# University of Florida/Eml4507/Team7 Report3
## Problem 1
### Find

For the 3-bar truss example in K.2008.2 on pg. 43, solve this problem, but with non-zero prescribed disp dofs, using your own FE matlab code(not CALFEM). Using the same "connectivity array" given in the same problem statement, provide all of the results, including all of the unknown global disp dofs, force components(reactions), and member forces. Plot the deformed shape superposed on the undeformed shape. Verify the results with CALFEM.

### Given

d$_3$= 2 cm, d$_4$= -1 cm,d$_5$= -3 cm, d$_6$= 5 cm

### Solution

MATLAB script file:
MATLAB output:
>> report3
X =
For element 1
k =
For element 2
k =
For element 3
k =
F1=
F2=
F3=
F=

## Problem 2

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

1. Draw the FBDs of all components in the spring-mass-damper system in Figure 1 below, with the known displacement degree of freedoms being at the left and right supports. ($d_{3}=0$ and $d_{4}=0$)
2. Derive equations 1, 2, 3, and 4 from the figure above.
| $\mathbf {M} {\ddot {d}}+\mathbf {C} {\dot {d}}+\mathbf {K} d=\mathbf {F} (t)$ | (2.1) |
$\mathbf {d} ={\begin{Bmatrix}d_{1}\\d_{2}\end{Bmatrix}}$ $\mathbf {F} ={\begin{Bmatrix}F_{1}\\F_{2}\end{Bmatrix}}$
| $\mathbf {M} ={\begin{bmatrix}m_{1}&0\\0&m_{2}\end{bmatrix}}$ | (2.2) |
| $\mathbf {C} ={\begin{bmatrix}(c_{1}+c_{2})&-c_{2}\\-c_{2}&(c_{2}+c_{3})\end{bmatrix}}$ | (2.3) |
| $\mathbf {K} ={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\end{bmatrix}}$ | (2.4) |

### Solution
#### Free Body Diagrams

FBD of Mass 1:
FBD of Mass 2:

#### Equation Derivations

In order to derive Equation 3.2.1, we can first apply Newton's second law of motion in order to obtain the equations of motion from the two masses.
From the Free body diagram of Mass 1(Figure 2), we can use Newton's second law in order to obtain:
| $m_{1}{\ddot {d}}_{1}+(c_{1}+c_{2}){\dot {d}}_{1}-c_{2}{\dot {d}}_{2}+(k_{1}+k_{2})d_{1}-k_{2}d_{2}=F_{1}$ | (2.5) |
Applying Newton's second law again to Mass 2(Figure 3), we also obtain:
| $m_{2}{\ddot {d}}_{2}-c_{2}{\dot {d}}_{1}+(c_{2}+c_{3}){\dot {d}}_{2}-k_{2}d_{1}+(k_{2}+k_{3})d_{2}=F_{2}$ | (2.6) |
From equations 2.5 and 2.6, the motion of the masses will influence each other. Where, the motion of mass 1 will influence mass 2 and vice versa. Equations 2.5 and 2.6 can be rewritten in matrix form. Therefore, we can now write them as:
| $\mathbf {M} {\ddot {d}}+\mathbf {C} {\dot {d}}+\mathbf {K} d=\mathbf {F} (t)$ | (2.7) |
Which is the same as equation 2.1. Where $\mathbf {M}$ is the mass matrix, $\mathbf {C}$ is the damping matrix, and $\mathbf {K}$ is the stiffness matrix and $d(t)$ and $F(t)$ are the displacement and force vectors,respectively.Using equations 2.5 and 2.6 the mass, stiffness, and damping matrices can be determined.
The vectors for displacement and force can be represented by:
$\mathbf {d} ={\begin{Bmatrix}d_{1}\\d_{2}\end{Bmatrix}}$
and
| $\mathbf {F} ={\begin{Bmatrix}F_{1}\\F_{2}\end{Bmatrix}}$ | (2.8) |
As for the mass matrix, we see that in Equation 3.2.5 there is only mass 1 present and in equation 3.2.6 only mass 2 is pressent, therefore we have:
| $\mathbf {M} ={\begin{bmatrix}m_{1}&0\\0&m_{2}\end{bmatrix}}$ | (2.9) |
For the damping matrix, it is determined by the coefficients in front of${\dot {d}}_{1}$ and ${\dot {d}}_{2}$. So we now have:
| $\mathbf {C} ={\begin{bmatrix}(c_{1}+c_{2})&-c_{2}\\-c_{2}&(c_{2}+c_{3})\end{bmatrix}}$ | (2.10) |
Lastly we determine the stiffness matrix by the coefficients in front of $d_{1}$ and $d_{2}$. Which is:
| $\mathbf {K} ={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\end{bmatrix}}$ | (2.11) |
Essentially, from Equations 2.5 and 2.6 the mass, stiffness, and damping matrices were determined. Equations 2.8 through 2.11 are the same as Equations 2.1 through 2.4, respectively. Without the equation of motions for both masses, neither the simplified matrix form of the equation nor the matrices can be derived.

## Problem 3

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

FE formulation leads to a system of coupled ODEs in time of the form:
$Mx''+Cx'+Kx=F(t)$
Integrate the governing system of L2-ODEs-CC in order to generate the time histories for the displacement degrees of freedom. Once the displacement matrix has been found, plot the time histories for the displacement degrees of freedom.

### Given

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

### Solution

The CALFEM toolbox is used to solve the system of equations and integrate the equations of motion. First the CALFEM command “eigen” is used to solve for the eigenvalues, $\lambda$, and eigenvectors, $x$, of the generalized eigenvalue problem:
$Kx=\lambda Mx$
After entering the matrices into MATLAB, the “eigen” function is run. The MATLAB script file for “eigen” is:
The function takes in the K and M matrices and outputs the eigenvalues, L, and eigenvectors, X, shown below:
The two eigenvalues are listed in increasing order, the smaller of which is used to find the fundamental circular frequency of the system, $\omega _{1}$ , using the formula:
$\lambda _{1}=(\omega _{1})^{2}$ $\omega _{1}=2.88rad/s$
The fundamental period of the system, T1, is found through the relation:
$T_{1}=2\pi /\omega _{1}=2.18s$
The fundamental period is then used to find the time step size, dt:
$dt=T_{1}/10=0.218s$
The CALFEM command “step2” is then utilized to numerically integrate the equations of motion for time, t, in the following interval:
$0\leq t\leq 10.91s$
The function “step2” is an algorithm for dynamic solution of second-order finite elemental equations considering boundary conditions. Its script is shown here:
After plugging in all the corresponding values and running the "step2" function, a plot was made for each mass. The plots of the displacement degrees of freedom, $d_{1}(t)$ and $d_{2}(t)$, over the time interval are shown below:
$d_{1}(t)$
$d_{2}(t)$

## Problem 4

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

Using MATLAB find the displacement at the center and reactions $R_{L}$ and $R_{R}$

### Given

$E=100GPa,F=10,000N,$ area of cross sections of the three portions shown are, starting from the left,$A_{1}=10^{-4}m^{2},A_{2}=2*10^{-4}m^{2},$ and$A_{3}=10^{-4}m^{2}$

### Solution

Since forces can only be applied at nodes, divide the middle bar into two different elements. There are 4 elements and 5 nodes.
MATLAB script file:
Variables are created for the constants f changes for each case:
MATLAB output:
Center_displacement =
R_L =
R_R =
The displacements were so insignificant that the displaced lines didn't appear to render:

## Problem 5

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

Determine displacements at Node 1 and axial forces in Element 1 and 2. Use Von Mises' theory to determine if the elements will yield or not. Use Euler buckling load (Pcr=pi^2*E*I/L^2) to determine if the elements under compressive loads will buckle.

### Given

Members are made of Aluminum
Outer Dimension= 0.012 meters
Inner Dimension= 0.009 meters
E=70GPa
Force at Node 1= 1000N

### Solution

EDU>> FEA_R3p5
Pcr1 =
Pcr2 =
Displacements_at_Node1_y_x_in_Meters =
Forces_at_Node2_and_Node3_in_Newtons =

## Problem 6

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

Using MATLAB find the deflections and element forces and plot the displacement for the following load cases:
    Load Case A: F$_{13}$ = F$_{27}$ = 10,000 N
    Load Case A: F$_{14}$ = F$_{28}$ = 10,000 N
    Load Case C: F$_{13}$ = 10,000 N and F$_{27}$ = -10,000 N

### Given

$E=100GPa,A=1.0cm^{2},L=0.3m$

### Solution
#### MATLAB for 2-D Truss System

MATLAB script file:
Variables are created for the constants f changes for each case:
Build the connectivity array:
ex and ey are matrices of the coordinates of the nodes:
Solve for the displacements, plot, and find the forces
Build the global stiffness matrix:
MATLAB output:
Load Case A:
Load Case B:
Load Case C:

#### CALFEM for 2-D Truss System

MATLAB script file:
MATLAB output: For Load Case A:
For Load Case B:
For Load Case C:

## Problem 7

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

| Verify the element stiffness transformation from local coordinates to global coordinates by: $\mathbf {k^{(e)}} =\mathbf {T} ^{(e)T}\,\mathbf {\hat {k}} ^{(e)}\mathbf {T} ^{(e)}$ | (7.1) |
| Verify the element stiffness transformation from local coordinates to global coordinates by: $\mathbf {k} ^{(e)}={\widetilde {\mathbf {T} }}^{{(e)}T}\,{\widetilde {\mathbf {k} }}^{(e)}\,{\widetilde {\mathbf {T} }}^{(e)}$ | (7.2) |

### Given

The local stiffness matrix:
| $\mathbf {{\hat {k}}^{(e)}} =k^{(e)}{\begin{bmatrix}1&-1\\-1&1\\\end{bmatrix}}$ | (7.3) |
| $\mathbf {T^{(e)}} ={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}$ | (7.4) |
| $\mathbf {T^{(e)^{T}}} ={\begin{bmatrix}l^{(e)}&0\\m^{(e)}&0\\0&l^{(e)}\\0&m^{(e)}\end{bmatrix}}$ | (7.5) |
| ${\widetilde {\mathbf {k} }}^{(e)}={\begin{bmatrix}1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{bmatrix}}$ | (7.6) |
| ${\widetilde {\mathbf {T} }}^{(e)}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\-m^{(e)}&l^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\\0&0&-m^{(e)}&l^{(e)}\end{bmatrix}}$ | (7.7) |
| ${\widetilde {\mathbf {T} }}^{{(e)}T}={\begin{bmatrix}l^{(e)}&-m^{(e)}&0&0\\m^{(e)}&l^{(e)}&0&0\\0&0&l^{(e)}&-m^{(e)}\\0&0&m^{(e)}&l^{(e)}\end{bmatrix}}$ | (7.8) |

### Solution

The element stiffness matrix $\mathbf {{\hat {k}}^{(e)}}$ is expressed in local coordinates. To transform the matrix to global coordinates: $\mathbf {k^{(e)}} =\mathbf {T^{(e)-1}} \mathbf {{\hat {k}}^{(e)}} \mathbf {T^{(e)}}$, where $\mathbf {T^{(e)}}$ is the transformation matrix.
It is shown that $\mathbf {T^{(e)-1}} =\mathbf {T^{(e)T}}$ because the interior vectors are orthogonal.
| $\mathbf {T^{(e)}} ={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}$ | (7.9) |
To prove orthogonality, the dot product of two vectors must equal zero. Using the dot product definition: The dot product of two vectors a = [a$_1$, a$_2$, ..., a$_n$] and b = [b$_1$, b$_2$, ..., b$_n$] is defined as:
| $\mathbf {a} \cdot \mathbf {b} =\sum _{i=1}^{n}a_{i}b_{i}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}$ | (7.10) |
The $\mathbf {T^{(e)}}$ matrix in illustrative form:
| $\mathbf {T^{(e)}} ={\begin{bmatrix}a_{1}&a_{2}&a_{3}&a_{4}\\b_{1}&b_{2}&b_{3}&b_{4}\end{bmatrix}}$ | (7.11) |
| Therefore, $\mathbf {a} \cdot \mathbf {b} =\sum _{i=1}^{2}a_{i}b_{i}=a_{1}b_{1}+a_{2}b_{2}=l^{(e)}*0+m^{(e)}*0+0*l^{(e)}+0*m^{(e)}=0$ | (7.11) |
Now we can say,
| $\mathbf {k} ^{(e)}=\mathbf {T} ^{(e)T}\mathbf {\hat {k}} ^{(e)}\mathbf {T} ^{(e)}$ | (7.12) |
$\mathbf {T} ^{(e)T}\mathbf {\hat {k}} ^{(e)}={\begin{bmatrix}l^{(e)}&0\\m^{(e)}&0\\0&l^{(e)}\\0&m^{(e)}\end{bmatrix}}k^{(e)}{\begin{bmatrix}1&-1\\-1&1\\\end{bmatrix}}$ $\mathbf {T} ^{(e)T}\mathbf {\hat {k}} ^{(e)}=k^{(e)}{\begin{bmatrix}l^{(e)}*1+0*-1&l^{(e)}*-1+0*1\\m^{(e)}*1+0*-1&m^{(e)}*-1+0*1\\0*1+l^{(e)}*-1&0*-1+l^{(e)}*1\\0*1+m^{(e)}*-1&0*-1+m^{(e)}*1\end{bmatrix}}=k^{(e)}{\begin{bmatrix}l^{(e)}&-l^{(e)}\\m^{(e)}&-m^{(e)}\\-l^{(e)}&-l^{(e)}\\-m^{(e)}&m^{(e)}\end{bmatrix}}$ $\mathbf {k} ^{(e)}=[\mathbf {T} ^{(e)T}\mathbf {\hat {k}} ^{(e)}]\mathbf {T} ^{(e)}=k^{(e)}{\begin{bmatrix}l^{(e)}&-l^{(e)}\\m^{(e)}&-m^{(e)}\\-l^{(e)}&-l^{(e)}\\-m^{(e)}&m^{(e)}\end{bmatrix}}\mathbf {T} ^{(e)}=k^{(e)}{\begin{bmatrix}l^{(e)}&-l^{(e)}\\m^{(e)}&-m^{(e)}\\-l^{(e)}&-l^{(e)}\\-m^{(e)}&m^{(e)}\end{bmatrix}}{\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}$
The element stiffness matrix in global coordinates is:
| $\mathbf {k^{(e)}} =k^{(e)}{\begin{bmatrix}(l^{(e)})^{2}&l^{(e)}m^{(e)}&-(l^{(e)})^{2}&-l^{(e)}m^{(e)}\\l^{(e)}m^{(e)}&(m^{(e)})^{2}&-l^{(e)}m^{(e)}&-(m^{(e)})^{2}\\-(l^{(e)})^{2}&-l^{(e)}m^{(e)}&(l^{(e)})^{2}&l^{(e)}m^{(e)}\\-l^{(e)}m^{(e)}&-(m^{(e)})^{2}&l^{(e)}m^{(e)}&(m^{(e)})^{2}\end{bmatrix}}$ | (7.13) |
Proving
| $\mathbf {k} ^{(e)}={\widetilde {\mathbf {T} }}^{{(e)}T}\,{\widetilde {\mathbf {k} }}^{(e)}\,{\widetilde {\mathbf {T} }}^{(e)}$ | (7.14) |
${\widetilde {\mathbf {T} }}^{{(e)}T}{\widetilde {\mathbf {k} }}^{(e)}={\begin{bmatrix}l^{(e)}&-m^{(e)}&0&0\\m^{(e)}&l^{(e)}&0&0\\0&0&l^{(e)}&-m^{(e)}\\0&0&m^{(e)}&l^{(e)}\end{bmatrix}}k^{(e)}{\begin{bmatrix}1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{bmatrix}}$ ${\widetilde {\mathbf {T} }}^{{(e)}T}{\widetilde {\mathbf {k} }}^{(e)}=k^{(e)}{\begin{bmatrix}l^{(e)}*1+-m^{(e)}*0+0*-1+0*0&l^{(e)}*0+-m^{(e)}*0+0*0+0*0&l^{(e)}*-1+-m^{(e)}*0+0*1+0*0&l^{(e)}*0+-m^{(e)}*0+0*0+0*0\\m^{(e)}*1+l^{(e)}*0+0*-1+0*0&m^{(e)}*0+l^{(e)}*0+0*0+0*0&m^{(e)}*-1+l^{(e)}*0+0*1+0*0&m^{(e)}*0+l^{(e)}*0+0*0+0*0\\0*1+0*0+l^{(e)}*-1+-m^{(e)}*0&0*0+0*0+l^{(e)}*0+-m^{(e)}*0&0*-1+0*0+l^{(e)}*1+-m^{(e)}*0&0*0+0*0+l^{(e)}*0+-m^{(e)}*0\\0*1+0*0+m^{(e)}*-1+l^{(e)}*0&0*0+0*0+m^{(e)}*0+l^{(e)}*0&0*-1+0*0+m^{(e)}*1+l^{(e)}*0&0*0+0*0+m^{(e)}*0+l^{(e)}*0\\\end{bmatrix}}$ $k^{(e)}{\begin{bmatrix}l^{(e)}&0&-l^{(e)}&0\\m^{(e)}&0&-m^{(e)}&0\\l^{(e)}&0&l^{(e)}&0\\-m^{(e)}&0&m^{(e)}&0\end{bmatrix}}{\widetilde {\mathbf {T} }}^{(e)}=k^{(e)}{\begin{bmatrix}l^{(e)}&0&-l^{(e)}&0\\m^{(e)}&0&-m^{(e)}&0\\l^{(e)}&0&l^{(e)}&0\\-m^{(e)}&0&m^{(e)}&0\end{bmatrix}}{\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\-m^{(e)}&l^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\\0&0&-m^{(e)}&l^{(e)}\end{bmatrix}}$ ${\widetilde {\mathbf {T} }}^{{(e)}T}\,{\widetilde {\mathbf {k} }}^{(e)}\,{\widetilde {\mathbf {T} }}^{(e)}=k^{(e)}{\begin{bmatrix}(l^{(e)})^{2}&l^{(e)}m^{(e)}&-(l^{(e)})^{2}&-l^{(e)}m^{(e)}\\l^{(e)}m^{(e)}&(m^{(e)})^{2}&-l^{(e)}m^{(e)}&-(m^{(e)})^{2}\\-(l^{(e)})^{2}&-l^{(e)}m^{(e)}&(l^{(e)})^{2}&l^{(e)}m^{(e)}\\-l^{(e)}m^{(e)}&-(m^{(e)})^{2}&l^{(e)}m^{(e)}&(m^{(e)})^{2}\end{bmatrix}}$
Once again, it is shown that: $\mathbf {k^{(e)}} =k^{(e)}{\begin{bmatrix}(l^{(e)})^{2}&l^{(e)}m^{(e)}&-(l^{(e)})^{2}&-l^{(e)}m^{(e)}\\l^{(e)}m^{(e)}&(m^{(e)})^{2}&-l^{(e)}m^{(e)}&-(m^{(e)})^{2}\\-(l^{(e)})^{2}&-l^{(e)}m^{(e)}&(l^{(e)})^{2}&l^{(e)}m^{(e)}\\-l^{(e)}m^{(e)}&-(m^{(e)})^{2}&l^{(e)}m^{(e)}&(m^{(e)})^{2}\end{bmatrix}}$

## Problem 8

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

### Find

The system natural circular frequency $\omega$ for the given problem, expression for the particular solution $y_{p}(t)$, the complete solution $y(t)$, the amplification factor A and then plot the homogeneous, particular and complete solution in separate figures and then on the same figure for comparison.

### Given

Consider the spring-mass damper system on p.53-2:
| $\lambda _{1}=-0.5+i2,\lambda _{2}=-0.5-i2$ | (8.1) |
Excitation:
| $f(t)=f_{0}\cos {\bar {\omega }}t$ | (8.2) |
| $f_{0}/m=r_{0}=1$ | (8.3) |
| ${\bar {\omega }}=0.9\omega$ | (8.4) |
Initial Conditions:
$y(0)=1,y'(0)=0$

### Solution

Case: Underdamped
| $\Delta =a^{2}-4b<0$ | (8.5) |
$\alpha =-0.5,\beta =2$
| $(\lambda -(-0.5+2i))(\lambda -(-0.5-2i))=0$ | (8.6) |
| $\lambda ^{2}+1\lambda +4.25=0$ | (8.7) |
$a=1,b=4.25$
Since,
| $\omega ^{2}=b$ | (8.8) |
$\omega =2.06$
| $\zeta ={\frac {a}{2\omega }}$ | (8.9) |
$\zeta =0.243$
Therefore L2-ODE-CC:
| $y_{h}''+y_{h}'+2.25y_{h}=0$ | (8.10) |
| $y_{p}''+y_{p}'+4.25y_{p}=f_{0}\cos {\bar {\omega }}t$ | (8.11) |
The particular solution is as follows:
| $y_{p}(t)=de^{i({\bar {\omega }}t-\phi )}$ | (8.12) |
Where,
| $d={\frac {f_{0}/m}{\sqrt {(\omega ^{2}-{\bar {\omega }}^{2})^{2}+(2\zeta \omega {\bar {\omega }})^{2}}}}$ | (8.13) |
| $\tan \phi ={\frac {2\zeta \omega {\bar {\omega }}}{\omega ^{2}-{\bar {\omega }}^{2}}}$ | (8.14) |
Therefore,
| $y_{p}(t)={\frac {f_{0}/m}{\sqrt {(\omega ^{2}-{\bar {\omega }}^{2})^{2}+(2\zeta \omega {\bar {\omega }})^{2}}}}e^{i({\bar {\omega }}t-\arctan {\frac {2\zeta \omega {\bar {\omega }}}{\omega ^{2}-{\bar {\omega }}^{2}}})}$ | (8.15) |
Calculating,
| $y_{p}(t)=0.494e^{i(1.854t-1.161)}$ | (8.16) |
Taking only the real part, $e^{it}=\cos(t)+i\sin(t)$
To calculate the amplification factor A the following equations are used:
| $\rho ={\frac {\bar {\omega }}{\omega }}$ | (8.17) |
| $A={\frac {1}{\left[(1-\rho ^{2})^{2}+(2\zeta \rho )^{2}\right]^{1/2}}}$ | (8.18) |
$A=2.1$
The homogeneous solution is as follows:
| $y_{h}(t)=C_{1}e^{(-0.5t)}\cos(2t)+C_{2}e^{(-0.5t)}\sin(2t)$ | (8.19) |
The complete solution:
| $y(t)=y_{h}(t)+y_{p}(t)$ | (8.20) |
| $y(t)=C_{1}e^{(-0.5x)}\cos(2x)+C_{2}e^{(-0.5x)}\sin(2x)+0.494\cos(1.854t-1.161)$ | (8.21) |
| $y'(t)=-0.5C_{1}e^{(-0.5t)}\cos(2t)-2C_{1}e^{(-0.5t)}\sin(2t)-0.5C_{2}e^{(-0.5t)}\sin(2t)+2C_{2}e^{(-0.5t)}\cos(2t)-0.916\sin(1.854t-1.161)$ | (8.21) |
Using the two initial conditions and solving equations the coefficients are as follows:
$C_{1}=0.803,C_{2}=-0.22$
Therefore,