User:EGM4313.f14.Team1.Linehan/Report 1

=Report 1=

Problem Statement


Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force f(t).

Solution
Start with a FBD:



Internal force at node B (right cut).


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$$ f_B = k_1(d_B - d_A) + c_1(d'_B - d'_A) $$     (1)
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Internal force at node A (left cut).


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$$ f_A = -f_B = k_1(-d_B + d_A) + c_1(-d'_B + d'_A) $$     (2)
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Combine (1) and (2) in matrix form.


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$$ \begin{Bmatrix} f_A \\ f_B \end{Bmatrix} = k_1\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + c_1\begin{bmatrix}  1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$     (3)
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Apply boundary conditions to (3).


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$$ d_A = d'_A = 0 $$
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$$ \begin{Bmatrix} f_A \\ f_B \end{Bmatrix} = k_1\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} \cancelto{0}{d_A} \\ d_B \end{Bmatrix} + c_1\begin{bmatrix}  1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} \cancelto{0}{d'_A} \\ d'_B \end{Bmatrix} $$
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Break (3) out of matrix form.


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$$ f_A = k_1(-d_B) + c_1(-d'_B) = -(k_1d_B + c_1d'_B) $$     (4)
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$$ f_B = -f_A = k_1d_B + c_1d'_B $$     (5)
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$$ m_1d_1 = m_1d_B $$
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Sum the forces about node B.


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$$ \sum_{} f = 0 = f_1(t) - f_B - m_1d''_B = 0 $$     (6)
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Substitute (5) into (6).


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$$ f_1(t) - (k_1d_B + c_1d'_B) - m_1d''_B = 0 $$
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Subtract $$f_1(t)$$ from both sides of the equation and divide by $$m_1$$.


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$$ d''_B + \frac{c_1}{m_1}d'_B + \frac{k_1}{m_1}d_B = \frac{f_1(t)}{m_1} $$
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Problem Statement


Derive the equation of motion of the spring-mass-dashpot in Fig.53, in Kreyszig 2011 p.85, with an applied force r(t) on the ball.

Solution
Because the system is fixed on both ends, the displacements will be equal in magnitude:


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$$ y=y_k=y_c $$
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Differentiate twice to get all the necessary terms:


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$$ y'=y'_k=y'_c $$
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$$ y=y_k=y''_c $$
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As shown in the FBD, sum the forces in the x-direction:
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$$ f_c=f_B^{(2)} ; f_k=f_B^{(1)} $$
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$$ my''+f_c+f_k=F(t) $$     (1)
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Equation of the force due to the damper:


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$$ f_c=cy' $$     (2)
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Equation of the force due to the spring:


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$$ f_k=ky $$     (3)
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Plugging in Equations (2) and (3) to (1):


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$$ my''+cy'+ky=F(t) $$
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Problem Statement


For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2) p.1-4.

Solution
When a spring and dashpot are in series, their displacement contributions can be added:


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$$ y = y_k + y_c $$
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Differentiate twice in order to get the necessary terms to substitute into the equation of motion:


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$$ y' = y'_k + y'_c $$
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$$ y = y_k + y''_c $$     (1)
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Since the spring and dashpot are in series, we can assume that their force contributions are equal in magnitude:


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$$ f_k = f_c = f_B^{(1)} $$
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$$ k y_k = c y'_c $$
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Solving for the velocity term yields:


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$$ \Rightarrow y'_c = \frac{k}{c} y_k $$     (2)
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From FBD of $$ m_1 $$ the equation of motion can be derived in terms of forces.


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$$ m y'' + f_B^{(1)} = f (t) $$
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From (1) we can recover the acceleration term by plugging in the solution for (2).


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$$
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y = y_k + y_c  = y_k + \left( y'_c \right)'   = y''_k + \left( \frac{k}{c} y_k \right)' $$
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Plugging in this solution into eq. of motion we retrieve:


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$$ \Rightarrow m \left( y_k'' + \frac{k}{c} y_k' \right) + k y_k = f (t) $$
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Simplifying and placing ODE in standard form:


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$$ y_k'' + \frac{k}{c} y_k' + \frac{k}{m} y_k = \frac{f (t)}{m} $$
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Problem Statement
Solve the homogeneous L2-ODE-CC with initial conditions: $$y(0)= 1 ; y'(0)= 0.5$$ (Kreyszig 2011, Sec.2.2, p.59, Pbs.3,12)

Problem 3:
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$$  \displaystyle y''+6y'+8.96y=0 $$
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Problem 12:


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$$  \displaystyle y''+9y'+20y=0 $$
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Problem 3

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$$  \displaystyle y''+6y'+8.96y=0 $$
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Initial Conditions: $$y(0)= 1 ; y'(0)= 0.5$$

Make sure the ODE is in the form (1):


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$$ \lambda''+a\lambda'+b\lambda=0 $$     (1)
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Use the quadratic equation (2) to find the roots of the ODE.


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$$ \lambda_1,\lambda_2= {-a\pm\sqrt{a^2-4b}\over2} $$     (2)
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$$	\lambda_1={-6+\sqrt{6^2-4(8.96)}\over2}=-3+0.2=-2.8 $$
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$$ \lambda_2={-6-\sqrt{6^2-4(8.96)}\over2}=-3-0.2=-3.2 $$
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Now, we must write an equation relating the solutions $$y_1,y_2 $$ to two real roots of the ODE.


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$$ y_1=e^{\lambda_1 x}=e^{-2.8x} $$
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$$ y_2=e^{\lambda_2 x}=e^{-3.2x} $$
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The general solution (3) is:


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$$ y=c_1 e^{\lambda_1 x}+c_2 e^{\lambda_2 x} $$ (3)
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$$ y=c_1 e^{-2.8x}+c_2 e^{-3.2x} $$
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Use the initial conditions to derive a system of equations necessary to solve for $$c_1$$ and $$c_2$$.


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$$ y=c_1 e^{-2.8x}+c_2 e^{-3.2x} $$
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$$ 1=c_1 e^{-2.8(0)}+c_2 e^{-3.2(0)} $$
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$$ 1=c_1+c_2 $$
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Take the first derivative of the general solution and plug in the initial conditions.


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$$ y'=-2.8c_1 e^{-2.8x} - 3.2c_2 e^{-3.2x} $$
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$$ {1\over2}=-2.8c_1 e^{-2.8(0)} - 3.2c_2 e^{-3.2(0)} $$
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$$ {1\over2}=-2.8c_1 - 3.2c_2 $$
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Solve the system of equations (1) and (2)


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$$ 0.5=-2.8(1-c_2)- 3.2c_2 $$
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$$ 0.5=-2.8- 0.4c_2 $$
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$$ c_2=-8.25 $$
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$$ c_1=9.25 $$
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The final solution is:


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$$ y=9.25e^{-2.8x}-8.25e^{-3.2x} $$
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Problem 12

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$$  \displaystyle y''+9y'+20y=0 $$
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Initial Conditions: $$ y(0)= 1 ; y'(0)= 0.5 $$

Make sure the ODE is in the form (1):
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$$ \lambda''+a\lambda'+b\lambda=0 $$     (1)
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Use the quadratic equation (2) to find the roots of the ODE.


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$$ \lambda_1,\lambda_2= {-a\pm\sqrt{a^2-4b}\over2} $$     (2)
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$$	\lambda_1={-9+\sqrt{9^2-4(20)}\over2}=-4.5+0.5=-4 $$
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$$ \lambda_2={-9-\sqrt{9^2-4(20)}\over2}=-4.5-0.5=-5 $$
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Now, we must write an equation relating the solutions $$y_1,y_2 $$ to two real roots of the ODE.


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$$ y_1=e^{\lambda_1 x}=e^{-4x} $$
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$$ y_2=e^{\lambda_2 x}=e^{-5x} $$
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The general solution (3) is:


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$$ y=c_1 e^{\lambda_1 x}+c_2 e^{\lambda_2 x} $$ (3)
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$$ y=c_1 e^{-4x}+c_2 e^{-5x} $$
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Use the initial conditions to derive a system of equations necessary to solve for $$c_1$$ and $$c_2$$.


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$$ y=c_1 e^{-4x}+c_2 e^{-5x} $$
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$$ 1=c_1 e^{-4(0)}+c_2 e^{-5(0)} $$
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$$ 1=c_1+c_2 $$
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Take the first derivative of the general solution and plug in the initial conditions.


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$$ y'=-4c_1 e^{-4x} - 5c_2 e^{-5x} $$
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$$ {1\over2}=-4c_1 e^{-4(0)} - 5c_2 e^{-5(0)} $$
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$$ {1\over2}=-4c_1 - 5c_2 $$
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Solve the system of equations (1) and (2)


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$$ 0.5=-4(1-c_2)- 5c_2 $$
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$$ 0.5=-4- c_2 $$
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$$ c_2=-4.5 $$
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$$ c_1=5.5 $$
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The final solution is:


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$$ y=5.5e^{-4x}-4.5e^{-5x} $$
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Problem Statement
Derive (2) and (3) from (1).


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$$  \displaystyle V = LC \frac{d^2 v_C}{dt^2} + RC \frac{d v_C}{dt} + v_C $$     (1)
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$$  \displaystyle L I'' + RI' + \frac{1}{C} I = V' $$ (2)
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$$  \displaystyle L Q'' + RQ' + \frac{1}{C} Q = V $$ (3)
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Derivation of (2)

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$$  \displaystyle V = LC \frac{d^2 v_C}{dt^2} + RC \frac{d v_C}{dt} + v_C $$     (1)
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$$  \displaystyle Q = Cv_C $$     (4)
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$$  \displaystyle Cv_C = \int i \, dt $$ (5)
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Differentiate both sides of (5).


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$$   \displaystyle i = C \frac{d v_C}{dt} $$
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Differentiate (5) again.


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$$   \displaystyle \frac{di}{dt} = C \frac{d^2 v_C}{dt^2} $$
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Substitute the first term of (5) and rewrite.


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$$   \displaystyle LC \frac{d^2 v_C}{dt^2} = L \frac{di}{dt} = LI' $$
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Substitute the second term of (1).


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$$   \displaystyle RC \frac{d v_C}{dt} = RI $$
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Divide both sides of (5) by C.


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$$   \displaystyle v_C = \frac{1}{C} \int i \, dt $$
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Substitute back into (1) to form (6).


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$$   \displaystyle V = LI' + RI + \frac{1}{C} \int i \, dt $$
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Differentiate both sides of the equation.


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$$   \displaystyle \frac{d}{dt}[V = LI' + RI + \frac{1}{C} \int i \, dt] $$
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$$   \displaystyle V' = LI'' + RI' + \frac{1}{C}I $$     (2)
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Derivation of (3)
Divide both sides of (4) by C.


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$$   \displaystyle Q = Cv_C \Rightarrow \frac{1}{C}Q = v_C $$
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Differentiate both sides of (4)


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$$   \displaystyle \frac{d}{dt}[Q = \int i \, dt] \Rightarrow I = \frac{dq}{dt} $$
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Differentiate (4) again


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$$   \displaystyle \frac{d}{dt}[I = \frac{dq}{dt}] \Rightarrow I' = \frac{d^2q}{dt^2} $$
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Substitute into (6) to form (3).


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$$   \displaystyle V = LI' + RI + \frac{1}{C}\int i \, dt = LQ'' + RQ' + \frac{1}{C}Q $$     (3)
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Problem Statement
Find the equation of motion in matrix form for the following 2-dof system.

Spring Constants
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$$ k_1 = k_2 = k_3 = 5 $$     (1)
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Mass Coefficients
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$$ m_1 = m_2 = 1.3 $$     (2)
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Dampening Coefficients
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$$ c_1 = c_2 = c_3 = 2 $$     (3)
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Solution
Element 1




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$$ f_A^{(1)} = k_1 ( -d_B + d_A ) + c_1 ( -d'_B + d'_A ) = 5 ( -d_B + d_A ) + 2 ( -d'_B + d'_A ) $$
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$$ f_B^{(1)} = k_1 ( d_B - d_A ) + c_1 ( d'_B - d'_A ) = 5 ( d_B - d_A ) + 2 ( d'_B - d'_A ) $$
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Rearrange Terms in Equations to Prepare to Change to Matrix Form


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$$ f_A^{(1)} = k_1 ( d_A - d_B ) + c_1 ( d'_A - d'_B ) = 5 ( d_A - d_B ) + 2 ( d'_A - d'_B ) $$
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$$ f_B^{(1)} = k_1 ( -d_A + d_B ) + c_1 ( -d'_A + d'_B ) = 5 ( -d_A + d_B ) + 2 ( -d'_A + d'_B ) $$
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Equations in Matrix Form


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$$ \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} = \begin{bmatrix} 5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$
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Element 2

Start with FBD:



Force matrix follows same topology as in element 1.


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$$ \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}   k_2 & - k_2 \\ -k_2 &   k_2 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} $$
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Element 3

Start with FBD:



Force matrix follows same topology as in elements 1 and 2.


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$$ \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix}   k_3 & - k_3 \\ -k_3 &   k_3 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  c_3 & - c_3 \\ -c_3 & c_3 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} $$
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Mass B



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


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$$ f_B^{(1)} = \begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix} = \begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix} $$
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$$ f_B^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
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Sub in For Forces


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$$ F_B(t) = m_Bd''_B + (\begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix}) + (\begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix}) $$  (1)
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$$ = 1.3d''_B + (\begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix}) + (\begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_B - 2d'_C \end{bmatrix}) $$  (1)
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Mass C



$$ \sum_{} f = 0 = F_C(t)- m_Cd''_C- f_C^{(2)} - f_C^{(3)} $$


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$$ f_C^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
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$$ f_C^{(3)} = \begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix} = \begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix} $$
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Sub in For Forces


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$$ F_C(t) = m_Cd''_C + (\begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix}) + (\begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix}) $$ (2)
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$$ = 1.3d''_C + (\begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix}) + (\begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix}) $$ (2)
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Boundary Conditions for Both Equations of Motion


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$$ d_A(t) = d'_A(t) = 0 $$
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$$ d_D(t) = d'_D(t) = 0 $$
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Equations of Motion After Substituting in Boundary Conditions


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$$ F_B(t) = m_Bd_B + k_1d_B + c_1d'_b + \begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_b - c_2d'_C \end{bmatrix} = 1.3d_B + 5d_B + 2d'_b + \begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_b - 2d'_C \end{bmatrix} $$   (3)
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$$ F_C(t) = m_Cd_C + \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + k_3d_C + c_3d'_C = 1.3d_C + \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + 5d_C + 2d'_C $$   (4)
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Into Matrix Form


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$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} d_b \\ d_c \end{bmatrix} $$
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Sub In Values Below into Equation of Motion
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$$ k_1 = k_2 = k_3 = 5 $$
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$$ m_1 = m_2 = 1.3 $$
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$$ c_1 = c_2 = c_3 = 2 $$
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$$ m_B = m_1 $$
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$$ m_C = m_2 $$
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Final Equation of Motion with All Values Substituted In


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$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
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