User:EGM4313.f14.Team9.Giannattasio.G/EGM4313.f14.Team9.Giannattasio.G/R1

=Report 1=

Problem Statement
Consider a spring-damper in parallel, with mass free at one end, and the other end fixed. Derive the equations of motion and draw the free body diagrams.



Solution
The first FBD shows




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$$  \displaystyle F_b(t)=-F_a(t) $$     (1)
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Expanding to show that $$F_a(t)$$ is the reaction of $$F_b(t)$$


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$$  \displaystyle F_b(t)= k_1(y_b(t)-y_a(t))+c_1(y_b'(t)-y_a'(t)) $$     (2)
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$$  \displaystyle F_a(t)= k_1(y_a(t)-y_b(t))+c_1(y_a'(t)-y_b'(t)) $$     (3)
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Knowing that d_a(t)=d_a'(t)=0 because it is at the wall reduces Equation 2 and 3 and further proves that $$F_a(t)$$ is the reaction of $$F_b(t)$$


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$$  \displaystyle F_b(t)= k_1y_b(t)+c_1y_b'(t) $$     (4)
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$$  \displaystyle F_a(t)= -k_1y_b(t)-c_1y_b'(t) $$     (5)
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Look at the mass FBD and sum the forces




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$$  \displaystyle F(t)= my''(t)+F_b(t) $$     (6)
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Replace $$F_b(t)$$ in Equation 6 with Equation 4 to determine the solution, which is


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$$  \displaystyle F(t)= my''(t)+c_1y_b'(t)+k_1y_b(t) $$     (7)
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Problem Statement
Consider a spring-mass-damper in series with both ends fixed. Derive the equations of motion and draw the free body diagrams.



Solution
Summing the forces in the FBD




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$$  \displaystyle F(t)= my''(t)+cy_c'(t)+ky_k(t) $$     (1)
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Since the system is bounded by two fixed walls


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$$  \displaystyle y(t)=y_k(t)=y_c(t) $$     (2)
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And therefore the first and second derivative of Equation 2 are


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$$  \displaystyle y'(t)=y_k'(t)=y_c'(t) $$     (3)
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$$  \displaystyle y(t)=y_k(t)=y_c''(t) $$     (4)
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Rewrite Equation 1 to simply the solution


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$$  \displaystyle F(t)=my''(t)+cy'(t)+ky(t) $$     (5)
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Problem Statement
Consider a spring-damper-mass in series, with mass free at one end, and the other end fixed. Derive the equations of motion and draw the free body diagrams.



Solution
Kinematics


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$$  \displaystyle y(t)=y_k(t)+y_c(t) $$     (1)
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Kinetics




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$$  \displaystyle F(t)= my''(t)+F_I(t) $$     (2)
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Internal Force F_I(t)


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$$  \displaystyle F_I(t)=F_k(t)=F_c(t) \Rightarrow F_I(t)=ky(t)=cy'(t) $$     (3)
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Rearrange Equation 1 to isolate y'(t)


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$$  \displaystyle y_c'(t)=\frac{k}{c}y_k(t) $$     (4)
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Differentiate Equation 1 twice


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$$  \displaystyle y'(t)=y_k'(t)+y_c'(t) $$     (5)
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$$  \displaystyle y(t)=y_k(t)+y_c''(t) $$     (6)
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Differentiate Equation 4


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$$  \displaystyle y_c''(t)=\frac{k}{c}y_k'(t) $$     (7)
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Substitute Equation 7 into Equation 6


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$$  \displaystyle y(t)=y_k(t)+\frac{k}{c}y_k'(t) $$     (8)
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Combining Equations 2 and 3 and substituting equation 8


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$$  \displaystyle F(t)= my(t)+F_I(t) \Rightarrow F(t)= my(t)+F_k \Rightarrow  F(t)= my''(t)+ky_k(t) $$     (9)
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$$  \displaystyle y(t)=y_k(t)+\frac{k}{c}y_k'(t) $$     (8)
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After combing Equation 8 with Equation 9, Equation 10 becomes the solution


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$$  \displaystyle F(t)=m(y_k''(t)+\frac{k}{c}y_k'(t))+ky_k(t) $$     (10)
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Problem Statement
Solve the two homogenous 2nd Order ODE-CC (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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With given initial conditions


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$$  \displaystyle y(0)=1 $$
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$$  \displaystyle y'(0)=0.5 $$
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Plot Problem 3 and 12 on Separate Graphs on the Intervals


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$$  \displaystyle 0 \le x \le 1 $$
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$$  \displaystyle 0 \le x \le 5 $$
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Solution to Problem 3
Using the general equation


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$$  \displaystyle a\lambda''+b\lambda'+c\lambda=0 $$   (1)
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Given a L2-ODE two solution exists, these can be determined by the quadratic equation and comparing Problem 3 to the Equation 1


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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
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$$  \displaystyle \lambda_1,\lambda_2=\frac{-6\pm\sqrt{36-35.84}}{2} $$
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$$  \displaystyle \lambda_1=\frac{-6+\sqrt{36-35.84}}{2}\Rightarrow\lambda_1=-2.8 $$
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$$  \displaystyle \lambda_2={-6-\sqrt{36-35.84}\over2}\Rightarrow\lambda_2=-3.2 $$
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Since $$\lambda_1$$ and $$\lambda_2$$ are real roots, the solution must be in the form of


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$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$
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Plugging in the known variables


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$$  \displaystyle y=c_1e^{-2.8x}+c_2e^{-3.2x} $$ (2)
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Differentiate Equation 2


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$$  \displaystyle y'=-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x} $$ (3)
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Now there are two equations and two unknowns: $$c_1, c_2$$. Solve for the unknowns using the initial conditions: $$y(0)=1, y'(0)=0.5$$


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$$  \displaystyle y=c_1e^{-2.8x}+c_2e^{-3.2x} $$ (2)
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$$  \displaystyle y'=-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x} $$ (3)
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$$  \displaystyle y(0)=1\Rightarrow 1=c_1e^{-2.8(0)}+c_2e^{-3.2(0)}\Rightarrow 1=c_1+c_2 $$ (4)
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$$  \displaystyle y'(0)=0.5\Rightarrow 0.5=-2.8c_1e^{-2.8(0)}-3.2c_2e^{-3.2(0)}\Rightarrow 0.5=-2.8c_1-3.2c_2 $$ (5)
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$$  \displaystyle 1-c_2=c_1 $$ (6)
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Substitute Equation 5 into Equation 6


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$$  \displaystyle 0.5=-2.8(1-c_2)-3.2c_2 \Rightarrow c_2=-8.25 $$ (7)
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Plug $$c_2$$ back into Equation 6


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$$  \displaystyle 1-c_2=c_1 \Rightarrow 1-(-8.25)=c_1 \Rightarrow c_1=9.25 $$ (8)
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Using the general solution and plugging in the known constants, the solution is


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$$  \displaystyle y=9.25e^{-2.8x}-8.25e^{-3.2x} $$
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Plots Over Different Intervals
$$0 \le x \le 1$$



$$0 \le x \le 5$$



Solution to Problem 12
Using the general equation


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$$  \displaystyle a\lambda''+b\lambda'+c\lambda=0 $$   (1)
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Given a L2-ODE two solution exists, these can be determined by the quadratic equation and comparing Problem 3 to the Equation 1


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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
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$$  \displaystyle \lambda_1,\lambda_2=\frac{-9\pm\sqrt{81-80}}{2} $$
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$$  \displaystyle \lambda_1=\frac{-9+\sqrt{81-80}}{2}\Rightarrow\lambda_1=-4 $$
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$$  \displaystyle \lambda_2={-9-\sqrt{81-80}\over2}\Rightarrow\lambda_2=-5 $$
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Since $$\lambda_1$$ and $$\lambda_2$$ are real roots, the solution must be in the form of


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$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$
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Plugging in the known variables


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$$  \displaystyle y=c_1e^{-4x}+c_2e^{-5x} $$ (2)
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Differentiate Equation 2


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$$  \displaystyle y'=-4c_1e^{-4x}-5c_2e^{-5x} $$ (3)
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Now there are two equations and two unknowns: $$c_1, c_2$$. Solve for the unknowns using the initial conditions: $$y(0)=1, y'(0)=0.5$$


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$$  \displaystyle y=c_1e^{-4x}+c_2e^{-5x} $$ (2)
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$$  \displaystyle y'=-4c_1e^{-4x}-5c_2e^{-5x} $$ (3)
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$$  \displaystyle y(0)=1\Rightarrow 1=c_1e^{-4(0)}+c_2e^{-5(0)}\Rightarrow 1=c_1+c_2 $$ (4)
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$$  \displaystyle y'(0)=0.5\Rightarrow 0.5=-4c_1e^{-4(0)}-5c_2e^{-5(0)}\Rightarrow 0.5=-4c_1-5c_2 $$ (5)
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Solve for $$c_1$$ in Equation 4


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$$  \displaystyle 1-c_2=c_1 $$ (6)
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Substitute Equation 5 into Equation 6


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$$  \displaystyle 0.5=-4(1-c_2)-5c_2 \Rightarrow c_2=-4.5 $$ (7)
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Plug $$c_2$$ back into Equation 6


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$$  \displaystyle 1-c_2=c_1 \Rightarrow 1-(-4.5)=c_1 \Rightarrow c_1=5.5 $$ (8)
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Using the general solution and plugging in the known constants, the solution is


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$$  \displaystyle y=5.5e^{-4x}-4.5e^{-5x} $$
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Plots Over Different Intervals
$$0 \le x \le 1$$



$$0 \le x \le 5$$



Problem Statement
Derive Equation (2) and Equation (3) from Equation (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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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (3)
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Using known identities


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$$  \displaystyle Q = Cv_c $$     (4)
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$$  \displaystyle I=C\frac{dv_c}{dt} $$     (5)
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Derivation of Equation (2)

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$$  \displaystyle L I'' + RI' + \frac{1}{C} I = V' $$ (2)
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Solve Equation 5 for $$\frac{dv_c}{dt}$$ to match Equation 1 form


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$$  \displaystyle \frac{dv_c}{dt}=\frac{1}{C}I $$     (6)
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Differentiate Equation 6 two times


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$$  \displaystyle \frac{d^2v_c}{dt^2}=\frac{1}{C}\frac{dI}{dt}\Rightarrow \frac{d^2v_c}{dt^2}=\frac{1}{C}I' $$     (7)
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$$  \displaystyle \frac{d^3v_c}{dt^3}=\frac{1}{C}\frac{d^2I}{dt^2}\Rightarrow \frac{d^3v_c}{dt^3}=\frac{1}{C}I'' $$     (8)
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Differentiate Equation 1


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$$  \displaystyle \frac{d}{dt}(V = LC \frac{d^2 v_C}{dt^2} + RC \frac{d v_C}{dt} + v_C)\Rightarrow \frac{dV}{dt} = LC \frac{d^3 v_C}{dt^3} + RC \frac{d^2v_C}{dt^2}+\frac{dv_c}{dt} $$     (9)
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Substitute in Equation 6, 7 and 8 into 9


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$$  \displaystyle L I'' + RI' + \frac{1}{C} I = V' $$ (2)
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Derivation of Equation (3)

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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (3)
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Use Equation 4


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$$  \displaystyle Q = Cv_c $$     (4)
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Divide by C in Equation 4 to isolate $$v_c$$


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$$  \displaystyle \frac{1}{C}Q = v_c $$     (10)
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Differentiate Equation 5


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$$  \displaystyle \frac{1}{C}\frac{dQ}{dt} = \frac{dv_c}{dt}\Rightarrow \frac{dv_c}{dt}=\frac{1}{C}Q' $$     (11)
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Differentiate Equation 6


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$$  \displaystyle \frac{1}{C}\frac{d^2Q}{dt^2} = \frac{d^2v_c}{dt^2}\Rightarrow \frac{d^2v_c}{dt^2}=\frac{1}{C}Q'' $$     (12)
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Substitute Equation 10, 11, 12 into Equation 1


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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (13)
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Problem Statement
Consider the same 2-dof system as on page 53.21, with the following parameters.



Stiffness coefficients

•	$$k_1 = k_2 = k_3 =5$$

Mass coefficients

•	$$m_1 = m_2 =1.3$$

Damping coefficients

•	$$c_1 = c_2 = c_3 =2$$

For simplicity, all the FBD for the spring and damper system have the same variable since in this problem the different resistances and damping coefficients are the same.

Find the equation of motion in matrix form.

===Solution===

FBD for Element 1 and summing the forces




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$$  \displaystyle f_B^{(1)}(t)=-f_A^{(1)}(t) $$     (1)
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Expanding to show that $$F_a(t)$$ is the reaction of F_b(t)


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$$  \displaystyle f_B^{(1)}(t)= k_1(d_B(t)-d_A(t))+c_1(d_B'(t)-d_A'(t)) $$     (2)
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$$  \displaystyle f_A^{(1)}(t)= k_1(d_A(t)-d_B(t))+c_1(d_A'(t)-d_B'(t)) $$     (3)
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Plugging in known constants


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$$  \displaystyle f_B^{(1)}(t)= 5(d_B(t)-d_A(t))+2(d_B'(t)-d_A'(t)) $$     (4)
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$$  \displaystyle f_A^{(1)}(t)= 5(d_A(t)-d_B(t))+2(d_A'(t)-d_B'(t)) $$     (5)
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Set up matrix in the form   $$\mathbf f^{(1)} = \mathbf k^{(1)} \mathbf d^{(1)} + \mathbf c^{(1)} {\mathbf d}'^{(1)}$$ and set $$d_A(t)=d'_A(t)=0$$. Plug in known values.


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$$  \displaystyle \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix}  \Rightarrow \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix}  5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} 0 \\ d_B \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} 0 \\ d'_B \end{Bmatrix} $$     (6)
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FBD for Element 2 and summing the forces is the same as FBD 1 except the neither distance is zero. Plug in known values.




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$$  \displaystyle \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} \Rightarrow \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}  5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} $$     (7)
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FBD for Element 3 is the same as FBD for Element 1 but flipped with $$d_D=d_D'=0$$. Plug in known values.




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$$  \displaystyle \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} \Rightarrow \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix}  5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_C \\ 0 \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ 0 \end{Bmatrix} $$     (8)
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Look at $$m_1$$ FBD and sum forces




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$$  \displaystyle F_1(t)= m_1d''_1+f_B^{(1)}(t)+f_B^{(2)}(t) $$     (9)
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From the matrix of element 2 (Equation 7), $$ F_B^{(2)}(t)$$ can be determined


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$$  \displaystyle f_B^(t){(2)} = [ k_2 d_B - k_2 d_C ] + [ c_2 d'_B - c_2 d'_C ] $$     (10)
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Replacing $$ F_B^{(1)}(t)$$ and $$ F_B^{(2)}(t)$$ in Equation 9 with Equations 4 and 10


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$$  \displaystyle F_1(t)=m_1d''_1 + [ -c_1 d'_A + (c_1 + c_2) d'_B -c_2 d'_C ] + [ -k_1 d_A + (k_1 + k_2) d_B -k_2 d_C ] $$     (10)
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$$  \displaystyle \Rightarrow F_1(t)=1.3 d''_1 + [ -2 d'_A + (2 + 2) d'_B -2 d'_C ] + [ -5 d_A + (5 + 5) d_B -5 d_C ] $$     (10)
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Simplify Equation 10


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$$  \displaystyle F_1(t)=1.3 d''_1 + [ -2 d'_A + 4 d'_B -2 d'_C ] + [ -5 d_A + 10 d_B -5 d_C ] $$     (11)
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Look at $$m_2$$ FBD and sum forces




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$$  \displaystyle F_2(t)= m_2d_2''(t)+f_C^{(2)}(t)+f_C^{(3)}(t) $$     (12)
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From the matrix of element 2 (Equation 7) $$ F_C^{(2)}(t)$$ can be determined and from matrix of element 3 (Equation 8) $$ F_C^{(3)}(t)$$ can be determined and plug in all constants.


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$$  \displaystyle f_C^{(2)}(t) = [ -k_2 d_B + k_1 d_C ] + [ -c_2 d'_B + c_1 d'_C ] \Rightarrow f_C^{(2)}(t) = [ -5 d_B + 5 d_C ] + [ -2 d'_B + 2 d'_C ] $$     (13)
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$$  \displaystyle f_B^{(3)}(t) = [ -k_3 d_D + k_3 d_C ] + [ -c_3 d'_D + c_3 d'_C ] \Rightarrow f_B^{(3)}(t) = [ -5 d_D + 5 d_C ] + [ -2 d'_D + 2 d'_C ] $$     (14)
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Replacing $$ F_C^{(2)}(t)$$ and $$ F_C^{(3)}(t)$$ in Equation 9 with Equations 13 and 14 respectively


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$$  \displaystyle F_2(t)=m_2 d''_ + [ -c_2 d'_B + (c_2 + c_3) d'_C -c_3 d'_D ] + [ -k_2 d_B + (k_2 + k_3) d_C -k_3 d_D ] $$     (15)
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$$  \displaystyle \Rightarrow F_C(t)=1.3 d''_C + [ -2 d'_B + (2 + 2) d'_C -2 d'_D ] + [ -5 d_B + (5 + 5) d_C -5 d_D ] $$
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Simplify Equation 15


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$$  \displaystyle F_2(t)=1.3 d''_C + [ -2 d'_B + 4 d'_C -2 d'_D ] + [ -5 d_B + 10 d_C -5 d_D ] $$     (16)
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General form equations of motion in matrix form


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$$  \displaystyle \mathbf M {\mathbf d}'' + \mathbf C {\mathbf d}' + \mathbf K {\mathbf d} = \mathbf F (t) $$     (17)
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The equation of motion for this problem is


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$$  \displaystyle \mathbf F (t) = \mathbf M {\mathbf d}'' + \mathbf C {\mathbf d}' + \mathbf K {\mathbf d} $$ (17)
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$$  \displaystyle \Rightarrow \mathbf F = \begin{Bmatrix} F_1 \\ F_2 \end{Bmatrix}= \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \mathbf \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} + \begin{bmatrix} (c_1 + c_2) & - c_2 \\ - c_2 & (c_2 + c_3) \end{bmatrix} \mathbf \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} + \begin{bmatrix} (k_1 + k_2) & - k_2 \\ - k_2 & (k_2 + k_3) \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} $$     (18)
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Recall Equation 11 and 15 and apply boundary conditions: $$ d_A=d'_A=0, d_D=d'_D=0$$ and $$ d_B=d_1, d_C=d_2$$ (just a variable change for consistency with general form)


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$$  \displaystyle F_1(t)=1.3 d''_1 + [ 4 d'_1 -2 d'_2 ] + [ 10 d_1 -5 d_2 ] $$     (11)
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$$  \displaystyle F_2(t)=1.3 d''_C + [ -2 d'_1 + 4 d'_2 ] + [ -5 d_1 + 10 d_2 ] $$     (16)
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$$  \displaystyle \Rightarrow \Rightarrow \mathbf F = \begin{Bmatrix} F_1 \\ F_2 \end{Bmatrix}= \begin{bmatrix} 1.3 & 0 \\ 0 & 1.3 \end{bmatrix} \mathbf \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} + \begin{bmatrix} 4 & - 2 \\ - 2  & 4 \end{bmatrix} \mathbf \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} + \begin{bmatrix} 10 & - 5 \\ -5 & 10 \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} $$     (19)
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