User:Egm4313.s12.team18/R1

=Report 1= Team 18

Problem
Derive the equation of motion of the spring-dashpot system in parallel.

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



Solution
Solution:

Using methods of kinematics to note that the displacement of the dashpot is equivalent to that of the spring:
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$$  \displaystyle y=y_k=y_c\! $$     (1)
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Using methods of kinetics, specifically Newton's 2nd Law:
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$$  \displaystyle my''+f_I=f(t)\! $$     (2)
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$$  \displaystyle f_I=f_k+f_c\! $$     (3)
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Constitutive Relations of forces for spring and dashpot:
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$$  \displaystyle f_k=ky_k=ky\! $$     (4)
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$$  \displaystyle f_c=cv=cy_c'\!=cy' $$     (5)
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From (3), (4), and (5):
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$$  \displaystyle f_I=ky+cy' $$     (6)
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Inserting (2) into (6):
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$$  \displaystyle my''+cy'+ky=f(t) $$
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Author
Solved and typed by - Egm4313.s12.team18.crete 16:25, 29 January 2012 (UTC)

Problem
Derive the equation of motion of the spring-mass-dashpot in fig. 5.3 in K 2011 p 85 with an applied force r(t) on the ball.



Solution
Find: The equation of motion with the applied force, r(t).




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$$  \displaystyle {F_c = cy'} $$     (1)
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$$  \displaystyle {F_k = ky} $$     (2)
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Using Newton's Second Law:
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$$  \displaystyle {\Sigma\vec{F}=m\vec{a}} $$
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$$  \displaystyle {\vec{a}=y''} $$
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From the Free Body Diagram above:


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$$  \displaystyle -ma=F_c+F_k-r(t) $$
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Or


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$$  \displaystyle r(t)=F_c+F_k+ma $$     (3)
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Plug Equation 1 and Equation 2 into Equation 3.


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$$  \displaystyle {r(t)=cy'+ky+my} $$
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Or


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$$  \displaystyle {r(t)=my''+cy'+ky} $$
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Author
Solved and typed by - Egm4313.s12.team18.gibson.mg 21:49, 30 January 2012 (UTC)

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

Spring-Mass-Dashpot System:

Solution
Spring FBD: Dashpot FBD: Mass FBD: Given:
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$$ \displaystyle f_k=f_c=f_I $$     (1) Using Newton's Second Law:
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$$ \displaystyle \sum \mathbf{F} = m\frac{\partial V}{\partial t} $$ (2)
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$$ \displaystyle \frac{\partial V}{\partial t}=a(t)=y{}'' $$     (3) Therefore
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$$ \displaystyle \sum \mathbf{F} = my{}'' $$     (4)
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Analyzing Mass FBD with Equation (1):
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$$ \displaystyle \sum \mathbf{F} = f(t)-f_I $$     (5) Combining Equation (4) & (5):
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$$ \displaystyle f(t)-f_I= my{}''
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$$     (6) Therefore:
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$$ \displaystyle f(t)= my{}''+f_I $$
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Author
Solved and typed by - Egm4313.s12.team18.alvinb 17:13, 29 January 2012 (UTC)

Problem
The goal of this exercise is to derive the alternative formulations of the circuit equation seen in (3) and (4) using the circuit equation from (2). Where:


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$$  \displaystyle Q=Cv_C \Rightarrow \int i\,dt=Cv_c \Rightarrow i = C \frac{dv_C}{dt} $$     (1a, 1b, 1c)
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$$  \displaystyle V=LC\frac{d^2v_C}{dt^2}+RC\frac{dv_C}{dt}+v_C $$     (2)
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$$  \displaystyle LI''+RI'+\frac{1}{C}I=V' $$     (3)
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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
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Deriving (3) from (2)
In order to derive first alternate version, (3), of the circuit equation, (2), we begin by taking the derivative of (2) with respect to time, the result of which can be seen below in (5)


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$$  \displaystyle V'=LC \frac{d^3 v_C}{dt^3} + RC \frac{d^2 v_C}{dt^2} + \frac{d v_C}{dt} $$     (5)
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Next, by rearranging the right hand equation from (1) we obtain the following relation:


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$$  \displaystyle \frac{I}{C}=\frac{d v_C}{dt} $$     (6)
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And by taking subsequent derivatives, we also get:
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$$  \displaystyle \frac{I'}{C}=\frac{d^2 v_C}{dt^2} \, \, \& \, \, \frac{I''}{C}=\frac{d^3 v_C}{dt^3} $$     (7) & (8)
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By plugging the relations from (6), (7), and (8) into equation (2) and reducing the fractions we obtain the variant seen in (3)
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$$  \displaystyle LI'' + RI' + \frac{1}{C}I=V' $$     (3)
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Deriving (4) from (2)
We can derive the second alternate equation, (4), from circuit equation (2) using a method similar to the one used to derive (3) as shown above.

By taking the derivatives of (1a) we obtain the following relations:


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$$  \displaystyle Q=Cv_C \, \, \& \, \, Q'=C \frac{dv_C}{dt} \, \, \& \, \, Q''=C \frac{d^2 v_C}{dt^2} $$     (9a, 9b, 9c)
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Once again, plugging (9a, 9b, 9c) into (2) and reducing the results we can obtain the second variation of the circuit equation: (4)
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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
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Author
Solved and typed by - Egm4313.s12.team18.hollon 19:22, 28 January 2012 (UTC)

Problem
K 2011 p.59 pb 4.

Find a general solution. Check your answer by substitution.


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$$  \displaystyle y''+4y'+(\pi^2+4)y=0 $$     (1)
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Solution
Note This is a homogenous linear 2nd order ordinary differential equation (ODE) with constant coefficients. Trial Solution:
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$$  \displaystyle y=e^{\lambda x} $$ (2)
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$$  \displaystyle y'=\lambda e^{\lambda x} $$ (3)
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$$  \displaystyle y''=\lambda^2 e^{\lambda x} $$ (4) Plugging (2),(3), and (4) into (1)
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$$  \displaystyle (\lambda^2 + 4\lambda + (\pi^2+4))e^{\lambda x} = 0 $$     (5)
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The characteristic equation is
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$$  \displaystyle \lambda^2 + 4\lambda + (\pi^2+4) = 0 $$     (6)
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Note The coefficients of this quadratic equation are
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$$  \displaystyle a=1 $$
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$$  \displaystyle b=4 $$
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$$  \displaystyle c=\pi^2+4 $$
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The discriminant can be found as follows
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$$  \displaystyle \Delta = b^2 - 4ac $$     (7)
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$$  \displaystyle \Delta = 4^2 - 4(1)(\pi^2+4) $$
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Note The discriminant is negative. Therefore, Eq (6) has complex roots of the form


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$$  \displaystyle \lambda = -1/2a \pm i\omega $$     (8) where
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$$  \displaystyle \omega=\sqrt{c- b^2/4} $$     (9)
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$$  \displaystyle y(x)=e^{-bx/2} (Acos\omega x + Bsin\omega x) $$ (10) Plugging known quantities into (9) and (10)
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$$  \displaystyle y(x)=e^{-2x} (Acos\sqrt{(\pi^2+4)-4}\,\,x + Bsin\sqrt{(\pi^2+4)-4}\,\,x) $$
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Now The general solution to the ODE is
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$$  \displaystyle y(x)=e^{-2x} (Acos\pi\,x + Bsin\pi\,x) $$
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Problem
K 2011 p.59 pb 5. Find a general solution. Check your answer by substation.


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$$  \displaystyle y'' + 2{\pi}y' +{\pi ^2}y = 0
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$$     (1)
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Solution
This is a homogeneous linear ODE with constant coefficients.
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$$  \displaystyle a=2\pi $$     (2)
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$$  \displaystyle b=\pi ^2 $$     (3)
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$$  \displaystyle a^2 -4b $$     (4)
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Substitution equations 2 and 3 into equation 4 we see that
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$$  \displaystyle a^2 -4b = (2 \pi) ^2 - 4( \pi ^2)= 4 \pi ^2 -4 \pi ^2 = 0 $$     (5)
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Because equation 5 is equal to 0, the solution is a real double root. Therefore the solution will be in the form
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$$  \displaystyle y = (c_1 + c_2 x)e^ {\lambda x}
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$$     (6)
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Where
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$$  \displaystyle \lambda = -\frac{a}{2}= -\frac{2 \pi}{2} = -\pi
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$$     (7)
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By combining equations 6 and 7 we see that the general solution is
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$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
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$$     (8)
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In order to check the solution, we take the first and second derivatives of equation 8


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$$  \displaystyle y' = c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}
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$$     (9)
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$$  \displaystyle y'' = -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x}
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$$     (10)
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After taking the derivates, we substitute equations 8, 9, and 10 back into equation 1


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$$  \displaystyle -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x} + 2\pi (c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}) + {\pi} ^2 ( (c_1 + c_2 x)e^ {\lambda x}) = 0
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$$     (11)
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Equation 11 is true. Therefore,
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$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
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$$
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Is a general solution to equation 1.

Authors
Solved and typed by - Egm4313.s12.team18.burgher.wb, Egm4313.s12.team18.roberts.jr13:56, 30 January 2012 (UTC)

Problem
For each ODE in Fig. 2 in K2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition applies.

Tower of Pisa

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$$ \displaystyle { y{}''=g } $$ (1)
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Order: Second Linear Principle of Superposition applies:


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$$ \displaystyle { y_{h}{}''=0} $$ (1.1)
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$$ \displaystyle { y_{p}{}''=g} $$ (1.2)
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$$ \displaystyle {(y_{h}+y_{p})=\bar{y}} $$ (1.3)
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$$ \displaystyle { (y_{h}+y_{p}){}''=g} $$ (1.4)
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$$ \displaystyle {\mathbf{\bar{ y}{}''=g}} $$ Is the same as $$ \displaystyle { \mathbf{y{}''=g}} $$ (1.5)
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Parachutist

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$$ \displaystyle {mv{}'+bv^{2}=mg} $$ (2)
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Order: First Non-Linear Principle of Superposition does not apply:


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$$ \displaystyle { mv_{h}{}'+bv_{h}^{2}=0} $$ (2.1)
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$$ \displaystyle { mv_{p}{}'+bv_{p}^{2}=mg} $$ (2.2)
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$$ \displaystyle {(v_{h}+v_{p})=\bar{v}} $$ (2.3)
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$$ \displaystyle { m(v_{h}+v_{p}){}'+b(v_{h}^{2}+v_{p}^{2})=mg} $$ (2.4)
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$$ \displaystyle {\mathbf{ m\bar{v}{}'+b(v_{h}^{2}+v_{p}^{2})=mg} } $$ Is not the same as $$ \displaystyle { \mathbf{mv{}'+bv^{2}=mg}} $$ (2.5)
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Out-flowing Water

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$$ \displaystyle {h{}'=-k\sqrt{h}} $$ (3)
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Order: First Non-Linear Principle of Superposition does not apply:


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$$ \displaystyle { h_{h}{}'+k\sqrt{h_{h}}=0} $$ (3.1)
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$$ \displaystyle { h_{p}{}'+k\sqrt{h_{p}}=0} $$ (3.2)
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$$ \displaystyle {(h_{h}+h_{p})=\bar{h}} $$ (3.3)
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$$ \displaystyle { (h_{h}+h_{p}){}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0} $$ (3.4)
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$$ \displaystyle {\mathbf{\bar{h}{}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0 }} $$ Is not the same as $$ \displaystyle {\mathbf{ h{}'+k\sqrt{h}=0}} $$ (3.5)
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Vibrating Mass on a String

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$$ \displaystyle {my{}''+ky=0} $$ (4)
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Order:2nd Linear Principle of Superposition applies:


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$$ \displaystyle {my_{h}{}''+ky_{h}=0} $$ (4.1)
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$$ \displaystyle {my_{p}{}''+ky_{p}=0} $$ (4.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (4.3)
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$$ \displaystyle {m(y_{p}+y_{h}){}''+k(y_{p}+y_{h})=0} $$ (4.4)
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$$  \displaystyle {\mathbf{m\bar{y}''+k\bar{y}=0 }} $$ is the same as $$ \displaystyle { \mathbf{my{}''+ky=0}} $$     (4.5)
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Beats of a Vibrating Sytem

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$$ \displaystyle {y{}''+\omega _{0}^{2}y=\cos( \omega t)} $$ (5)
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Order:2nd Linear Principle of Superposition applies:


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$$ \displaystyle {y_{h}{}''+\omega _{0}^{2}y_{h}=0} $$ (5.1)
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$$ \displaystyle {y_{p}{}''+\omega _{0}^{2}y_{p}=\cos( \omega t)} $$ (5.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (5.3)
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 * }
 * }


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$$ \displaystyle {(y_{p}+y_{h}){}''+\omega _{0}^{2}(y_{p}+y_{h})=\cos( \omega t)} $$ (5.4)
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 * }
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$$ \displaystyle {\mathbf{\bar{y}{}''+\omega _{0}^{2}\bar{y}=\cos( \omega t)}} $$ Is the same as $$ \displaystyle {\mathbf{ y{}''+\omega _{0}^{2}y=\cos( \omega t)}} $$ (5.5)
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 * }
 * }

Current I in RLC Circuit

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$$ \displaystyle {LI{}''+RI{}'+\frac{1}{c}I=E{}'} $$ (6)
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 * }
 * }

Order:2nd Linear Principle of Superposition applies:


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$$ \displaystyle {LI_{h}{}''+RI_{h}{}'+\frac{1}{c}I_{h}=0} $$ (6.1)
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 * }


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$$ \displaystyle {LI_{p}{}''+RI_{p}{}'+\frac{1}{c}I_{p}=E{}'} $$ (6.2)
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 * }
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$$ \displaystyle {(I_{p}+I_{h})=\bar{I}} $$ (6.3)
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 * }
 * }


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$$ \displaystyle {L(I_{p}+I_{h}){}''+R(I_{p}+I_{h}){}'+\frac{1}{c}(I_{p}+I_{h})=E{}'} $$ (6.4)
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 * }
 * }


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$$ \displaystyle {\mathbf{L\bar{I}{}''+R\bar{I}{}'+\frac{1}{c}\bar{I}=E{}' }} $$ is the same as $$ \displaystyle {\mathbf{ LI{}''+RI{}'+\frac{1}{c}I=E{}'}} $$ (6.5)
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 * <p style="text-align:right;">
 * }
 * }

Deformation of a Beam

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$$ \displaystyle {EIy^{iv}=f(x)} $$ (7)
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Order:Fourth Linear Principle of Superposition applies:


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$$ \displaystyle {EIy_{h}^{iv}=0} $$ (7.1)
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 * }
 * }


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$$ \displaystyle {EIy_{p}^{iv}=f(x)} $$ (7.2)
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 * }
 * }


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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (7.3)
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 * }
 * }


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$$ \displaystyle {EI(y_{h}+y_{p})^{iv}=f(x)} $$ (7.4)
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 * }
 * }


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$$ \displaystyle {\mathbf{EI\bar{y}^{iv}=f(x)} } $$ Is the same as $$ \displaystyle {\mathbf{ EIy^{iv}=f(x)}} $$ (7.5)
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 * }
 * }

Pendulum

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$$ \displaystyle {L\Theta {}''+gsin(\Theta )=0} $$ (8)
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 * }
 * }

Order:Second Non-Linear Principle of Superposition does not apply:


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$$ \displaystyle {L\Theta_{h} {}''+gsin(\Theta_{h} )=0} $$ (8.1)
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 * }
 * }


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$$ \displaystyle {L\Theta_{p} {}''+gsin(\Theta_{p} )=0} $$ (8.2)
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 * <p style="text-align:right;">
 * }
 * }


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$$ \displaystyle {(\Theta_{p}+\Theta_{h})=\bar{\Theta}} $$ (8.3)
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 * }
 * }


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$$ \displaystyle {L(\Theta_{p}+\Theta_{h}){}''+gsin(\Theta_{p}+\Theta_{h})=0} $$ (8.4)
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 * }
 * }


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$$ \displaystyle {\mathbf{L\bar{\Theta} {}+gsin(\Theta_{p}+\Theta_{h} )=0}} $$ Is not the same as $$ \mathbf{L\Theta {}+gsin(\Theta )=0} $$
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Author
Solved and typed by: Egm4313.s12.team18.windham 18:36, 30 January 2012 (UCT)