$ into (2) gives


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 14-1.3\beta & -\sqrt{6} \\ -\sqrt{6} & 3-1.3\beta \end{bmatrix} \boldsymbol \phi = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

(3) is now


 * {| style="width:100%" border="0"

$$ \begin{vmatrix} 14-1.3\beta & -\sqrt{6} \\ -\sqrt{6} & 3-1.3\beta \end{vmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which is equivalent to the equation


 * {| style="width:100%" border="0"

$$ (14-1.3\beta)(3-1.3\beta)-(-\sqrt{6})^2 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which simplifies to the quadratic equation


 * {| style="width:100%" border="0"

$$ 1.69\beta^2-22.1\beta+36= 0 $$     (11)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The two roots of (11) are $$\frac{-5(\sqrt{145}-17)}{13}$$ and $$\frac{5(\sqrt{145}+17)}{13}$$, which are the two eigenvalues of Case 4. To get the first eigenvector, we now rewrite (4) as


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} \frac{11}{2}+\frac{\sqrt{145}}{2} & -\sqrt{6} \\ -\sqrt{6} & -\frac{11}{2}+\frac{\sqrt{145}}{2}\end{bmatrix} \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For simplicity, we set $$\phi_{11}$$ equal to 1, although we could set either coefficient of the eigenvector to any number,

and we solve for $$\phi_{12}$$.


 * {| style="width:100%" border="0"

$$ (\frac{11}{2}+\frac{\sqrt{145}}{2})(1)-\sqrt{6}\phi_{12} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\phi_{12}=\frac{(\sqrt{145}+11)\sqrt{6}}{12}$$. Therefore, the first eigenpair of Case 4 is

$$\beta_1=\frac{-5(\sqrt{145}-17)}{13}, \boldsymbol \phi_1=\begin{Bmatrix} 1 \\ \frac{(\sqrt{145}+11)\sqrt{6}}{12} \end{Bmatrix}$$.

Now, we do the same for $$\beta_2$$ to get the second eigenpair.


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} \frac{11}{2}-\frac{\sqrt{145}}{2} & -\sqrt{6} \\ -\sqrt{6} & -\frac{11}{2}-\frac{\sqrt{145}}{2}\end{bmatrix} \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

As before, we set $$\phi_{21}$$ equal to 1 for simplicity and solve for $$\phi_{22}$$.


 * {| style="width:100%" border="0"

$$ (\frac{11}{2}-\frac{\sqrt{145}}{2})(1)-\sqrt{6}\phi_{22} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\phi_{22}=\frac{-(\sqrt{145}-11)\sqrt{6}}{12}$$. Therefore, the second eigenpair of Case 4 is

$$\beta_2=\frac{5(\sqrt{145}+17)}{13}, \boldsymbol \phi_2=\begin{Bmatrix} 1 \\ \frac{-(\sqrt{145}-11)\sqrt{6}}{12} \end{Bmatrix}$$

Using (6) to calculate the modal mass of the first eigenpair,


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & \frac{(\sqrt{145}+11)\sqrt{6}}{12} \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ \frac{(\sqrt{145}+11)\sqrt{6}}{12} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & \frac{(\sqrt{145}+11)\sqrt{6}}{12} \end{bmatrix} \begin{Bmatrix} 1.3 \\ \frac{13(\sqrt{145}+11)\sqrt{6}}{120} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = 1.3 + \frac{13(11\sqrt{145}+133)}{120} = \frac{377}{24}+\frac{143\sqrt{145}}{120} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the first modal mass is $$\frac{377}{24}+\frac{143\sqrt{145}}{120}$$. We calculate the second modal mass the same way.


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & -\frac{(\sqrt{145}-11)\sqrt{6}}{12} \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ -\frac{(\sqrt{145}-11)\sqrt{6}}{12} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & -\frac{(\sqrt{145}-11)\sqrt{6}}{12} \end{bmatrix} \begin{Bmatrix} 1.3 \\ \frac{-13(\sqrt{145}-11)\sqrt{6}}{120} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = 1.3 + \frac{-13(11\sqrt{145}-133}{120} = \frac{377}{24}-\frac{143\sqrt{145}}{120} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the second modal mass is $$\frac{377}{24}-\frac{143\sqrt{145}}{120}$$

We can now normalize each eigenvector by (7).


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ \frac{(\sqrt{145}+11)\sqrt{6}}{12} \end{Bmatrix}}{\frac{377}{24}+\frac{143\sqrt{145}}{120}} = \begin{Bmatrix} \frac{-(11\sqrt{145}-145)}{377} \\ \frac{2\sqrt{870}}{377} \end{Bmatrix} \approx \begin{Bmatrix} 0.0333 \\ 0.1565 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ \frac{-(\sqrt{145}-11)\sqrt{6}}{12} \end{Bmatrix}}{\frac{377}{24}-\frac{143\sqrt{145}}{120}} = \begin{Bmatrix} \frac{11\sqrt{145}+145}{377} \\ \frac{-2\sqrt{870}}{377} \end{Bmatrix} \approx \begin{Bmatrix} 0.7360 \\ -0.1565 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Mode 1
To summarize, Mode 1 is consisted of the normalized first eigenvectors, $$\overline {\boldsymbol \phi_1}$$, for each of the four cases.

Case 1:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \begin{Bmatrix} \frac{1}{\sqrt{5}} \\ \sqrt{\frac{6}{5}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 2:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \begin{Bmatrix} \frac{1}{\sqrt{2.6}} \\ \frac{1}{\sqrt{2.6}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 3:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \begin{Bmatrix} \frac{1}{\sqrt{-3\sqrt{13}+13}} \\ \frac{-3+\sqrt{13}}{\sqrt{-3\sqrt{13}+13}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 4:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \begin{Bmatrix} \frac{-(11\sqrt{145}-145)}{377} \\ \frac{2\sqrt{870}}{377} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Animation of Mode 1: https://www.desmos.com/calculator/f9ws1jjdud

Mode 2
As with Mode 1, Mode 2 is composed of the normalized second eigenvectors, $$\overline {\boldsymbol \phi_2}$$, for each of the four cases.

Case 1:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \begin{Bmatrix} \sqrt{\frac{10}{3}} \\ \frac{-2}{\sqrt{5}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 2:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \begin{Bmatrix} \frac{1}{\sqrt{2.6}} \\ \frac{-1}{\sqrt{2.6}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 3:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \begin{Bmatrix} \frac{1}{\sqrt{3\sqrt{13}+13}} \\ \frac{-3-\sqrt{13}}{\sqrt{3\sqrt{13}+13}} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Case 4:
 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \begin{Bmatrix} \frac{11\sqrt{145}+145}{377} \\ \frac{-2\sqrt{870}}{377} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Animation of Mode 2: https://www.desmos.com/calculator/mhiyhkbp9y

Comparison
As can be seen in the animations, both masses in Mode 1 are in phase with each other, but the the mass in Mode 2

are out of phase with each other. This is because, for Mode 1, both coefficients of each eigenvector have the same sign.

In Mode 2, however, the coefficients of each eigenvector have the opposite signs.

Mode 1:



Mode 2:



Problem Statement
Use the same structure as in R2.7, with the same stiffness / mass coefficients as in the example in the lecture notes on p.53c-25 (i.e., sec.53c, p.53-21). Also, use the parameters E,A,L are as defined in R2.7.


 * {| style="width:100%" border="0"

$$ k_1=14-\sqrt{6}, k_2=\sqrt{6}, k_3=3-\sqrt{6}, $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$m_1=2, m_2=\frac{1}{2}, $$ :{| style="width:100%" border="0" $$c_1=\frac{(76-11\sqrt{6})}{10}, c_2=\frac{11\sqrt{6}}{10}, c_3=\frac{(27-22\sqrt{6})}{20}, $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$E=2, A=\frac{1}{2}, L=5 $$ Part 1: Find the eigenpairs for this problem (set the 1st coefficient in each eigenvector to 1).
 * style="width:95%" |
 * style="width:95%" |
 * }

Do the desmos animation of the 2 modes, for both this problem and the example in the lecture notes.

Discuss in detail (similarities and differences) the results for these two cases.

The minimum is to observe the natural frequencies and the amplitudes of the modes, together with

explanations for these observations. Verify the orthogonality of the normalized eigenvectors with

respect to the mass matrix and stiffness matrix; do the same for the damping matrix.

Part 2: Using the same initial conditions, and apply the same “case 1” forces (sec.53c, Eq.(1)-(2) p.53-22,

h(t) = 1), as in the example in the lecture notes, find and solve the modal equations, satisfy the modal initial conditions,

then combine the modal displacements to obtain the displacements for the MDOF system of this problem.

Part 3: Compare the displacements for 2 cases: (1) the example in the lecture notes, (2) the structure in the current problem.

Plots and animations (if applicable) should be provided.

Part 1:
Referring to the solution obtained in P2.7, the EOM in matrix form is:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_1(t) \\ F_2(t) \end{Bmatrix} = \begin{bmatrix}   m_1 & 0 \\ 0 &  m_2 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix} \frac{EA}{L} & -\frac{EA}{L} \\ -\frac{EA}{L} & \frac{EA}{L} \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging in the new values:


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_1(t) \\ F_2(t) \end{Bmatrix} = \begin{bmatrix}   2 & 0 \\ 0 &  \frac{1}{2} \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \frac{1}{20} \begin{bmatrix}  152 & -22\sqrt{6} \\ -22\sqrt{6} & 27 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 14.2 & -\sqrt{6}-0.2 \\ -\sqrt{6}-0.2 & 3.2 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The Generalized Eigenvalue Problem (GEP) has the form:


 * {| style="width:100%" border="0"

$$ \mathbf K \phi = \beta \mathbf M \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

It can be rewritten as:
 * {| style="width:100%" border="0"

$$ [ \mathbf K - \beta \mathbf M ] \phi = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \det [ \mathbf K - \beta \mathbf M ] = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, we must determine the proper matrix to find the determinant of:


 * {| style="width:100%" border="0"

$$ \mathbf K - \beta \mathbf M = \begin{bmatrix} K_{11} & K_{12} \\ K_{21} & K_{22} \end{bmatrix} - \beta \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} = \begin{bmatrix} K_{11} - m_1 \beta & K_{12} \\ K_{21} & K_{22} - m_2 \beta \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging in the given values to the matrix:


 * {| style="width:100%" border="0"

$$ \mathbf K - \beta \mathbf M = \begin{bmatrix} 14.2 - 2 \beta & -\sqrt{6}-0.2 \\ -\sqrt{6}-0.2 & 3.2- 0.5 \beta \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, calculate the determinant:


 * {| style="width:100%" border="0"

$$ \det \begin{bmatrix} 14.2 - 2 \beta & -\sqrt{6}-0.2 \\ -\sqrt{6}-0.2 & 3.2- 0.5 \beta \end{bmatrix}  = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The following quadratic equation is produced:


 * {| style="width:100%" border="0"

$$ \beta^2 -13.5\beta + (39.4-0.4\sqrt{6}) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Using the quadratic equation, two eigenvalues can be computed:


 * {| style="width:100%" border="0"

$$ \beta_1 = 9.423 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \beta_2 = 4.077 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging these eigenvalues into the initial equation, we can solve for the eigenvector.

For $$ \beta_1 = 9.423 $$


 * {| style="width:100%" border="0"

$$ [ \mathbf K - \beta \mathbf M ] \phi = \begin{bmatrix} -4.645 & -\sqrt{6}-0.2 \\ -\sqrt{6}-0.2 & -1.511 \end{bmatrix} \begin{bmatrix} \phi_{11} \\ \phi_{12} \end{bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ -4.645\phi_{11} + (-\sqrt{6}-0.2)\phi_{12} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Setting $$ \phi_{11} = 1 $$:
 * {| style="width:100%" border="0"

$$ \phi_{12}= -1.751 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For $$ \beta_2 = 4.077 $$


 * {| style="width:100%" border="0"

$$ [ \mathbf K - \beta \mathbf M ] \phi = \begin{bmatrix} 5.845 & -\sqrt{6}-0.2 \\ -\sqrt{6}-0.2 & 1.161 \end{bmatrix} \begin{bmatrix} \phi_{21} \\ \phi_{22} \end{bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ 5.845\phi_{21} + (-\sqrt{6}-0.2)\phi_{22} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Setting $$ \phi_{21} = 1 $$:
 * {| style="width:100%" border="0"

$$ \phi_{22}= 2.206 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

And, thus, the eigenvectors are:


 * {| style="width:100%" border="0"

$$ \phi_1 = \begin{bmatrix} 1\\ -1.751 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \phi_2 = \begin{bmatrix} 1\\ 2.206 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \boldsymbol \phi= \begin{bmatrix} 1 & 1\\ -1.751 & 2.206 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Class example:


 * Mode 1: https://www.desmos.com/calculator/irresvzabu


 * Mode 2: https://www.desmos.com/calculator/cwtdvxe6nt

Homework problem:


 * Mode A: https://www.desmos.com/calculator/qkmoli9etk


 * Mode B: https://www.desmos.com/calculator/hcuwqavanq

''Note: Mode A corresponds to mode 1 of the class example. And, mode B corresponds to mode 2 of the class example.''

Comparing the class example to the homework problem:
 * Mode A has a higher amplitude than Mode 1.
 * Mode A has a slightly higher frequency than Mode 1.
 * Mode B has a higher amplitude than Mode 2.
 * Mode B has a slightly higher frequency than Mode 2.

Part 2:
Solving for the modal masses:


 * {| style="width:100%" border="0"

$$ \bar m_1 := \boldsymbol \phi_1^T \mathbf M \boldsymbol \phi_1 = \lfloor 1 \    -1.751 \rfloor \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ -1.751 \end{Bmatrix} = \lfloor 1 \     -1.751 \rfloor \begin{Bmatrix} 2 \\ -0.8755 \end{Bmatrix} = 2 + 1.533 = 3.533 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \bar m_2 := \boldsymbol \phi_2^T \mathbf M \boldsymbol \phi_2 = \lfloor 1  \   2.206 \rfloor \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ 2.206 \end{Bmatrix} = \lfloor 1   \    2.206 \rfloor \begin{Bmatrix} 2 \\ 1.103 \end{Bmatrix} = 2 + 2.433 = 4.433 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Creating normalized eigenvectors:


 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \phi}_1 = \frac{\boldsymbol \phi_1}{\sqrt{\bar m_1}} = \frac{1}{\sqrt{3.533}} \begin{Bmatrix} 1 \\ -1.751 \end{Bmatrix} = 0.532 \begin{Bmatrix} 1 \\ -1.751 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \phi}_2 = \frac{\boldsymbol \phi_2}{\sqrt{\bar m_2}} = \frac{1}{\sqrt{4.433}} \begin{Bmatrix} 1 \\ 2.206 \end{Bmatrix} = 0.475 \begin{Bmatrix} 1 \\ 2.206 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

This results in:
 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \Phi} := [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ] = \begin{bmatrix} 0.532 & 0.475 \\ -0.932 & 1.048 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now, expressing the displacement matrix in the basis of the normalized eigenvectors with the coordinates

being the modal coordinates:
 * {| style="width:100%" border="0"

$$ \mathbf d(t) = \sum_{j=1}^2 \bar{\boldsymbol \phi_j} \, y_j(t) = \bar{\boldsymbol \phi_1} y_1(t) + \bar{\boldsymbol \phi_2} y_2(t) = \lfloor \bar{\boldsymbol \phi_1} \ \bar{\boldsymbol \phi_2} \rfloor \begin{Bmatrix} y_1(t) \\ y_2(t) \end{Bmatrix} = \bar{\boldsymbol \Phi} \, \mathbf y(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Similarly, the velocity and acceleration matrices are:
 * {| style="width:100%" border="0"

$$ \mathbf d'(t) = \sum_{j=1}^2 \bar{\boldsymbol \phi_j} \, y'_j(t) = \bar{\boldsymbol \Phi} \, \mathbf y'(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf d(t) = \sum_{j=1}^2 \bar{\boldsymbol \phi_j} \, y_j(t) = \bar{\boldsymbol \Phi} \, \mathbf y''(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Recalling the EOM..
 * {| style="width:100%" border="0"

$$ \mathbf M {\mathbf d}'' + \mathbf C {\mathbf d}' + \mathbf K {\mathbf d} = \mathbf F (t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We can plug in for displacement, velocity, and acceleration:
 * {| style="width:100%" border="0"

$$ \mathbf M \bar{\boldsymbol \Phi} \, \mathbf y''(t) + \mathbf C \bar{\boldsymbol \Phi} \, \mathbf y'(t) + \mathbf K \bar{\boldsymbol \Phi} \, \mathbf y(t) = \mathbf F (t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Multiplying by the transpose of the normalized eigenvector:
 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \Phi}^T [ \mathbf M \bar{\boldsymbol \Phi} \, \mathbf y'' + \mathbf C \bar{\boldsymbol \Phi} \, \mathbf y' + \mathbf K \bar{\boldsymbol \Phi} \, \mathbf y ] = \bar{\boldsymbol \Phi}^T \mathbf F(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We get a new modal mass matrix:
 * {| style="width:100%" border="0"

$$ \mathbf M^\star := \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol \Phi} = \mathbf I = \text{Diag} [1, \, 1] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The damping matrix is not diagonizable because the normalized eigenvectors that would be used to diagonalize

it came from the generalized eigenvalue problem (GEP), which involved only the stiffness and mass matrices, NOT the damping matrix.


 * {| style="width:100%" border="0"

$$ \mathbf C^\star := \bar{\boldsymbol \Phi}^T \mathbf C \bar{\boldsymbol \Phi} = \begin{bmatrix} 0.532 & -0.932 \\ 0.475 & 1.048 \end{bmatrix} \begin{bmatrix} 6 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 0.532 & 0.475 \\ -0.932  & 1.048 \end{bmatrix} = \begin{bmatrix} 1.8110 & -2.7260 \\   -2.7260  & 5.7609 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Therefore, in order to develop the modal equations, we will use the values of $$ a_1 $$ and $$ a_2 $$ from the example in lecture.

New modal damping matrix:
 * {| style="width:100%" border="0"

$$ \mathbf C^\star := \bar{\boldsymbol \Phi}^T \mathbf C \bar{\boldsymbol \Phi} = \text{Diag} [a_1, \, a_2] = \text{Diag} [6, \, \frac{1}{2}] = \begin{bmatrix} 6 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

New modal stiffness matrix:
 * {| style="width:100%" border="0"

$$ \mathbf K^\star := \bar{\boldsymbol \Phi}^T \mathbf K \bar{\boldsymbol \Phi} = \text{Diag} [(\omega_1)^2, \, (\omega_2)^2] = \text{Diag} [b_1, \, b_2] = \text{Diag} [9.423 , \, 4.077] = \begin{bmatrix} 9.423 & 0 \\ 0 & 4.077 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Following the example given in class, the applied forces are:


 * {| style="width:100%" border="0"

$$ \mathbf F(t) = \begin{Bmatrix} F_1(t) \\ F_2(t) \end{Bmatrix} = \begin{Bmatrix} f_1 \\ f_2 \end{Bmatrix} h(t) = \frac{1}{\sqrt{5}} \begin{Bmatrix} 4 \\ \sqrt{6} \end{Bmatrix} h(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal applied forces:
 * {| style="width:100%" border="0"

$$ \mathbf r(t) = \begin{Bmatrix} r_1(t) \\ r_2(t) \end{Bmatrix} = \bar{\boldsymbol \Phi}^T \mathbf F(t) = {\begin{bmatrix} 0.532 & 0.475 \\ -0.932 & 1.048 \end{bmatrix}}^T \frac{1}{\sqrt{5}} \begin{Bmatrix} 4 \\ \sqrt{6} \end{Bmatrix} h(t)=\begin{Bmatrix} -0.0693 \\ 1.9977 \end{Bmatrix} h(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The "Case 1" condition is:


 * {| style="width:100%" border="0"

$$ h(t) = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Applying Case 1:
 * {| style="width:100%" border="0"

$$ \mathbf r(t) = \begin{Bmatrix} r_1(t) \\ r_2(t) \end{Bmatrix} = \bar{\boldsymbol \Phi}^T \mathbf F(t) = \begin{Bmatrix} -0.0693 \\ 1.9977 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The new EOM is...
 * {| style="width:100%" border="0"

$$ \mathbf M^\star \mathbf y'' + \mathbf C^\star \mathbf y' + \mathbf K^\star {\mathbf y} = \mathbf r(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} y_1 \\ y_2 \end{Bmatrix} + \begin{bmatrix} 6 & 0 \\ 0& \frac{1}{2}\end{bmatrix} \begin{Bmatrix} y'_1 \\ y'_2 \end{Bmatrix} + \begin{bmatrix} 9.423 & 0 \\ 0 & 4.077 \end{bmatrix} \begin{Bmatrix} y_1 \\ y_2 \end{Bmatrix} = \begin{Bmatrix} -0.0693 \\ 1.9977 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

And thus, modal equation 1 is:
 * {| style="width:100%" border="0"

$$ y_1''+6\,y_1'+9.423\,y_1 =-0.0693 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal equation 2 is:
 * {| style="width:100%" border="0"

$$ y_2'' + \frac{1}{2} \, y_2' + 4.077 \, y_2 = 1.9977 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving the second order ODE:
 * {| style="width:100%" border="0"

$$ \lambda''+6\lambda'+9.423\lambda=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The solution has two complex conjugate roots because


 * {| style="width:100%" border="0"

$$ b^2-4ac<0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$	\lambda_1=-3+0.65i $$
 * style="width:95%" |
 * style="width:95%" |
 * }


 * {| style="width:100%" border="0"

$$ \lambda_2=-3-0.65i $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Solving the other second order ODE:
 * {| style="width:100%" border="0"

$$ \lambda''+0.5\lambda'+4.077\lambda=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The solution has two complex conjugate roots because


 * {| style="width:100%" border="0"

$$ b^2-4ac<0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$	\lambda_3=-0.25+2i $$
 * style="width:95%" |
 * style="width:95%" |
 * }


 * {| style="width:100%" border="0"

$$ \lambda_4=-0.25-2i $$
 * style="width:95%" |
 * style="width:95%" |
 * }

The general solution is:
 * {| style="width:100%" border="0"

$$ y_1=c_1e^{-3x}sin(0.65x)+c_2e^{-3x}cos(0.65x)+c_3 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$ y_2=c_4e^{-0.25x}sin(2x)+c_5e^{-0.25x}cos(2x)+c_6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Taking the first derivative in order to solve for the constants:
 * {| style="width:100%" border="0"

$$ y'_1=c_1e^{-3x}(-3sin(0.65x)+0.65cos(0.65x))+c_2e^{-3x}(-3cos(0.65x)-0.65sin(0.65x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$ y'_2=c_4e^{-0.25x}(-0.25sin(2x)+2cos(2x))+c_5e^{-0.25x}(-0.25cos(2x)-2sin(2x)) $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Modal initial displacements:
 * {| style="width:100%" border="0"

$$ \mathbf d(0) = \begin{Bmatrix} d_{10} \\ d_{20} \end{Bmatrix} = \frac{1}{\sqrt{5}} \begin{Bmatrix} 1 \\ \sqrt{6} \end{Bmatrix} = \bar{\boldsymbol \phi}_1 \Rightarrow \mathbf y(0) = \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol \phi}_1 = \begin{Bmatrix} -0.0346 \\ 0.9989 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal initial velocities:
 * {| style="width:100%" border="0"

$$ \mathbf d'(0) = \begin{Bmatrix} v_{10} \\ v_{20} \end{Bmatrix} = \sqrt{\frac{3}{10}} \begin{Bmatrix} 1 \\ \frac{-4}{\sqrt{6}} \end{Bmatrix} = \bar{\boldsymbol \phi}_1 \Rightarrow \mathbf y'(0) = \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol \phi}_1 = \begin{Bmatrix} -1.207 \\ -0.563 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_1(0) = -0.0346, \ y'_1(0) = -1.207 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2(0) = 0.9989, \ y'_2(0) = -0.563 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging initial conditions into equation 1:
 * {| style="width:100%" border="0"

$$ -0.0346=c_2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ -1.207=0.65c_1+-3c_2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ c_1=-2.017 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plugging initial conditions into equation 2:
 * {| style="width:100%" border="0"

$$ 0.9989=c_5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ -0.563=2c_4-0.25c_5 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ c_4=-0.157 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Thus, the displacements are:


 * {| style="width:100%" border="0"

$$ y_1=-2.017e^{-3x}sin(0.65x)-0.0346e^{-3x}cos(0.65x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_2=-0.157e^{-0.25x}sin(2x)+0.9989e^{-0.25x}cos(2x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Part 3:
Plot of Modal Displacements:

https://www.desmos.com/calculator/ederqzmymi

As you can see from the graph of the modal displacements, $$y_1$$ stabilizes much faster than $$y_2$$

and has less oscillations. Both displacements remain very close to 0.

In the class example, the oscillating solution remained above the x-axis. This is different than the solution to the

homework problem in that the oscillating solution crossed the x-axis and at $$t=\infty $$, $$y=0$$.

The solution for $$y_2$$ in class looks very similar to the homework solution except that it is reflected

about the x-axis. Both approach $$y=0$$ as $$t=\infty $$.

Problem Statement
generalized eigenvalue problem; continuation of R2.8.

use the same structure as in R2.8, with the same stiffness / mass coefficients as in the example in my lecture notes on p.53c-25 (i.e., sec.53c, p.53-21).

the parameters E,A,L for each of the 2 elastic bars are as defined in R2.8.

find the eigenpairs for this problem (set the 1st coefficient in each eigenvector to 1).

do the desmos animation of the 2 modes for both this problem.

discuss in detail (similarities and differences) the results for 3 cases: (1) the example in my lecture notes, (2) the structure in R3.4,

and (3) the structure in the current problem.

the minimum is to observe the natural frequencies and the amplitudes of the modes, together with explanations for these observations (just like i did in class).

using the same initial conditions, and apply the same “case 1” forces (sec.53c, Eq.(1)-(2) p.53-22,

h(t) = 1), as in the example in my lecture notes, find and solve the modal equations,

satisfy the modal initial conditions, then combine the modal displacements to obtain the displacements for the MDOF system of this problem.

compare the displacements for 3 cases: (1) the example in my lecture notes, (2) the structure in R3.4, and (3) the structure in the

current problem. plots and animations (if applicable, screenshots and web address) should be provided.

Solution
Use FBD from Problem 2.8 of Report 2.



Stiffness Coefficients(Spring Constants)
 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

k_1 = 14 - \sqrt{6}, \ k_2 = \sqrt{6}, \ k_3 = 3 - \sqrt{6} $$
 * }
 * }

Mass Coefficients
 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

m_1 = 2, m_2 = \frac12 $$
 * }
 * }

Find the Eigenvalues using the GEP (Generalized Eigenvalue Problem)


 * {| style="width:100%" border="0"

$$
 * style="width:95%" |
 * style="width:95%" |

\mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
 * }
 * }

Rewrite as:
 * {| style="width:100%" border="0"

$$[ \mathbf K - \beta \mathbf M ] \boldsymbol \phi = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To find non-trivial (non-zero) solutions for column matrix Φ, we must have


 * {| style="width:100%" border="0"

$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * \mathbf K - \beta \mathbf M | = 0
 * }
 * }

Expand Determinant
 * {| style="width:100%" border="0"

$$ \begin{vmatrix} K_{11} - \beta m_1 & K_{12} \\ K_{21} & K_{22} - \beta m_2 \end{vmatrix} = 0 $$    (1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Find the Eigenvectors using the GEP
 * {| style="width:100%" border="0"

$$ \mathbf K \boldsymbol \phi = \beta \mathbf M \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Expand the GEP


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} K_{11} & K_{12} \\ K_{21} & K_{22} \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} =
 * style="width:95%" |
 * style="width:95%" |

\beta \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} $$
 * }
 * }

Plug in values for constants and values obtained through the GEP for the Eigenvalues to Solve for Phi


 * {| style="width:100%" border="0"

$$ ( \mathbf K - \beta_1 \mathbf M ) \boldsymbol \phi_1 = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} K_{11} - \beta_1 * m_1 & K_{12} \\ K_{21} & K_{22} - \beta_1 * m_2 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\
 * style="width:95%" |
 * style="width:95%" |

\phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$    (2)
 * 
 * }

Simplify to Solve for $$ \phi_{12} $$. Remember $$ \phi_{11} $$ is given:


 * {| style="width:100%" border="0"

$$ (K_{11} - \beta_1 * m_1)(\phi_{11}) + (K_{12})(\phi_{12}) = 0
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }


 * {| style="width:100%" border="0"

$$ \phi_{12} = \frac{-(K_{11} - \beta_1 * m_1)(\phi_{11})}{ (K_{12})}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

From R2.8, we know that the equation of motion is


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}   4 & -2 \\ -2 & 4  \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & \frac{91}{9} \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, $$\mathbf M = \begin{bmatrix}  1.3 & 0 \\ 0 &  1.3 \end{bmatrix} $$ and $$ \mathbf K = \begin{bmatrix} 10 & -5 \\ -5 & \frac{91}{9} \end{bmatrix}$$.

Substituting into (1) gives


 * {| style="width:100%" border="0"

$$ \begin{vmatrix} 10 - 1.3\beta & -5 \\ -5 & \frac{91}{9} - 1.3\beta \end{vmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Which is equivalent to the quadratic equation


 * {| style="width:100%" border="0"

$$ \frac{169\beta^2}{100} - \frac{2353\beta}{90} + \frac{685}{9} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving for $$\beta$$ gives two solutions: the two eigenvalues.


 * {| style="width:100%" border="0"

$$ \beta_1=\frac{-5(\sqrt{8101}-181)}{117} $$
 * style="width:95%" |
 * style="width:95%" |

$$ \beta_2=\frac{5(\sqrt{8101}+181)}{117} $$
 * }
 * }

Substituting $$\beta_1$$ into (2) gives


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} \frac{-1+\sqrt{8101}}{18} & -5 \\ -5 & \frac{1+\sqrt{8101}}{18} \end{bmatrix} \begin{Bmatrix} 1 \\ \phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving for $$ \phi_{12}$$ gives


 * {| style="width:100%" border="0"

$$ \frac{-1+\sqrt{8101}}{18} -5(\phi_{12}) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \phi_{12} = \frac{-1+\sqrt{8101}}{90} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the first eigenpair is $$\beta_1=\frac{-5(\sqrt{8101}-181)}{117}, \boldsymbol \phi_1=\begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix}$$.

Now we do the same with $$\beta_2$$ to get $$\boldsymbol \phi_2$$

Substituting $$\beta_2$$ into (2) gives


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} \frac{-1-\sqrt{8101}}{18} & -5 \\ -5 & \frac{1-\sqrt{8101}}{18} \end{bmatrix} \begin{Bmatrix} 1 \\ \phi_{22} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving for $$ \phi_{12}$$ gives


 * {| style="width:100%" border="0"

$$ \frac{-1-\sqrt{8101}}{18} -5(\phi_{22}) = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \phi_{22} = \frac{-1-\sqrt{8101}}{90} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the second eigenpair is $$\beta_2=\frac{5(\sqrt{8101}+181)}{117}, \boldsymbol \phi_2=\begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix}$$.

Now, to get the modes of the system, we need to normalize the eigenvectors. To normalize the eigenvectors, we need to find the

modal mass of each eigenvector. The modal mass is calculated by the equation


 * {| style="width:100%" border="0"

$$ \overline {m_i} = \boldsymbol \phi_i^T \mathbf M \boldsymbol \phi_i $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Where $$i$$ is $$1$$ or $$2$$, depending on which eigenvector is being normalized. Once the modal mass is

calculated, the eigenvector is normalized by the equation


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_i} = \frac{\boldsymbol \phi_i}{\sqrt{\overline{m_i}}} $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Starting by subsituting $$\boldsymbol \phi_1$$ into (3).


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & \frac{-1+\sqrt{8101}}{90} \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = \begin{bmatrix} 1 & \frac{-1+\sqrt{8101}}{90} \end{bmatrix} \begin{Bmatrix} 1.3 \\ \frac{13(\sqrt{8101}-1}{900} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_1} = 1.3 + \frac{-13(\sqrt{8101}-4051)}{40500} = \frac{105313-13\sqrt{8101}}{40500} \approx 2.5714 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the first modal mass is $$\frac{105313-13\sqrt{8101}}{40500}$$. We calculate the second modal mass the same way.


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & \frac{-1-\sqrt{8101}}{90} \end{bmatrix} \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = \begin{bmatrix} 1 & \frac{-1-\sqrt{8101}}{90} \end{bmatrix} \begin{Bmatrix} 1.3 \\ \frac{-13(\sqrt{8101}+1}{900} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {m_2} = 1.3 + \frac{13(\sqrt{8101}+4051)}{40500} = \frac{105313+13\sqrt{8101}}{40500} \approx 2.6292 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So the second modal mass is $$\frac{105313+13\sqrt{8101}}{40500}$$.

We can now normalize the eigenvectors by substituting each eigenvector and its respective modal mass into (4).


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313-13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \end{Bmatrix} \approx \begin{Bmatrix} 0.6236 \\ 0.6099 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313+13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{Bmatrix} \approx \begin{Bmatrix} 0.6167 \\ -0.6236 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

To summarize,

Mode 1

 * {| style="width:100%" border="0"

$$ \beta_1=\frac{-5(\sqrt{8101}-181)}{117} $$
 * style="width:95%" |
 * style="width:95%" |

$$ \overline {\boldsymbol \phi_1} = \frac{\begin{Bmatrix} 1 \\ \frac{-1+\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313-13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \end{Bmatrix} $$
 * }
 * }

Animation: https://www.desmos.com/calculator/ueb4vnare4



Mode 2

 * {| style="width:100%" border="0"

$$ \beta_2=\frac{5(\sqrt{8101}+181)}{117} $$
 * style="width:95%" |
 * style="width:95%" |

$$ \overline {\boldsymbol \phi_2} = \frac{\begin{Bmatrix} 1 \\ \frac{-1-\sqrt{8101}}{90} \end{Bmatrix}}{\sqrt{\frac{105313+13\sqrt{8101}}{40500}}} = \begin{Bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{Bmatrix} $$
 * }
 * }

Animation: https://www.desmos.com/calculator/3mj7z91soy



Projection
In order to solve for the modal displacement, we need to project the mass, stiffness, and damping matrices onto the eigendirections. This is done by the equation


 * {| style="width:100%" border="0"

$$ \mathbf M^\star = \overline {\boldsymbol \Phi^T} \mathbf M \overline {\boldsymbol \Phi} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Where $$\mathbf M^\star$$ is the modal mass matrix, $$\overline \boldsymbol \Phi$$ is the matrix of the eigenvectors, and $$\overline {\boldsymbol \Phi^T}$$ is the transposed matrix of the eigenvectors. The same process is done for the stiffness and damping matrices to get the respective modal matrices. If the original matrices are diagonalizable, this process will diagonalize the matrices, so the system can be uncoupled.

Modal mass matrix:


 * {| style="width:100%" border="0"

$$ \mathbf M^\star = \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} \begin{bmatrix}  1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf M^\star = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal stiffness matrix:


 * {| style="width:100%" border="0"

$$ \mathbf K^\star = \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} \begin{bmatrix} 10 & -5 \\ -5 & \frac{91}{9} \end{bmatrix} \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf K^\star = \begin{bmatrix} \frac{905-5\sqrt{8101}}{117} & 0 \\ 0 & \frac{905+5\sqrt{8101}}{117} \end{bmatrix} \approx \begin{bmatrix} 3.8887 & 0 \\ 0 & 11.5814 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal damping matrix:


 * {| style="width:100%" border="0"

$$ \mathbf C^\star = \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} \\ \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf C^\star = \begin{bmatrix} \frac{40}{13}-\frac{1800\sqrt{8101}}{105313} & \frac{2\sqrt{65618100}}{947817} \\ \frac{2\sqrt{65618100}}{947817} & \frac{40}{13}+\frac{1800\sqrt{8101}}{105313} \end{bmatrix} \approx \begin{bmatrix} 1.5386 & 0.01709 \\ 0.01709 & 4.6153 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Because $$\mathbf C^\star$$ is not a diagonal matrix, we will be using the diagonalized modal damping matrix from class.


 * {| style="width:100%" border="0"

$$ \mathbf C^\star = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 6 \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal force matrix:


 * {| style="width:100%" border="0"

$$ \mathbf r(t) = \begin{Bmatrix} r_1(t) \\ r_2(t) \end{Bmatrix} = \bar{\boldsymbol \Phi}^T \mathbf F(t) = \begin{bmatrix} \frac{90\sqrt{65}}{13\sqrt{8101-\sqrt{8101}}} & \frac{\sqrt{65}(\sqrt{8101}-1)}{13\sqrt{8101-\sqrt{8101}}} \\ \frac{90\sqrt{65}}{13\sqrt{8101+\sqrt{8101}}} & \frac{-\sqrt{65}(\sqrt{8101}+1)}{13\sqrt{8101+\sqrt{8101}}} \end{bmatrix} \begin{Bmatrix} \frac{4}{\sqrt{5}} \\ \sqrt{\frac{6}{5}} \end{Bmatrix} h(t) \approx \begin{Bmatrix} 1.7911 \\ 0.4201 \end{Bmatrix} h(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Modal Displacement
The new equation of motion is:


 * {| style="width:100%" border="0"

$$ \mathbf M^\star \mathbf y'' + \mathbf C^\star \mathbf y' + \mathbf K^\star {\mathbf y} = \mathbf r(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} y_1 \\ y_2 \end{Bmatrix} + \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 6 \end{bmatrix} \begin{Bmatrix} y'_1 \\ y'_2 \end{Bmatrix} + \begin{bmatrix} \frac{905-5\sqrt{8101}}{117} & 0 \\ 0 & \frac{905+5\sqrt{8101}}{117} \end{bmatrix} \begin{Bmatrix} y_1 \\ y_2 \end{Bmatrix} = \begin{Bmatrix} 1.7911 \\ 0.4201 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

So, the first modal displacement is defined by the equation


 * {| style="width:100%" border="0"

$$ y''_1(t)+\frac{1}{2}y'_1(t)+3.8887y_1(t)=1.7911 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving the ODE by first solving the homogeneous equation.


 * {| style="width:100%" border="0"

$$ \lambda_1^2+\frac{1}{2}\lambda_1+3.8887=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving for $$\lambda_1$$ gives two complex conjugates.


 * {| style="width:100%" border="0"

$$ \lambda_1=-0.25 \pm 1.9561i $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

From Euler's formula, we know that the homogeneous solution is


 * {| style="width:100%" border="0"

$$ y_{h1}(t)=e^{-0.25t}[C_{11}cos(1.9561t)+C_{12}sin(1.9561t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now we solve the particular solution


 * {| style="width:100%" border="0"

$$ y''_{p1}(t)+\frac{1}{2}y'_{p1}(t)+3.8887y_{p1}(t)=1.7911 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{p1}=0.4606 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The solution of the first modal equation is the homogeneous solution plus the particular solution.


 * {| style="width:100%" border="0"

$$ y_1(t)=e^{-0.25t}[C_{11}cos(1.9561t)+C_{12}sin(1.9561t)]+0.4606 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We now do the same to get the second modal equation of motion. The second modal displacement is defined by the equation


 * {| style="width:100%" border="0"

$$ y''_2(t)+6y'_2(t)+11.5814y_2(t)=0.4201 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving the ODE by first solving the homogeneous equation.


 * {| style="width:100%" border="0"

$$ \lambda_2^2+6\lambda_2+11.5814=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Solving for $$\lambda_1$$ gives two complex conjugates.


 * {| style="width:100%" border="0"

$$ \lambda_2=-3 \pm 1.6067i $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

From Euler's formula, we know that the homogeneous solution is


 * {| style="width:100%" border="0"

$$ y_{h2}(t)=e^{-3t}[C_{21}cos(1.6067t)+C_{22}sin(1.6067t)] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now we solve the particular solution


 * {| style="width:100%" border="0"

$$ y''_{p2}(t)+6y'_{p1}(t)+11.5814y_{p1}(t)=0.4201 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y_{p2}=0.03627 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The solution of the first modal equation is the homogeneous solution plus the particular solution.


 * {| style="width:100%" border="0"

$$ y_2(t)=e^{-3t}[C_{21}cos(1.6067t)+C_{22}sin(1.6067t)]+0.03627 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

In order to solve for the constants, we need to know the modal initial conditions.


 * {| style="width:100%" border="0"

$$ y(0) = \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol \phi_1} = \begin{Bmatrix} 1 \\ -0.005586 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ y'(0) = \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol \phi_2} = \begin{Bmatrix} -0.005586 \\ 1.0114 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We can now solve for the constants in both solutions.


 * {| style="width:100%" border="0"

$$ y_1(0)=1=e^{0}[C_{11}cos(0)+C_{12}sin(0)]+0.4606=C_{11}+0.4606 $$
 * style="width:95%" |
 * style="width:95%" |

$$ C_{11}=0.5394 $$

$$ y'_1(0)=-0.005586=-0.25e^{0}[0.5394cos(0)+C_{12}sin(0)]+1.9561e^{0}[-0.5394sin(0)+C_{12}cos(0)]=-0.1349+1.9561C_{12} $$

$$ C_{12}=0.6611 $$

$$ y_2(0)=-0.005586=e^{0}[C_{21}cos(0)+C_{22}sin(0)]+0.03627=C_{21}+0.03627 $$

$$ C_{21}=-0.04186 $$

$$ y'_2(0)=-0.005586=-3*e^{0}[-0.04186cos(0)+C_{22}sin(0)]+1.6067e^{0}[-0.5394sin(0)+C_{22}cos(0)]=-0.04186+1.6067C_{22} $$

$$ C_{22}=0.02257 $$
 * }
 * }

So, the equations for the modal displacements are


 * {| style="width:100%" border="0"

$$ y_1(t)=e^{-0.25t}[0.5394cos(1.9561t)+0.6611sin(1.9561t)]+0.4606 $$
 * style="width:95%" |
 * style="width:95%" |

$$ y_2(t)=e^{-3t}[-0.04186cos(1.6067t)+0.02257sin(1.6067t)]+0.03627 $$
 * }
 * }

Problem Statement


(1) Derive the equation of motion of the above scenario in matrix form in terms of the above variables.

(2) Derive and solve the generalized eigenvalue problem.

(3) Verify the orthogonality of the normalized eigenvectors with respect to the mass matrix and stiffness matrix; do same for damping matrix. Are the normalized eigenvectors orthogonal to the damping matrix?

(4) Obtain the modal equations using the same modal damping coefficients as of the support on the right was not removed. Derive the modal initial conditions.

(5) Reconstruct the damping matrix using these modal damping coefficients, compare this reconstructed damping matrix to the damping matrix obtained when the support on the right was removed (as obtained from your derivation of the equation of motion), and deduce the new values of the damping coefficients c1, c2, c3 that should be used to obtain the same reconstructed damping matrix when you derive the equation of motion.

(6) Solve the modal equations of motion, satisfy the modal initial conditions, and recombine to modal displacements to obtain the displacements of this 2-DOF system.

(7) Compare the displacements for 2 cases: (1) the example in my lecture notes, (2) the structure in the current problem.

Part 1
With right support removed there is no force generated by the mass-damper in parallel for element 3; thus it can be removed.

Element 1




 * {| Italic textstyle="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( -d_B + d_A ) + c_1 ( -d'_B + d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( d_B - d_A ) + c_1 ( d'_B - d'_A ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Rearrange Terms in Equations to Prepare to Change to Matrix Form


 * {| style="width:100%" border="0"

$$ f_A^{(1)} = k_1 ( d_A - d_B ) + c_1 ( d'_A - d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = k_1 ( -d_A + d_B ) + c_1 ( -d'_A + d'_B ) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Equations in Matrix Form


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }

Element 2 Start with FBD:



Force matrix follows same topology as in element 1.


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}   k_2 & - k_2 \\ -k_2 &   k_2 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Mass B



$$ \sum_{} f = F_B(t)- m_Bd''-f_B^{(1)}-f_B^{(2)} = 0 $$


 * {| style="width:100%" border="0"

$$ f_B^{(1)} = \begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ f_B^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in For Forces


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd''_B + (\begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix}) + (\begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix}) $$  (1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Mass C



$$ \sum_{} f = 0 = F_C(t)- m_Cd''_C- f_C^{(2)} $$


 * {| style="width:100%" border="0"

$$ f_C^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub in For Forces


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd''_C + (\begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix}) $$ (2) Boundary Conditions for Both Equations of Motion
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ d_A(t) = d'_A(t) = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * }

Equations of Motion After Substituting in Boundary Conditions


 * {| style="width:100%" border="0"

$$ F_B(t) = m_Bd''_B + k_1d_B + c_1d'_b + \begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_b - c_2d'_C \end{bmatrix} $$   (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ F_C(t) = m_Cd''_C + \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} $$   (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Into Matrix Form


 * {| style="width:100%" border="0"

$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \begin{bmatrix} d_b \\ d_c \end{bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Sub In Values Below into Equation of Motion


 * {| style="width:100%" border="0"

$$ k_1 = 14 - \sqrt{6}, \ k_2 = \sqrt{6} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_1 = 2, m_2 = \frac12 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ c_1 = \frac{(76 - 11 \sqrt{6})}{10}, \ c_2 = \frac{11 \sqrt6}{10} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_B = m_1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ m_C = m_2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Final Equation of Motion takes form:


 * {| style="width:100%" border="0"

$$ \mathbf M {\mathbf d}'' + \mathbf C {\mathbf d}' + \mathbf K {\mathbf d} = \mathbf F (t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

With All Values Substituted In


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} d_1(t) \\ d_2(t) \end{Bmatrix} + \frac1{20}
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 152 & -22 \sqrt6 \\ -22 \sqrt6 & 22 \sqrt6 \end{bmatrix} \begin{Bmatrix} d'_1(t) \\ d'_2(t) \end{Bmatrix} +

\begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & \sqrt6 + 3 \end{bmatrix} \begin{Bmatrix} d_1(t) \\ d_2(t) \end{Bmatrix} =

\frac{1}{\sqrt{5}} \begin{Bmatrix} 4 \\ \sqrt{6} \end{Bmatrix} h(t) $$
 * }
 * }

Initial Conditions

Displacements:


 * {| style="width:100%" border="0"

$$ \mathbf d(0) = \begin{Bmatrix} d_{10} \\ d_{20} \end{Bmatrix} = \frac{1}{\sqrt{5}} \begin{Bmatrix} 1 \\ \sqrt{6}
 * style="width:95%" |
 * style="width:95%" |

\end{Bmatrix} $$
 * }
 * }

Velocities:


 * {| style="width:100%" border="0"

$$ \mathbf d'(0) = \begin{Bmatrix} v_{10} \\ v_{20} \end{Bmatrix} = \sqrt{\frac{3}{10}} \begin{Bmatrix} 1 \\ \frac{-4}{\sqrt{6}}
 * style="width:95%" |
 * style="width:95%" |

\end{Bmatrix} $$
 * }
 * }

Part 2
To solve the generalized eigenvector problem we begin with the undamped free vibration equation.


 * {| style="width:100%" border="0"

$$ \mathbf M \mathbf d'' + \mathbf K \mathbf d = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} d_1(t) \\ d_2(t) \end{Bmatrix} +
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6} \end{bmatrix} \begin{Bmatrix} d_1(t) \\ d_2(t) \end{Bmatrix} =

\begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * }
 * }

Start by assuming a trial solution to the undamped free vibration equation. Such solution takes form


 * {| style="width:100%" border="0"

$$ \mathbf d (t) = e^{\alpha t} \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Where


 * {| style="width:100%" border="0"

$$ \mathbf d' (t) = \alpha e^{\alpha t} \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \mathbf d'' (t) = \alpha^2 e^{\alpha t} \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

substituting trial solution back into undamped free vibration equation yields


 * {| style="width:100%" border="0"

$$ [ \mathbf M \alpha^2 + \mathbf K ] \boldsymbol \phi = \mathbf 0 \Rightarrow \mathbf K \boldsymbol \phi = - \alpha^2 \mathbf M \boldsymbol \phi $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The generalized eigenvalue problem now reads


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6} \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} = \beta
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} \phi_1 \\ \phi_2 \end{Bmatrix} $$
 * }
 * }

We rewrite equation as


 * {| style="width:100%" border="0"

$$ [ \mathbf K - \beta \mathbf M ] \boldsymbol \phi = \mathbf 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

We note from this equation that a trivial solution will arise if $$ \phi = 0 $$. Thus for nontrivial solutions


 * {| style="width:100%" border="0"

$$ \det [ \mathbf K - \beta \mathbf M ] = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Thus


 * {| style="width:100%" border="0"

$$ \det \begin{bmatrix} 14 - \beta \, 2 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6} - \beta \, \frac12 \end{bmatrix} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \beta^2 - (2\sqrt{6}+7) \beta + 14\sqrt{6}-6 = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \beta_1 = (\omega_1)^2 = 3.2842 \Rightarrow \omega_2 = 1.812 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$ \beta_2 = (\omega_2)^2 = 8.6147 \Rightarrow \omega_2 = 2.935 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

recall general eigenvalue problem noting we have now two eigenvalues in which we would like to obtain the two eigenvectors


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6} \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix} =
 * style="width:95%" |
 * style="width:95%" |

\beta_i \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix} \begin{Bmatrix} \phi_{i1} \\ \phi_{i2} \end{Bmatrix}, \ i=1,2 $$
 * }
 * }

for i=1 the first eigenpair is


 * {| style="width:100%" border="0"

$$ \beta_1 = 3.2842, \ \boldsymbol \phi_1 = \begin{Bmatrix} \phi_{11} \\ \phi_{12} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Thus


 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 14 - 3.2842 * 2 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6} - 3.2842 * \frac12 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\
 * style="width:95%" |
 * style="width:95%" |

\phi_{12} \end{Bmatrix} = \begin{bmatrix} 7.4316 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6}-1.6421 \end{bmatrix} \begin{Bmatrix} \phi_{11} \\

\phi_{12} \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} $$
 * }
 * }

Since determinant is zero there is a non-trivial solution. Setting $$ \phi = 1 $$ therefore


 * {| style="width:100%" border="0"

$$ \phi_{12} = \frac{7.4316}{\sqrt{6}} = 3.0339 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

First eigenvector is thus


 * {| style="width:100%" border="0"

$$ \boldsymbol \phi_1 = \begin{Bmatrix} 1 \\ 3.0339 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Run through same procedure this time with second eigenpair


 * {| style="width:100%" border="0"

$$ \beta_2 = 8.6147, \ \boldsymbol \phi_2 = \begin{Bmatrix} \phi_{21} \\ \phi_{22} \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Thus second eigenvector


 * {| style="width:100%" border="0"

$$ \boldsymbol \phi_1 = \begin{Bmatrix} 1 \\ -1.3184 \end{Bmatrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Mass orthogonality of normalized eigenvector verification

Modal mass of mode 1:


 * {| style="width:100%" border="0"

$$ \bar m_1 := \boldsymbol \phi_1^T \mathbf M \boldsymbol \phi_1 = \lfloor 1 3.0339 \rfloor \begin{bmatrix} 2 & 0 \\ 0
 * style="width:95%" |
 * style="width:95%" |

& \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ 3.0339 \end{Bmatrix} = \lfloor 1 3.0339 \rfloor \begin{Bmatrix} 2 \\

\frac{3.0339}{2} \end{Bmatrix} = 2 + 3.0339^2/2 = 6.6023 $$     (1)
 * <p style="text-align:right">
 * }

1st normalized eigenvector


 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \phi}_1 = \frac{\boldsymbol \phi_1}{\sqrt{\bar m_1}} = \frac{1}{\sqrt{6.6023}} \begin{Bmatrix} 1 \\
 * style="width:95%" |
 * style="width:95%" |

3.0339 \end{Bmatrix} $$     (1)
 * <p style="text-align:right">
 * }

Modal mass of mode 2:


 * {| style="width:100%" border="0"

$$ \bar m_2 := \boldsymbol \phi_2^T \mathbf M \boldsymbol \phi_2 = \lfloor 1 -1.3184 \rfloor \begin{bmatrix} 2 & 0 \\ 0
 * style="width:95%" |
 * style="width:95%" |

& \frac12 \end{bmatrix} \begin{Bmatrix} 1 \\ -1.3184 \end{Bmatrix} = \lfloor 1 -1.3184 \rfloor \begin{Bmatrix} 2 \\

\frac{-1.3184}{2} \end{Bmatrix} = 2 + (-1.3184)^2/2 = 2.8691 $$     (1)
 * <p style="text-align:right">
 * }

2nd normalized eigenvector


 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \phi}_2 = \frac{\boldsymbol \phi_2}{\sqrt{\bar m_2}} = \frac{1}{\sqrt{2.8691}} \begin{Bmatrix} 1 \\
 * style="width:95%" |
 * style="width:95%" |

-1.3184 \end{Bmatrix} $$     (1)
 * <p style="text-align:right">
 * }

therefore we can express mass orthogonality of normalized eigenvectors as


 * {| style="width:100%" border="0"

$$ [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ]^T \mathbf M [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 &

\frac{1}{\sqrt{2.8691}} \cdot -1.3184 \end{bmatrix}^T \begin{bmatrix} 2 & 0 \\ 0 & \frac12 \end{bmatrix}

\begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 & \frac{1}{\sqrt{2.8691}}

\cdot -1.3184 \end{bmatrix}

= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$     (1)
 * <p style="text-align:right">
 * }

Now we want to verify orthogonality with the stiffness matrix.


 * {| style="width:100%" border="0"

$$ \bar{\boldsymbol \Phi}^T \mathbf K \bar{\boldsymbol \Phi} = \boldsymbol \Omega^2 = \begin{bmatrix} (\omega_1)^2 & 0 \\ 0 & (\omega_2)^2 \end{bmatrix} = \begin{bmatrix} 3.2842 & 0 \\ 0 & 8.6147 \end{bmatrix} = \begin{bmatrix} b_1 & 0 \\ 0 &
 * style="width:95%" |
 * style="width:95%" |

b_2 \end{bmatrix} $$     (1)
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ]^T \mathbf K [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 &

\frac{1}{\sqrt{2.8691}} \cdot -1.3184 \end{bmatrix}^T

\begin{bmatrix} 14 & -\sqrt{6} \\ -\sqrt{6} & \sqrt{6}

\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 & \frac{1}{\sqrt{2.8691}}

\cdot -1.3184 \end{bmatrix} = \begin{bmatrix} 3.2842 & 0 \\ 0 & 8.6147 \end{bmatrix} =

\begin{bmatrix} b_1 & 0 \\ 0 & b_2 \end{bmatrix} $$     (1)
 * <p style="text-align:right">
 * }

Since these eigenvectors were retrieved using a general eigenvalue problem with the Mass and Stiffness matrix we

would expect the Mass and Stiffness matrices to be diagonalizable and it has been proven above.

Since we did not include the damping matrix in the GEP we should not expect the damping matrix to be diagonalizable by

the retrieved eigenvectors in this new scenario. Lets verify this.


 * {| style="width:100%" border="0"

$$ [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ]^T \mathbf K [ \bar{\boldsymbol \phi}_1 \ \bar{\boldsymbol \phi}_2 ] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 &

\frac{1}{\sqrt{2.8691}} \cdot -1.3184 \end{bmatrix}^T \frac1{20}

\begin{bmatrix} 152 & -22\sqrt{6} \\ -22\sqrt{6} & 22\sqrt{6}

\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 & \frac{1}{\sqrt{2.8691}}

\cdot -1.3184 \end{bmatrix} = \begin{bmatrix} 2.4312 & -1.7921 \\ -1.7921 & 6.7576 \end{bmatrix} =

\begin{bmatrix} a_1 & 0 \\ 0 & a_2 \end{bmatrix} $$     (1)
 * <p style="text-align:right">
 * }

Recognize the damping matrix is not diagonizable by the retrieved eigenvectors.


 * {| style="width:100%" border="0"

$$ \mathbf M {\mathbf d}'' + \mathbf C {\mathbf d}' + \mathbf K {\mathbf d} = \mathbf F (t) $$
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 * }
 * }

Now we want to project our Equation of Motion unto the basis of the normalized eigenvectors.

This will allow us to change our displacements into modal coordinates.


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$$ \mathbf d(t) = \sum_{j=1}^2 \bar{\boldsymbol \phi_j} \, y_j(t) = \bar{\boldsymbol \Phi} \, \mathbf y(t) $$
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 * }
 * }


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$$ \mathbf d'(t) = \sum_{j=1}^2 \bar{\boldsymbol \phi_j} \, y'_j(t) = \bar{\boldsymbol \Phi} \, \mathbf y'(t) $$
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 * }
 * }


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$$ \mathbf d(t) = \sum_{j=1}^2 \bar{\boldsymbol \phi_j} \, y_j(t) = \bar{\boldsymbol \Phi} \, \mathbf y''(t) $$
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 * }
 * }

this yields the equation of motion in modal coordinates.


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$$ \bar{\boldsymbol \phi_i}^T \sum_{j=1}^{2} [ \mathbf M \bar{\boldsymbol \phi_j} \, y''_j(t) + \mathbf C \bar{\boldsymbol (4)
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\phi_j} \, y'_j(t) + \mathbf K \bar{\boldsymbol \phi_j} \, y_j(t) ] = \bar{\boldsymbol \phi_i}^T \mathbf F (t) $$
 * }
 * }

If we premultiply our EOM w=now in modal coordinates we can then use orthogonality of the normalized eigenvectors with respect to the mass, damping, and stiffness matrixes.

thus


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$$ \bar{\boldsymbol \phi}_i^T \mathbf M \bar{\boldsymbol \phi}_j = \delta_{ij} $$
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 * }
 * }


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$$ \bar{\boldsymbol \phi}_i^T \mathbf C \bar{\boldsymbol \phi}_j = a_j \delta_{ij} $$
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 * }
 * }


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$$ \bar{\boldsymbol \phi}_i^T \mathbf K \, \bar{\boldsymbol \phi}_j = (\omega_j)^2 \delta_{ij} $$
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 * }
 * }

therefore


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$$ \sum_{j=1}^{2} [ \delta_{ij} \, y''_j(t) + a_j \delta_{ij} \, y'_j(t) + (\omega_j)^2 \delta_{ij} \, y_j(t) ] = r_i(t) $$
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 * }
 * }

or in a more tangible form.


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$$ y''(t) + a \, y'(t) + b \, y(t) = r(t), \text{ for } i = 1,2 $$
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 * }
 * }

Solving now for the projected applied force r(t)


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$$ \mathbf F(t) = \begin{Bmatrix} F_1(t) \\ F_2(t) \end{Bmatrix} = \begin{Bmatrix} f_1 \\ f_2 \end{Bmatrix} h(t) = \frac{1}{\sqrt{5}} \begin{Bmatrix} 4 \\ \sqrt{6} \end{Bmatrix} h(t) $$
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 * style="width:95%" |
 * }
 * }


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$$ \mathbf r(t) = \begin{Bmatrix} r_1(t) \\ r_2(t) \end{Bmatrix} = \bar{\boldsymbol \Phi}^T \mathbf F(t) = \begin{bmatrix} \frac{1}{\sqrt{6.6023}} & \frac{1}{\sqrt{2.8691}} \\ \frac{1}{\sqrt{6.6023}} \cdot 3.0339 &
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\frac{1}{\sqrt{2.8691}} \cdot -1.3184 \end{bmatrix}^T \frac{1}{\sqrt{5}} \begin{Bmatrix} 4 \\ \sqrt{6} \end{Bmatrix} = \begin{Bmatrix} 1.990 \\ 0.2035 \end{Bmatrix}

$$
 * }
 * }

Now we have all parameters ready to generate our two modal equations. Replacing a, b, and r(t) we retrieve


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$$ y_1'' + \frac12 \, y_1' + 3.2842 \, y_1 = 1.990 $$
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 * }
 * }


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$$ y_2'' + 6 \, y_2' + 8.6147 \, y_2 = 0.2035
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 * style="width:95%" |

$$
 * }
 * }

now we must define the modal initial conditions


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$$ \mathbf d(0) = \begin{Bmatrix} d_{10} \\ d_{20} \end{Bmatrix} = \frac{1}{\sqrt{5}} \begin{Bmatrix} 1 \\ \sqrt{6}
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\end{Bmatrix} = \bar{\boldsymbol \phi}_1 \Rightarrow \mathbf y(0) = \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol

\phi}_1 = \begin{Bmatrix} 0.9948 \\ 0.10173 \end{Bmatrix}

$$
 * }
 * }


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$$ \mathbf d'(0) = \begin{Bmatrix} v_{10} \\ v_{20} \end{Bmatrix} = \sqrt{\frac{3}{10}} \begin{Bmatrix} 1 \\ \frac{-4}{\sqrt{6}}
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\end{Bmatrix} = \bar{\boldsymbol \phi}_2 \Rightarrow \mathbf y(0) = \bar{\boldsymbol \Phi}^T \mathbf M \bar{\boldsymbol

\phi}_2 = \begin{Bmatrix} -0.10173 \\ 0.9948 \end{Bmatrix}

$$
 * }
 * }

Part 3
Now we must construct the new damping matrix using the used modal damping coefficients

Thus we recall our equation projecting our damping matrix for the modal damping coefficients


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$$ \bar{\boldsymbol \Phi}^T \mathbf C \bar{\boldsymbol \Phi} = \begin{bmatrix} \frac12 & 0 \\ 0 & 6 \end{bmatrix} $$
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 * }
 * }

Rearranging this equation to solve for the unknown damping matrix C


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$$ \mathbf C = \begin{bmatrix} \frac12 & 0 \\ 0 & 6 \end{bmatrix} (\bar{\boldsymbol \Phi}^T)^{-1}(\bar{\boldsymbol \Phi})^{-1} $$
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 * }
 * }

We retrieve the following relation


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$$ \mathbf C = \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix} $$
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 * }
 * }

We know that the damping matrix for this problem takes the style
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$$ \mathbf C = \begin{bmatrix} c_1+c_2 & -c_2 \\ -c_2 & c_2 \end{bmatrix} $$
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 * style="width:95%" |
 * }
 * }

The retrieved solution for the damping matrix for this system doesn't make sense. We should be seeing a the top right

value and the bottom left value in the matrix being a non-zero answer which would indicate the negative of the

damping on spring two. Further we should be seeing the same magnitude of value for the bottom right spot in the damping matrix. Due to the fact that we are not seeing

this indicated that there must be a miscalculation in the eigen vector matrix.

Part 4
Starting with the modal equations previously derived


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$$ y_1'' + \frac12 \, y_1' + 3.2842 \, y_1 = 1.990 $$
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 * }
 * }


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$$ y_2'' + 6 \, y_2' + 8.6147 \, y_2 = 0.2035
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$$
 * }
 * }

Solving the second order ODE:
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$$ \lambda''+\frac12\lambda'+3.2842 \lambda=0 $$
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 * }
 * }

The solution has two complex conjugate roots because


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$$ b^2-4ac<0 $$
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 * }
 * }


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$$	\lambda_1=-0.25+1.7949i $$
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 * }


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$$ \lambda_2=-0.25-1.7949i $$
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 * }

Solving the other second order ODE:
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$$ \lambda''+6\lambda'+8.6147\lambda=0 $$
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 * }
 * }

The solution has two real conjugate roots because


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$$ b^2-4ac>0 $$
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 * }
 * }


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$$	\lambda_3=-3.621 $$
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 * }


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$$ \lambda_4=-2.379 $$
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 * }

The general solution is:
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$$ y_1=c_1e^{-0.25x}sin(1.7949x)+c_2e^{-0.25x}cos(1.7949x)+c_3 $$
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 * }
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$$ y_2=c_4e^{-3.621x}+c_5e^{-2.379x}+c_6 $$
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 * }

Taking the first derivative in order to solve for the constants:
 * {| style="width:100%" border="0"

$$ y'_1=c_1e^{-0.25x}(-0.25sin(1.7949x)+1.7949cos(1.7949x))+c_2e^{-0.25x}(-0.25cos(1.7949x)-1.7949sin(0.65x) $$
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 * }
 * {| style="width:100%" border="0"

$$ y'_2=-3.621c_4e^{-3.621x}-2.379c_5e^{-2.379x} $$
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 * }

Modal initial displacements: Previously we calculated our modal displacement and velocity initial conditions


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$$ y_1(0) = 0.9948, \ y'_1(0) = -0.10173 $$
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 * }
 * }


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$$ y_2(0) = 0.10173, \ y'_2(0) = 0.9948 $$
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 * }
 * }

Therefore plugging in initial conditions the coefficients become


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$$ c_1=0.08188 ,c_2=0.9948 ,c_4=-0.9958,c_5=1.0976 $$
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 * }
 * }

Therefore the displacements of the system become


 * {| style="width:100%" border="0"

$$ y_1=0.08188e^{-0.25x}sin(1.7949x)+0.9948e^{-0.25x}cos(1.7949x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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$$ y_2=-0.9958e^{-3.621x}+1.0976e^{-2.379x} $$
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 * style="width:95%" |
 * }
 * }

Part 5
Desmos plot can be seen here:

Link: Here

Zooming in on the plot one can see that the plot satisfies the initial conditions.

Comparing to the system completed in the notes it can be seen that the period of damped oscillation is longer in

this case vs the case done in the notes. Further it can be seen that in mode one for the notes case, the displacements

settle to 0.5 as time goes to infinity; whereas in the scenario solved for in this problem the displacements fall toward zero.