User:EGM4313.f14.Team1.Linehan/EAS 4400: HW 2

=HW 3=

Problem Statement
Derive eigenvalues and eigenvectors for $$ \mathbf A= \begin{Bmatrix} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & 0\\ 0 & 0 & a_{33} \end{Bmatrix}$$

Solution
To solve for eigenvalues and eigenvectors one must solve the standard eigenvalue problem


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$$ 0=\boldsymbol v(\mathbf A- \boldsymbol \lambda \mathbf I) $$
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where $$\boldsymbol \phi$$ and $$\lambda \phi$$ are eigenvectors and eigenvectors respectively.

For non-trivial solutions,
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$$ 0=\det(\mathbf A- \boldsymbol \lambda \mathbf I) $$
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therefore


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$$ 0=\det\begin{bmatrix} a_{11}-\lambda & a_{12} & 0 \\ a_{21} & a_{22}-\lambda & 0\\ 0 & 0 & a_{33}-\lambda \end{bmatrix} $$
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Eigenvalue solutions are therefore,


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$$ \boldsymbol \lambda= \begin{bmatrix} a_{33} \\ \frac{1}{2}(a_{11} + a_{22} + (a_{11}^2 - 2 a_{11} a_{22} + a_{22}^2 + 4 a_{12} a_{21})^{\frac{1}{2}} \\ \frac{1}{2}(a_{11} + a_{22} - (a_{11}^2 - 2 a_{11} a_{22} + a_{22}^2 + 4 a_{12} a_{21})^{\frac{1}{2}} \end{bmatrix} $$
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First eigenvector must satisfy the following equation,


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$$ 0=\begin{bmatrix} v_{11} \\ v_{12} \\ v_{13} \end{bmatrix} \begin{bmatrix} a_{11}-\lambda_1 & a_{12} & 0 \\ a_{21} & a_{22}-\lambda_1 & 0\\ 0 & 0 & a_{33}-\lambda_1 \end{bmatrix} $$
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Therefore first eigenvector is


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$$ \boldsymbol v_1 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$
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Second eigenvector must satisfy the following equation,


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$$ 0=\begin{bmatrix} v_{21} \\ v_{22} \\ v_{23} \end{bmatrix} \begin{bmatrix} a_{11}-\lambda_2 & a_{12} & 0 \\ a_{21} & a_{22}-\lambda_2 & 0\\ 0 & 0 & a_{33}-\lambda_2 \end{bmatrix} $$
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Therefore second eigenvector is


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$$ \boldsymbol v_2 = \begin{bmatrix} \frac{1}{2 a_{21}} ( a_{11} + a_{22} - (a_{11}^2 - 2 a_{11} a_{22} + a_{22}^2 + 4 a_{12} a_{21})^{\frac{1}{2}} - 2 a_{22}) \\ 1 \\ 0 \end{bmatrix} $$
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Third eigenvector must satisfy the following equation,


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$$ 0=\begin{bmatrix} v_{31} \\ v_{32} \\ v_{33} \end{bmatrix} \begin{bmatrix} a_{11}-\lambda_3 & a_{12} & 0 \\ a_{21} & a_{22}-\lambda_3 & 0\\ 0 & 0 & a_{33}-\lambda_3 \end{bmatrix} $$
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Therefore Third eigenvector is


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$$ \boldsymbol v_3 = \begin{bmatrix} \frac{1}{2 a_{21}} ( a_{11} + a_{22} + (a_{11}^2 - 2 a_{11} a_{22} + a_{22}^2 + 4 a_{12} a_{21})^{\frac{1}{2}} - 2 a_{22}) \\ 1 \\ 0 \end{bmatrix} $$
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Problem Statement
Derive the eigenvalues and eigenvectors of $$ \mathbf X = \begin{Bmatrix} 0 & -x & 0 \\ x & 1 & x\\ 0 & -x & 0 \end{Bmatrix}$$ for some real scalar x.

Solution

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$$ 0=\boldsymbol v(\mathbf X- \boldsymbol \lambda \mathbf I) $$
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$$ 0=\det\begin{bmatrix} -\lambda & -x & 0 \\ x & 1-\lambda & x\\ 0 & -x & -\lambda \end{bmatrix} $$
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Eigenvalue solutions are therefore,


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$$ \boldsymbol \lambda= \begin{bmatrix} 0 \\ 1/2 - (1 - 8*x^2)^{1/2}/2 \\ (1 - 8*x^2)^{1/2}/2 + 1/2 \end{bmatrix} $$
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First eigenvector must satisfy the following equation,


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$$ 0=\begin{bmatrix} v_{11} \\ v_{12} \\ v_{13} \end{bmatrix} \begin{bmatrix} -\lambda_1 & -x & 0 \\ x & 1-\lambda_1 & x\\ 0 & -x & -\lambda_1 \end{bmatrix} $$
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Therefore first eigenvalue is


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$$ \boldsymbol v_1= \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} $$
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Second eigenvector must satisfy the following equation,


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$$ 0=\begin{bmatrix} v_{21} \\ v_{22} \\ v_{23} \end{bmatrix} \begin{bmatrix} -\lambda_2 & -x & 0 \\ x & 1-\lambda_2 & x\\ 0 & -x & -\lambda_2 \end{bmatrix} $$
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Therefore second eigenvalue is


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$$ \boldsymbol v_2= \begin{bmatrix} 0 \\ 1/2 - (1 - 8*x^2)^{1/2}/2 \\ 0 \end{bmatrix} $$
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Third eigenvector must satisfy the following equation,


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$$ 0=\begin{bmatrix} v_{31} \\ v_{32} \\ v_{33} \end{bmatrix} \begin{bmatrix} -\lambda_3 & -x & 0 \\ x & 1-\lambda_3 & x\\ 0 & -x & -\lambda_3 \end{bmatrix} $$
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Therefore Third eigenvalue is


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$$ \boldsymbol v_3= \begin{bmatrix} 0 \\ 1/2 - (1 - 8*x^2)^{1/2}/2 \\ 0 \end{bmatrix} $$
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Problem Statement
If the matrix of $$\begin{bmatrix} 7 & 6 \\ x & y \end{bmatrix}$$ has the eigen values of j and -j then what are x and y.

Solution
For the eigenvalues to be as follows the determinant of the standard eigenvalue problem must equal the following.
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$$ \det(\begin{bmatrix} 7-\lambda & 6 \\ x & y-\lambda \end{bmatrix}) = \lambda^2+1 ) $$
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therefore
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$$ \lambda^2 + (-y-7) \lambda + 7y-6x = \lambda^2+1 $$
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therefore in order for this expression to be correct


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$$ y = -7 $$
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$$ x = \frac{-25}{3} $$
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Problem Statement
If a Matrix, A, has a multiplication property that $$ AA=-I $$ then what are the eigenvectors of A.

Solution

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$$ AA=-I $$
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premultiplying each side by eigenvector v


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$$ AAv=-Iv=-v $$
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eigen values must agree with standard eigen value definition.
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$$ (A-\lambda)v=0 $$
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Rearrangement unveils


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$$ Av=v\Lambda $$
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therefore plugging back into starting equation


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$$ AAv=Av\lambda=(v\lambda)\lambda=-v $$
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therefore


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$$ v\lambda\lambda=-v $$
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multiplying each side by $$ v^{-1} $$


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$$ \lambda\lambda=-vv^{-1}=-I $$
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for the following to be true the eigenvalues must be equal to one with opposite sign.

Problem Statement
Given a state-space system with $$ \begin{Bmatrix} \dot{x_1} \\ \dot{x_2} \end{Bmatrix} = \begin{Bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{Bmatrix} \begin{Bmatrix} {x_1} \\ {x_2} \end{Bmatrix} $$, find a state transformation, T, such that $$ \begin{Bmatrix} \dot{\nu_1} \\ \dot{\nu_2} \end{Bmatrix} = \begin{Bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{Bmatrix} \begin{Bmatrix} {\nu_1} \\ {\nu_2} \end{Bmatrix} $$ with the requirement $$ \nu=Tx $$.

Solution
Since matrix \nu is the eigenvalues of matrix A in order for


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$$ \nu=TX $$ therefore
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$$ \dot{\nu}=T\dot{X} $$
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thus


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$$ \dot{\nu}=T\dot{X} $$
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$$ \lambda \nu= T\dot{X} $$
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$$ \lambda TX= T\dot{X} $$
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$$ T_{-1} \lambda TX= \dot{X} $$
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$$ T_{-1} \lambda TX= AX $$
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Problem Statement
Augment the standard system of 2-masses and 3-springs that we used during class to include a damper (with values of c1 affecting y1 and c2 affecting y2) in line with each spring. Also, include a force that may be commanded to the first mass.

a) Derive the state-space representation using the state vector of $$ x= \begin{Bmatrix} y_1+y_2 \\  y_1-y_2 \\ \dot{y_1} \\ \dot{y_2} \end{Bmatrix} $$

(b) Using k1 = k2 = k3 = 20N/m and m1 = m2 = 2kg and c1 = c2 = 1Ns/m, compute (in matlab) the eigenvalues and eigenvectors using the eig command.

(c) For the modes in part b, compute (in matlab) the values of ωn, ωd, ζ, t1/2

(d) For the system in part b, compute (in matlab) and plot the impulse response using the impulse command.

(e) For the responses in part d, compute (in matlab) and plot the FFT using the fft command.

(f) For the system in part b, compute (in matlab) and plot the step response using the step command.

(g) For the responses in part f, compute (in matlab) and plot the FFT using the fft command.

Solution
Equation of motion for system in matrix form


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$$ \begin{bmatrix} F_1(t) \\ 0 \end{bmatrix} = \begin{bmatrix}   m_1 & 0 \\ 0 &  m_2 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} + \begin{bmatrix}  c_1 & 0 \\ 0 & c_2 \end{bmatrix} \begin{bmatrix} y'_1 \\ y'_2 \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} $$
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$$ m_1y''_1+c_1y'_1+(k_1+k_2)y_1-k_2y_2= F_1 $$
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$$ m_2y''_2+c_2y'_2+(k_2+k_3)y_2-k_2y_1=0 $$
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Part A
Create state vector


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$$ x= \begin{Bmatrix} y_1+y_2 \\ y_1-y_2 \\ \dot{y_1} \\ \dot{y_2} \end{Bmatrix} $$
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since


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$$ x_1=y_1+y_2 $$
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and


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$$ x_2=y_1-y_2 $$
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therefore in terms of y_1 and y_2


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$$ y_1=\frac{x_1+x_2}{2} $$
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$$ y_2=\frac{x_1-x_2}{2} $$
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Finding state derivatives


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$$ \dot{x_1}=y'_1+y'_2 $$
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in terms of state variables


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$$ \dot{x_1}=x_3+x_4 $$
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$$ \dot{x_2}=y'_1-y'_2 $$
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in terms of state variables


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$$ \dot{x_2}=x_3-x_4 $$
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To find now to find $$ \dot{x_3} $$


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$$ \dot{x_3}=y''_1=\frac{-k_2(y_1-y_2) -k_1y_1-c_1y'_1+F_1}{m_1} $$
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now purely in terms of state variables


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$$ \dot{x_3}=y''_1=\frac{-k_2(x_2) -k_1(\frac{x_1+x_2}{2})-c_1x_3+F_1}{m_1} $$
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now to find $$ \dot{x_4} $$


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$$ \dot{x_4}=y''_2=\frac{k_2(y_1-y_2) -k_3y_2-c_2y'_2}{m_2} $$
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in terms of only state variables


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$$ \dot{x_4}=y''_2=\frac{k_2(x_2) -k_3(\frac{x_1-x_2}{2})-c_2x_4}{m_2} $$
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The state-space matrix can now be produced


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$$ \begin{bmatrix} \dot{x_1} \\ \dot{x_2} \\ \dot{x_3} \\ \dot{x_4}\end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ -\frac{k_1}{2m_1} & -(\frac{2k_2+k_1}{2m_1}) & -\frac{c_1}{m_1} & 0\\ -\frac{k_3}{2m_2} &  (\frac{k_3+2k_2}{2m_2})& 0 & -\frac{c_2}{m_2} \end{bmatrix} \begin{bmatrix} {x_1} \\ {x_2} \\ {x_3} \\ {x_4}\end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \frac{1}{m_1} \\ 0\end{bmatrix} F_1 $$
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Part B-G
Attached Matlab Documents.

Problem Statement
a) Derive the nonlinear equations of motion

b) Derive the linearized equations of motion

c) Assume the equilibrium condition is such that each spring is uncompressed. Compute (in matlab) the eigenvalues of the state-space representation of the linearized dynamics using kl = 100 and kr = 200 and k0 = 20.

d) Assume the equilibrium condition is such that the left spring is compressed 0.1 m and the right spring is compressed 0.2 m while the middle spring is uncompressed. Compute (in matlab) the eigenvalues of the state-space representation of the linearized dynamics using kl = 100 and kr = 200 and k0 = 20.

Part A
EOM in matrix form


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$$ \begin{bmatrix} F_1(t) \\ 0 \end{bmatrix} = \begin{bmatrix}   m_1 & 0 \\ 0 &  m_2 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} + \begin{bmatrix}  c_1 & 0 \\ 0 & c_2 \end{bmatrix} \begin{bmatrix} y'_1 \\ y'_2 \end{bmatrix} + \begin{bmatrix} k_0 + k_0 & -k_0 \\ -k_0 & k_0 + k_0 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} +\begin{bmatrix} k_l & 0 \\ 0 & k_r \end{bmatrix} \begin{bmatrix} y^3_1 \\ y^3_2 \end{bmatrix} $$
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$$ m_1y''_1+c_1y'_1+2k_0y_1+k_ly^3_1-k_0y_2= F_1 $$
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$$ m_2y''_2+c_2y'_2+2k_0y_2+k_ry^3_2-k_0y_1= 0 $$
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Part B
Starting with the force of the spring on the left.


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$$ m_1y''_1+c_1y'_1+(k_0+k_0)y_1+k_ly^3_1-k_0y_2= F_1 $$
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we can rearrange our first EOM


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$$ m_1y''_1+c_1y'_1-k_0(y_2-y_1)+k_0y_1+k_ly^3_1= F_1 $$
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$$ F_{left}=-k_0y_1-k_ly^3_1 $$
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therefore linearization about arbitrary pt.


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$$ F_{left}(a)=b $$
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the linearization would take the form


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$$ F_{left}(y_1)=F_{left}(a) + \left. \frac{\partial F_{left}}{\partial y_1} \right|_{y_1=a} (y_1-a) $$
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where


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$$ \left. \frac{\partial F_{left}}{\partial y_1} \right|_{y_1=a} = -(k_0+3k_la^2) $$
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Thus linearized force equation for right spring becomes


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$$ F_{left}(y_1)=F_{left}(a) + -(k_0+3k_la^2)(y_1-a) $$
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Therefore 1st linearized EOM


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$$ m_1y''_1+c_1y'_1-k_0(y_2-y_1) + F_{left}(a) + (k_0+3k_la^2)(y_1-a)= F_1 $$
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Following same procedures for linearizing the right non-linear spring force about arbitrary pt.


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$$ F_{right}(c)=d $$
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yields


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$$ F_{right}(y_2)=F_{right}(c) + \left. \frac{\partial F_{right}}{\partial y_2} \right|_{y_2=c} (y_2-c) $$
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where


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$$ \left. \frac{\partial F_{right}}{\partial y_2} \right|_{y_2=a} = -(k_0+3k_rc^2) $$
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Thus linearized force equation for right spring becomes


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$$ F_{right}(y_2)=F_{right}(c) + -(k_0+3k_rc^2)(y_2-c) $$
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Therefore 2nd linearized EOM


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$$ m_2y''_2+c_2y'_2+k_0(y_2-y_1) + F_{right}(c) + (k_0+3k_rc^2)(y_2-c)= 0 $$
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Part C
For uncompressed springs we would linearize about zero. Thus $$ F_{left}(0)=F_{right}(0)=0  $$

Thus our EOMs become


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$$ m_1y''_1+c_1y'_1+k_0(y_1-y_2)+k_0y_1= F_1 $$
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$$ m_2y''_2+c_2y'_2-k_0(y_1-y_2)+k_0y_2= 0 $$
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the state space representation is representative of the one found for problem 6a.


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$$ \begin{bmatrix} \dot{x_1} \\ \dot{x_2} \\ \dot{x_3} \\ \dot{x_4}\end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ -\frac{k_0}{2m_1} & -(\frac{-2k_0+k_0}{2m_1}) & -\frac{c_1}{m_1} & 0\\ -\frac{k_0}{2m_2} &  (\frac{k_0+2k_0}{2m_2})& 0 & -\frac{c_2}{m_2} \end{bmatrix} \begin{bmatrix} {x_1} \\ {x_2} \\ {x_3} \\ {x_4}\end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \frac{1}{m_1} \\ 0\end{bmatrix} F_1 $$
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since $$ k_0=k_1=k_2=k_3=20 $$ as in problem 6 an identical state-space representation would be generated. Thus the same eigenvalues would be produced.

Part D
Now linearizing about 0.1 for the left spring and about 0.2 for the right spring.


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$$ F_{left}(0.1)=-k_0(0.1)-k_l(0.1)^3 $$
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Therefore 1st linearized EOM
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$$ m_1y''_1+c_1y'_1-k_0(y_2-y_1) + k_0(0.1)+k_l(0.1)^3 + (k_0+3k_l(0.1)^2)(y_1-0.1)= F_1 $$
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for right spring
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$$ F_{right}(0.2)=-k_0(0.2)-k_r(0.2)^3 $$
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Therefore 2nd linearized EOM


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

$$ m_2y''_2+c_2y'_2+k_0(y_2-y_1) + k_0(0.2)+k_r(0.2)^3 + (k_0+3k_r(0.2)^2)(y_2-0.2)= 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }