text:
User:Egm4313.s12.team3.R1

Report 1 =Problem 1: Spring-dashpot system in parallel=

Problem Statement
For a spring-dashpot system in parallel with mass and an applied force f(t), derive the equation of motion. See [[media:iea.s12.sec1.djvu| Sec. 1 notes]] for a similar example. Figure of a spring-dashpot system in parallel:

Problem Solution
Force Body Diagram for the mass The kinematics for this problem:
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$$\displaystyle y=y_k=y_c $$      (1.1)
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The kinetics for this problem:
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$$ \displaystyle ma+{{f}_{c}}+{{f}_{k}}=f(t) $$     (1.2)
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For a spring:
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$$\displaystyle f_k=ky_k $$     (1.3) For a dashpot:
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$$\displaystyle f_c=c{y_c}' $$     (1.4)
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Taking equation 1.3 and 1.4 and plugging it back into equation 1.2 yields:
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$$\displaystyle f(t)=my''+c{y_c}'+ky_k $$     (1.5)
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Take the time derivative of Equation 1.1 to get:
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$$\displaystyle {y}'={y_k}'={y_c}' $$     (1.6)
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Substituting Equation 1.1 and 1.6 into Equation 1.5 you get:
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$$ \displaystyle my''+cy'+ky=f(t) $$
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=Problem 2: Spring-mass-dashpot system=

Problem Statement
Derive the equation of motion for the spring-mass-dashpot system with an applied force referred to on pages 1-4 of [[media:iea.s12.sec1.djvu| section 1 notes]].

Problem Solution
Force Body Diagram Kinematics:
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$$\displaystyle y=y_c=y_k $$     (2.1) Kinetics:
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$$\displaystyle ma+{{f}_{c}}+{{f}_{k}}=f(t) $$     (2.2)
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Hookes Law:
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$$  \displaystyle {{f}_{k}}=k{{y}_{k}} $$     (2.3) Dashpot:
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$$  \displaystyle {{f}_{c}}=c{{y}_{c}}' $$     (2.4)
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$$\displaystyle ma=my'' $$     (2.5) Combining equations 2.2-2.5:
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$$\displaystyle my''+c{{y}_{c}}'+k{{y}_{k}}=f(t) $$     (2.6)
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Take the time derivative of Equation 2.1 to get:
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$$\displaystyle {y}'={y_k}'={y_c}' $$     (2.7) Final equation of motion:
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$$ \displaystyle my''+cy'+ky=f(t) $$
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=Problem 3: Spring-Dashpot-Mass System in Series=

Problem Statement
For the following system (as can be seen in the figure below) comprised of a spring, a dashpot, and a mass in series solve for the equation of motion and draw the free body diagrams.

Problem Solution
For information regarding the order and linearity of ordinary differential equations, see [[media:iea.s12.sec1.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 1]]. Also, for a similar problem look into [[media:iea.s12.sec1.djvu| Sec 1 of Loc Vu-Quoc's class EGM4313 p.1-4]]. Force Body Diagrams Kinematics

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$$\displaystyle {y}=y_k+y_c $$     (3.1)
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Where:
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$$\displaystyle {y_k}=Spring\,Displacement $$
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&
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$$\displaystyle {y_c}=Dashpot\,Displacement $$
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Kinetics
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$$\displaystyle {a}=d^2y/dt^2={y}'' $$
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From the previous equation for acceleration and the mass (represented by the variable m) the sum of the forces can represented as follows:
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$$\displaystyle m{y}''=f(t)-f_I $$     (3.2) Where:
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$$\displaystyle f_I=The\,Internal\,Forces\,on\,the\,Mass $$     (3.3)
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Taking the second derivative of Equation 3.1:
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$$\displaystyle {(y=y_k+y_c)}'' $$
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Equals
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$$\displaystyle {y}={y_k}+{y_c}'' $$     (3.4)
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From the free body diagrams above through geometry :
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$$\displaystyle f_I=f_k $$     (3.5)
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And:
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$$\displaystyle f_k=f_c $$     (3.6)
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Which finally means that:
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$$\displaystyle f_I=f_c=f_k $$     (3.7)
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With:
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$$\displaystyle f_k=k{y_k} $$     (3.8)
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&
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$$\displaystyle f_c=C{y_c}' $$     (3.9)
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From Equation 3.7 we can solve for $$\displaystyle {y_c}'$$
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$$\displaystyle k{y_k}=C{y_c}' $$     (3.10)
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$$\displaystyle {y_c}'=(k/C){y_k} $$     (3.11) Then solving for $$\displaystyle {y_c}''$$ equals:
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$$\displaystyle {(y_c')}'=((k/C)y_k)'\,\,==>{y_c}''=(k/C){y_k}' $$     (3.12)
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Then plugging Equation 3.12 into Equation 3.4:
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$$\displaystyle {y}=y_k+(k/C){y_k}' $$     (3.13)
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Substituting Equation 3.13 for $$\displaystyle {y}''$$
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$$\displaystyle m({y_k''+(k/C){y_k}'})=f(t)-f_I $$     (3.14)
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And solving for $$\displaystyle f(t)$$ with $$\displaystyle f_I=$$ Equation 3.5
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$$\displaystyle m({y_k''+(k/C){y_k}'})+f_k=f(t) $$     (3.15)
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Finally with Equation 3.8 substituted into $$\displaystyle f_k$$ in Equation 3.15 gives the final answer of:
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$$ \displaystyle m({y_k''+(k/C){y_k}'})+k{y_k}=f(t) $$
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=Problem 4: Deriving circuit equations for RLC circuit in series=

Problem Statement
Derive equations (4.1) and (4.2) from equation (4.3)

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$$\displaystyle V = LQ'' + RQ' + \frac{1}{C} Q $$ (4.1)
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$$\displaystyle V' = LI'' + RI' + \frac{1}{C} I $$ (4.2)
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$$\displaystyle V = LC \frac{d^2 v_c}{dt^2} + RC \frac{dv_c}{dt} +v_c $$     (4.3) For more information and notes relating to this problem, see [[media:iea.s12.sec2.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 2]]
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Problem Solution
Given:
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$$\displaystyle V = v_R +v_L+v_C $$     (4.4)
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$$\displaystyle v_R = RI $$ (4.5)
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$$\displaystyle v_L = L \frac{dI}{dt} = LI' $$     (4.6)
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$$\displaystyle Q = C v_C \rightarrow v_C = \frac{Q}{C} $$     (4.7)
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$$\displaystyle I = \frac{dQ}{dt} = Q' \therefore Q = \int I dt $$ (4.8)
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Part A
Derivation of Equation (4.1) From Equation (4.7)
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$$\displaystyle v_C = \frac{Q}{C} $$     (4.9) Therefore, taking the first derivative...
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$$\displaystyle \frac{dv_c}{dt} = \frac{d}{dt} (\frac{Q}{C}) = \frac{Q'}{C} $$     (4.10) and taking the second derivative...
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$$\displaystyle \frac{d^2 v_c}{dt^2} = \frac{d}{dt} (\frac{Q'}{C}) = \frac{Q''}{C} $$     (4.11) Substituting Equations (4.7), (4.10), and (4.11) into Equation (4.3)...
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$$\displaystyle V = LC (\frac{Q''}{C}) + RC (\frac{Q'}{C}) + (\frac{Q}{C}) $$     (4.12) which simplifies to...
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$$\displaystyle V = LQ'' + RQ' + \frac{1}{C}Q $$     (4.1)
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Part B
Derivation of Equation (4.2) Taking the derivative from Equation (4.1) (Previously derived from Equation (4.3))
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$$\displaystyle V' = LQ' + RQ + \frac{1}{C}Q' $$     (4.13) Taking the first derivative of Equation (4.8)...
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$$\displaystyle Q' = I $$ (4.14) and taking the second derivative...
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$$\displaystyle Q'' = I' $$ (4.15) and taking the third derivative...
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$$\displaystyle Q' = I $$     (4.16) Then substituting Equations (4.14), (4.15), and (4.16) into Equation (4.1)...
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$$\displaystyle V' = LI'' + RI' + \frac{1}{C}I $$     (4.2)
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=Problem 5: Solving Ordinary Differential Equations by find the general solution=

Problem Statement
Given ODE's of this kind find the general solution and check the results using substitution. Similar ODE's and classification are shown in [[media:Iea.s12.sec2d.djvu|Sec 2 p. 2-5]].
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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ $$ \displaystyle y''+2\pi y'+\pi^2 y=0 $$
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Part A

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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ (5.1) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$
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Equation (5.1) becomes
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$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)=0 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{16-4(\pi^2+4)})}2 \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{16-4(\pi^2+4)})}2 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{-4\pi^2)}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{-4\pi^2)}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+2\pi i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-2\pi i}{2} $$ $$   \displaystyle \lambda_1=-2+\pi i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-\pi i $$
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$$  \displaystyle y_1=e^{-2x} \cos{\pi x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y_2=e^{-2x} \sin{\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=Ae^{-2x} \cos{\pi x}+Be^{-2x} \sin{\pi x} $$ $$ \displaystyle y'=[-2Ae^{-2x} \cos{\pi x}-A\pi e^{-2x} \sin{\pi x}]+[\pi Be^{-2x} \cos{\pi x}-2Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle y''=[4Ae^{-2x} \cos{\pi x}\;+2\;\pi Ae^{-2x} \sin{\pi x}\;+\;2\pi Ae^{-2x} \sin\pi x\;-\;\pi^2 Ae^{-2x} \cos{\pi x}]\;+\;[-2B\pi e^{-2x} \cos{\pi x}\;-\;\pi^2 Be^{-2x} \sin{\pi x}\;+\;4Be^{-2x} \sin{\pi x}\;-\;2\pi Be^{-2x} \cos{\pi x}] $$ Simplifying some of the terms:
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$$  \displaystyle 4y'=[-8Ae^{-2x} \cos{\pi x}-4A\pi e^{-2x} \sin{\pi x}]+[4\pi Be^{-2x} \cos{\pi x}-8Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle (\pi^2 +4)y=(\pi^2 +4)Ae^{-2x} \cos{\pi x}+(\pi^2 +4)Be^{-2x} \sin{\pi x} $$
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Finally subbing the terms into the original equation:
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$$  \displaystyle y''\;+\;4y'\;+\;(\pi^2\;+\;4)y=[e^{-2x} \cos{\pi x}]\cdot[4A\;-\;\pi^2 A\;-\;2\pi B\;-\;2\pi B\;-\;8A\;+\;4\pi B\;+\;\pi^2 A\;+\;4A]\;+\;[e^{-2x} \sin{\pi x}]\cdot[2\pi A\;+\;2\pi A\;-\;\pi^2 B\;+\;4B\;-\;4\pi A\;-\;8B\;+\;\pi^2 B\;+\;4B] $$ $$ \displaystyle [e^{-2x} \cos{\pi x}]\cdot[0]+[e^{-2x} \sin{\pi x}]\cdot[0]=0 $$
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Part B

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$$  \displaystyle y''+2\pi y'+\pi^2 y=0 $$ (5.2) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$ Equation (5.2) becomes
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$$  \displaystyle \lambda^2+2\pi \lambda+\pi^2=0 $$ $$   \displaystyle \lambda_{1,2}=\frac{-2\pi \pm \sqrt{4\pi^2 - 4\pi^2}}{2} $$ $$   \displaystyle \lambda_{1,2}=-\pi $$
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$$  \displaystyle y_{1,2}=(c_1+c_2x)e^{-\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=(c_1+c_2x)e^{-\pi x} $$ $$ \displaystyle y'=c_2e^{-\pi x}-\pi (c_1+c_2x)e^{-\pi x} $$ $$ \displaystyle y''=-\pi c_2e^{-\pi x}\;+\;[(-\pi (c_1\;+\;c_2x)\;\cdot\; -\pi e^{-\pi x})(-\pi c_2e^{-\pi x})] =-2\pi c_2e^{-\pi x}\;+\;\pi^2 (c_1\;+\;c_2x)e^{-\pi x} $$ Finally subbing the terms into the original equation:
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$$  \displaystyle y''+2\pi y'+\pi^2 y=\pi^2 (c_1+c_2x)e^{-\pi x}-2\pi c_2e^{-\pi x}+2\pi c_2e^{-\pi x}-2\pi^2 (c_1+c_2x)e^{-\pi x}+\pi^2(c_1+c_2x)e^{-\pi x} = 0 $$
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=Problem 6: Classifying ODE's and Applying the Superposition Principle=

Problem Statement
For each of the following ODE's, determine the order and linearity. Then, determine if the superposition principle can be applied.

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$$\displaystyle {y}''=g $$     (6.1)
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$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
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$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
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$$\displaystyle m{y}''+ky=0 $$     (6.4)
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$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
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$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
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$$\displaystyle EI{y}=f(x) $$     (6.7)
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$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
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Problem Solution
For information regarding the order and linearity of ordinary differential equations, see [[media:iea.s12.sec1.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 1]]. For information regarding the application of the superposition principle, see [[media:iea.s12.sec2.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 2]].

Part A

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$$\displaystyle {y}''=g $$     (6.1)
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This is a second order, linear ODE.
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For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
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$$\displaystyle {y_h}''=0 $$     (1)
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Particular problem:
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$$\displaystyle {y_p}''=g $$     (2)
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We define the sum of the homogeneous solution and particular solution:
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$$\displaystyle \bar{y}=y_h+y_p $$     (3)
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Summing (1) and (2), we obtain the following:

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$$\displaystyle {y_h}+{y_p}=g $$     (4)
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$$\displaystyle {({y_h}'+{y_p}')}'=g $$     (5)
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$$\displaystyle {(y_h+y_p)}''=g $$     (6)
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$$\displaystyle {\bar{y}}''=g $$     (7)
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 * style="width:95%" |
 *

 * }

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

From equation (7), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part B

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

$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a second order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle m{v_h}'+bv_h^{2}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle m{v_p}'+bv_p^{2}=mg $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{v}=v_h+v_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle m({v_h}'+{v_p}')+b(v_h^{2}+v_p^{2})=mg $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle m{(v_h+v_p)}'+b(v_h^{2}+v_p^{2})=mg $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle m{\bar{v}}'+b(v_h^{2}+v_p^{2})=mg $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We cannot substitute any further since $$ b(v_h^{2}+v_p^{2}) \neq b(v_h+v_p)^{2} $$.

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

$$ \bar{v} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part C

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

$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a first order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$h_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle {h_1}'+k\sqrt{h_1}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And $$h_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle {h_2}'+k\sqrt{h_2}=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{h}=c_1h_1+c_2h_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle {h_1}'+{h_2}'+k\sqrt{h_1}+k\sqrt{h_2}=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {(h_1+h_2)}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {\bar{h}}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We cannot substitute any further since $$ k(\sqrt{h_1}+\sqrt{h_2})) \neq k\sqrt{h_1+h_2} $$.

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

$$ \bar{h} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part D

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

$$\displaystyle m{y}''+ky=0 $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$y_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle m{y_1}''+ky_1=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And $$y_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle m{y_2}''+ky_2=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=c_1y_1+c_2y_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle m({y_1}+{y_2})+k(y_1+y_2)=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle m{(y_1+y_2)}''+k(y_1+y_2)=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle m{\bar{y}}''+k\bar{y}=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part E

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

$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle {y_h}''+\omega_0^{2}y_h=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {y_p}''+\omega_0^{2}y_p=cos(\omega t) $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle {y_h}+{y_p}+\omega_0^{2}y_h+\omega_0^{2}y_p=cos(\omega t) $$ (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {(y_h+y_p)}''+\omega_0^{2}(y_h+y_p)=cos(\omega t) $$ (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle {\bar{y}}''+\omega_0^{2}(\bar{y})=cos(\omega t) $$ (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part F

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

$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle L{I_h}''+R{I_h}'+\frac{1}{C}I_h=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle L{I_p}''+R{I_p}'+\frac{1}{C}I_p={E}' $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{I}=I_h+I_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle L({I_h}+{I_p})+R({I_h}'+{I_p}')+\frac{1}{C}(I_h+I_p)={E}' $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle L{(I_h+I_p)}''+R{(I_h+I_p)}'+\frac{1}{C}(I_h+I_p)={E}' $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle L{\bar{I}}''+R{\bar{I}}'+\frac{1}{C}\bar{I}={E}' $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

From equation (6), we can see that $$ \bar{I} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part G

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

$$\displaystyle EI{y}=f(x) $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a fourth order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle EI{y_h}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle EI{y_p}=f(x) $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle EI({y_h}'+{y_p}')=f(x) $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle EI{(y_h+y_p)}=f(x) $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle EI{\bar{y}}=f(x) $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part H

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

$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

This is a second order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$\theta_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle L{\theta_1}''+gsin\theta_1 =0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And $$\theta_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle L{\theta_2}''+gsin\theta_2 =0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{\theta}=c_1\theta_1+c_2\theta_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

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

$$\displaystyle L({\theta_1}+{\theta_2})+g(sin\theta_1+sin\theta_2)=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle L{(\theta_1+\theta_2)}''+g(sin\theta_1+sin\theta_2)=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle L{\bar{\theta}}''+g(sin\theta_1+sin\theta_2)=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We cannot substitute any further since $$ sin\theta_1+sin\theta_2 \neq sin(\theta_1+\theta_2) $$.

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

$$ \bar{\theta} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=Contributing Members for This Project=

title: User:Egm4313.s12.team3.R1


text:
User:Egm4313.s12.team3.guzy

 * 1) Report 1
 * 2) Report 2
 * 3) Lecture Plan
 * 4) Exams
 * 5) Report Table

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm4313.s12.team3.guzy


text:
Chemistry (A-Level)/Transition elements

Defining a transition element
A transition metal is a d-block element that can form at least one stable ion with an incomplete d sub-shell. D-block elements are so-called because their sub-shell with the highest energy level is a d sub-shell. In A2-level chemistry, you only need to know of the first row of the d-block, elements Scandium(Sc) - Zinc(Zn). Of these ten metals, all but Scandium(Sc) and Zinc(Zn) are transition elements.

Electronic configuration
The electronic configuration for Scandium(Sc) is: 1s22s22p63s23p64s23d1. As you move across the first row of the d-block, electrons are added onto the 3d sub-shells, up to Zinc(Zn): 1s22s22p63s23p64s23d10. The 4s sub-shells are generally written before the 3d sub-shells.This is because 4s sub-shells have a lower energy level than 3d sub-shells and a therefore filled before 3d sub-shells. There are two exceptions to this in the top row of the d-block, Copper(Cu) and Chromium(Cr). The electronic configuration for Copper(Cu) is 1s22s22p63s23p64s13d5, and for Chromium(Cr) it is 1s22s22p63s23p64s13d10. These elements act like this as it is more energetically favored for the electrons to half-fill all of the orbitals, or fill all the d-orbitals leaving the s-orbital half-filled. When forming ions, electrons are taken from the 4s sub-shell before the 3d sub-shell. e.g. Nickel(Ni): 1s22s22p63s23p64s23d8 Nickel(II)(Ni2+): 1s22s22p63s23p63d8 Nickel(III)(Ni3+): 1s22s22p63s23p63d7

Properties
All of the transition metals have the properties of metals. This includes: high melting and boiling points, high density, shiny surface, forms giant metallic lattices and coducts electricity. As well as these there are properties that only apply to transition metals. These are: The transition elements you need to know (Ti to Cu) can all form two or more ions with oxidation numbers +2 and above. When light passes though a solution, some of the wavelengths are absorbed. The colour you see is a result of wavelengths that were not absorbed by the ions. e.g. A Cu2+ solution absorbed light at red/orange wavelengths, therefore we see it as blue. Generally, if a compound does not contain a transition metal ion (one with incompletely filled d-orbitals) it will be colourless.
 * Variable oxidation states
 * Form coloured compounds
 * Catalytic properties
 * From complex ions

Catalytic Uses
Transition metals are good catalysts. There are two main ways in which they are used. There are some well known uses of transition metals as catalyst that you should know.
 * They can provide a surface for a reactant to adsorb onto. The reactant can then be desorbed to take part in the reaction.
 * They can change their oxidation state and then bond with the reactants to give a lower activation energy.
 * Iron (Fe) - used in the Haber process in the production of ammonia (NH3)
 * Vanadium Oxide (V2O5) - used in the contact process in the formation of sulphur trioxide(SO3) for the production of sulpuric acid(H2SO4)
 * Nickel (Ni) - used for hydrogenation of alkenes, breaking the double bond to form alkanes
 * Manganese Oxide (MnO2) - in the decomposition of hydrogen peroxide(H2O2) which is a simple way to produce oxygen gas(O2)

Complex Ions
A complex ion contains a central transition metal ion surrounded by (usually 4 or 6) ligands. A ligand is a molecule or ion which can donate a lone pair of electrons to form a coordinate bond with a central ion in a complex ion. A complex ion is given a coordination number which indicates the number of coordinate bonds to the central ion. 6 and 4 are common coordination numbers. Some common ligands include: An example of a complex ion would be [Cu((H2O)6]2+. We can work out the charge and oxidation number by looking at the overall charge and charge of the ligands. In this example the overall charge is 2+ and the six ligands are neutral. From that we can see that the central Cu ion must have a charge of 2+ and therefore an oxidation number of +2. [FeCl4]2- contains four chloride(Cl-) ligands that add up to a total charge of 4-. Using the same technique, we can work out that the central Fe ion has a charge of 2+. [FeCl4]2- only has four ligands because of the size of the chloride(Cl-) ions. They are too big to fit six into the complex ion.
 * Water(H2O)
 * Ammonia(NH3)
 * Chloride(Cl-)
 * Hydroxide(OH-)
 * Thiocyanate(SCN-)
 * Cyanide(CN-)

Back to Chemistry (A-Level)

title: Chemistry (A-Level)/Transition elements


text:
User:Egm4313.s12.team16.matth

I dont hurt small animals for fun. Often.

Team 16

title: User:Egm4313.s12.team16.matth


text: File:R.1.1 drawing.jpg

title: File:R.1.1 drawing.jpg


text:
User talk:MSB

 Hello MSB, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 20:03, 28 January 2012 (UTC)

E-learning course
Looking forward to your contributions on Wikiversity! How did you hear about the project? I thought you might be interested in these links...User:Egm4313.s12 are both User:Areil314 running courses right now. A mixture of user subpages and main topic pages has worked very well for them. There is also a large project at the University of Canberra and a table of featured projects here. Cheers, --HappyCamper 07:41, 29 January 2012 (UTC)

 * Hi there, thanks for the recommendations and links. Followed one already and found a lot of inspiration there. Found out about this via German Wikiversity, a colleague, Christian Spannagel, uses Wikiversity a lot. I'll start with a few "private" courses (E-Learning for lecturers) and I plan to move on to "proper" university course in my field, Business Informatics--MSB 11:00, 29 January 2012 (UTC)

title: User talk:MSB


text:
Chemistry (A-Level)/Practical skills in chemistry

Overview
For A-level chemistry, you will have to partake in certain practical experiments of three types: quantitative, qualitative and evaluative. You will be graded for one of each in both you AS and A2 parts of the course, adding up to six graded experiments in the two years. You will usually have two or three attempts at each category.

Candidates should carry out experimental and investigative activities in order to develop their practical skills. Experimental and investigative activities should be set in contexts appropriate to, and reflect the demand of, the AS content. These activities should allow candidates to use their knowledge and understanding of Chemistry in planning, carrying out, analysing and evaluating their work.

Back to Chemistry (A-Level)


text:
User talk:MrPtheScienceT

 Hello Proco123, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 20:49, 28 January 2012 (UTC)

Nuclear Physics
Hey there! I noticed your Nuclear Physics learning project. Looks very interesting to me, a subject I've read a lot about on wikipedia as a sort of subject of interest to me. Let me know if I can help out at all with your new project. For starters, I could definitely whip up a nav box for it, currently I couldn't really find any of the resources except from the Recent Changes page (where I first noticed them.) If you're interested, let me know and provide links to all the pages (current and planned) and I'll put a template together. MyNameWasTaken 21:52, 1 February 2012 (UTC)
 * I like your edits here too. Keep up the good work! --HappyCamper 18:18, 8 February 2012 (UTC)

 * Just a note that I replied to your message on my talk page, apparently there's no talkback template here as on WP (yet). MyNameWasTaken 18:23, 9 February 2012 (UTC)

Copyright problems on photos
Hello. There is a problem with one or more files you have uploaded. Every file you upload must have a clear statement of the copyright position. If you took the photos yourself, add PD-self if you are placing it unreservedly in the public domain. If you wish to retain minimal rights, add Cc-by-sa-3.0. If you did not take the photo, please add an explanation of why it can be hosted here. I am afraid that if you do not do this, the file wil be deleted. Let me know if you have any problems.--Collingwood (talk) 12:56, 5 July 2012 (UTC)

title: User talk:MrPtheScienceT


text: Literature/1967/Stafford Beer

title: Literature/1967/Stafford Beer


text:
Literature/1972/Stafford Beer

Excerpts

 * "Now it is easier to think about the brain as a computer than to think about the electronic computer itself as some kind of brain." (p 127)

title: Literature/1972/Stafford Beer


title: File:R1a3a1.jpg


text:
User:Egm4313.s12.team6.berthoumieux

Statement
For each ODE in Fig.2 in K 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show wether the principle of superposition can be applied.

Given
$$(1) {y}''=g=constant$$

$$(2) m{v}'=mg-bv^{2}$$

$$(3) {h}'=-k\sqrt{h}$$

$$(4) m{y}''+ky=0$$

$$(5) y''+\omega_{0}^2=cos(\omega t)$$,  $$\omega_{0}=\omega$$

$$(6) L{I}''+R{I}'+\frac{1}{c}I={E}'$$

$$(7) EIy^{iv}=\mathit{f(x)}$$

$$(8) \mathit{L{\theta}''+gsin{\theta}=0}$$

Solution
$$(1) {y}''=g=constant$$

Order:2nd order. The highest derivative is a 2nd derivative on the y.

Linearity:Linear.

Superposition:Yes.

$$(2) m{v}'=mg-bv^{2}$$

Order:1st order. The highest derivative is a 1st on the v.

Linearity:Non-linear.

Superposition:No.

$$(3) {h}'=-k\sqrt{h}$$

Order:1st order

Linearity:Non-linear.

Superposition:No.

$$(4) m{y}''+ky=0$$

Order:2nd order.

Linearity:Linear.

Superposition:Yes.

$$(5) y''+\omega_{0}^2=cos(\omega t)$$,  $$\omega_{0}=\omega$$

Order:2nd order

Linearity:Linear.

Superposition:Yes.

$$(6) L{I}''+R{I}'+\frac{1}{c}I={E}'$$

Order:2nd order

Linearity:Linear.

Superposition:Yes.

$$(7) EIy^{iv}=\mathit{f(x)}$$

Order:4th oder

Linearity:Linear.

Superposition:Yes.

$$(8) \mathit{L{\theta}''+gsin{\theta}=0}$$

Order:2nd order

Linearity:Non-linear.

Superposition:Yes.

Statement
Realize spring-mass-dashpot systems in series as shown in the figure.

With characteristic: $${\lambda}^2-10{\lambda}+25=0$$

And with double real root at  $${\lambda}=-3$$ Find the values of k, c, m.

Solution
First, we write down the equations we can figure out from kinematics and from kinetics.

Kinematics: $$y=y_{k}+y_{c} $$

Kinetics: $$m{y}''+f_{l}=f(t)$$

$$f_{l}=f_{k}=f_{c}$$

Next, we can find any relations that exist:

$$f_{k}=ky_{k} $$

$$f_{c}=c{y_{c}}' $$

Since we already established that $$y=y_{k}+y_{c} $$  , we can say that $${y}={y_{k}} + {y_{c}}'' $$

We also established that $$f_{k}=f_{c}$$  so we can say that $$ky_{y}=c{y_{c}}'n \to {y_{c}}'=\frac{k}{c}y_{k}$$

Next, we plug in the value for $${y_{c}}'$$  :

$${y}={y_{k}}+ {({y_{c}}')}'={y_{k}}''+\frac{k}{c}{y_{k}}'$$

Finally, plug $${y}''$$  back into the kinetics equation:

$$m({y_{k}}''+\frac{k}{c}{y_{k}}')+ky_{k}=f(t)$$

$$m{y_{k}}''+m\frac{k}{c}{y_{k}}'+ky_{k}=f(t)$$

We are given that $${\lambda}=-3$$ so the characteristic equation must be:

$$(\lambda-(-3))^{2}=0$$

$$\lambda^{2}+6\lambda+9=0$$

This relates to the second order ODE:

$${y}''+6{y}'+9y=0$$

Comparing this to the following equation:

$$m{y_{k}}''+m\frac{k}{c}{y_{k}}'+ky_{k}=f(t)$$

We find that $$m=1$$ $$k=9$$ $$c=\frac{9}{6}=\frac{3}{2}$$

title: User:Egm4313.s12.team6.berthoumieux


text:
Introduction aeroelasticity

Introduction
Aeroelasticity is the interaction between aerodynamic flows and elastic structures - such that the aerodynamic forces are dependent on the structural deformation, and the structural deformation is dependent on the aerodynamic forces. This interaction depends on the flexibility of the aerodynamic object.

 * Example: A flexible wing experiences an aerodynamic force and bends. The change in shape will cause the forces to also change. The change in forces will then cause a different deformation...

Specializations
Aeroelastic problems that also concern high-gain control systems belong to the field aeroservoelasticity.

Aeroelastic problems that also include thermal effects belong to the field aerothermoelasticity.

Problem Characteristics
Aeroelastic problems tend to belong to one of four types:
 * Instability Boundary Problems which find the threshold between stability and instability
 * Static Instability Boundary Problems which don't include inertial effects. These problems frequently concern "divergence"
 * Dynamic Instability Boundary Problems which include inertial effects, and frequently have oscillatory characteristics. These problems typically concern "flutter"
 * Response Problems which predict the response of a system to some input signal.
 * Static Response Problems which don't include inertial effects
 * Dynamic Response Problems which do include inertial effects.

Simplified Aeroelastic Problems
There exist some thought-experiments which are too simplified for real-world application, but demonstrate concepts.

Steady Aerodynamics
$$M_y$$ moment about the elastic axis (positive nose up)

$$M_{AC}$$ moment about the aerodynamic center (positive nose up)

$$C_{MAC}$$ coefficient of moment about the aerodynamic center

$$q$$ dynamic pressure

$$S$$ planform wing area

$$L$$ lift force (positive up)

$$C_L$$ lift coefficient

$$C_{L_0}$$ lift coefficient at 0 angle of attack

$$\alpha$$ angle of attack

$$e$$ distance from aerodynamic center to elastic axis (positive aft)

$$M_{AC}$$ moment about the aerodynamic center

From the parallel axis theorem, we know:

$$M_y = M_{AC} + Le $$

$$= C_{MAC}qS + C_L qSe $$

$$= C_{MAC_0}qS + \left(\Complex_{L_0} + \frac{\partial{C_L}}{\partial{\alpha}} \alpha\right) qSe $$

title: Introduction aeroelasticity


text:
File:EducationalTechnologist.ConceptMap.jpg

The Educational Technologist Knowledge Domain


text:
User:Egm4313.s12.team11.gooding

Report 1: R1.2

title: User:Egm4313.s12.team11.gooding


text:
University of Florida/Egm4313/s12.team11.gooding.R1.r12

Problem Statement
Derive the equation of motion of the spring-mass-dashpot, with an applied force $$ r(t)\!$$ on the ball.

Kinematics
$$ y=y_{k}+y_{c}+y_{m}+r(t)\!$$

Kinetics
Spring $$F_{k}=-k*y\!$$

Dashpot $$F_{c}=-c*y'\!$$ Mass $$F_{m}=m*y''\!$$ Forcing Function $$F_{applied}=r(t)\!$$

Sum of Forces $$\sum F_{applied} +F_{m}+ F_{c}+ F_{k}=0\!$$

Substitution $$0=r(t)-m*y''-c*y'-k*y\!$$

Final Answer
$$r(t)=m*y''+c*y'+k*y\!$$

Created By:Kyle Gooding 01:46, 29 January 2012 (UTC)


text:
User:Egm4313.s12.team8.colocar/R1.6

Falling Stone (Sec. 1.1)
$$ \displaystyle y''=g=const $$

Parachutist (Sec. 1.2)
$$ \displaystyle mv'=mg-bv^2 $$

Out-flowing Water (Sec.1.3)
$$ \displaystyle h'=-k\sqrt{h} $$

Vibrating Mass on a Spring (Sec. 2.4,2.8)
$$ \displaystyle my''+ky=0 $$

Beats of a vibrating system (Sec. 2.8)
$$ \displaystyle y''+w_0^2y= \cos wt$$

Current I in RLC circuit (Sec. 2.9)
$$ \displaystyle LI''+RI'+ \frac 1 c I=E' $$

Deformation of a Beam (Sec. 3.3)
$$ \displaystyle EIy^{iv} = f(x) $$

Pendulum (Sec. 4.5)
$$ \displaystyle L \theta''+g \sin \theta = 0 $$

Falling Stone (Sec. 1.1)
Order: 2nd Linearity: Linear Superposition: Yes The Homogeneous Solution, $$ \displaystyle y_h''=0 $$ plus the Particular Solution, $$ \displaystyle y_p''=g $$ yields $$ \displaystyle y_h+y_p=g $$ which is the same as $$ \displaystyle (y_h+y_p)= \bar y=g $$. Therefore, the Superposition Principle can be applied.

Parachutist (Sec. 1.2)
Order: 1st Linearity: Nonlinear Superposition: No

Rewriting the original equation with $$ \displaystyle \mathit{v}$$ on one side yields $$ \displaystyle mv'+bv^2=mg $$

The Homogeneous Solution is $$ \displaystyle mv_h'+bv_h^2=0 $$.

The Particular Solution is $$ \displaystyle mv_p'+bv_p^2=mg $$. Summing both solutions yields $$ \displaystyle mv_h'+bv_h^2 + mv_p'+bv_p^2 = m(v_h+v_p)' + b(v_h^2+v_p^2) = mg $$.

$$ \bar v = v_h+v_p $$, however, $$ v_h^2+v_p^2 \neq (v_h+v_p)^2 $$

Thus, the Superposition Principle CANNOT be applied.

Out-flowing Water (Sec.1.3)
Order: 1st Linearity: Nonlinear Superposition: No

The Homogeneous Solution $$ \displaystyle h_h'=-k\sqrt{h_h} $$

plus the Particular Solution $$ \displaystyle h_p'=-k\sqrt{h_p} $$

yields $$ \displaystyle h_h'+h_p'=-k\sqrt{h_h}-k\sqrt{h_p}$$

or $$ \displaystyle (h_h+h_p)'= \bar h' = -k(h_h+h_p)^{1/2}$$.

Yet, $$ \displaystyle -k(h_h^{1/2}+h_p^{1/2}) \neq -k(h_h+h_p)^{1/2}$$.

Therefore, The Superposition Principle CANNOT be applied.

Vibrating Mass on a Spring (Sec. 2.4,2.8)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution $$ \displaystyle my_h''+ky_h=0 $$

plus the Particular Solution $$ \displaystyle my_p''+ky_p=0 $$

yields $$ \displaystyle m(y_h+y_p)''+k(y_h+y_p)=0 $$.

or $$ \displaystyle m \bar y''+k \bar y=0 $$.

Therefore, the Superposition Principle applies.

Beats of a vibrating system (Sec. 2.8)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution is $$ \displaystyle y_h''+w_0^2y_h=0 $$

The Particular Solution is $$ \displaystyle y_p''+w_0^2y_p= \cos wt $$ Summing both solutions yields $$ \displaystyle (y_h+y_p)'' + w_0^2(y_h+y_p) = \cos wt $$

We know $$ \displaystyle \bar y = y_h+y_p $$,

so $$ \displaystyle \bar y'' + w_0^2 \bar y = \cos wt $$

Therefore, the Superposition Principle can be applied.

Current I in RLC circuit (Sec. 2.9)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution is $$ \displaystyle LI_h''+RI_h'+ \frac 1 c I_h=0 $$

The Particular Solution is $$ \displaystyle LI_p''+RI_p'+ \frac 1 c I_p=E' $$ Summing both solutions yields $$ \displaystyle L(I_h+I_p)''+R(I_h+I_p)'+ \frac 1 c (I_h+I_p)=E' $$

We know $$ \displaystyle \bar I = I_h+I_p $$,

so $$ \displaystyle L \bar I''+R \bar I'+ \frac 1 c \bar I=E' $$

Therefore, the Superposition Principle can be applied.

Deformation of a Beam (Sec. 3.3)
Order: 4th Linearity: Linear Superposition: Yes

The Homogeneous Solution $$ \displaystyle EIy_h^{iv} = 0 $$

plus the Particular Solution $$ \displaystyle EIy_p^{iv} = f(x) $$

yields $$ \displaystyle EI(y_h^{iv}+y_p^{iv}) = EI(y_h+y_p)^{iv} = f(x) $$.

Knowing $$\bar y = y_h+y_p$$, we have

$$ \displaystyle EI(\bar y)^{iv} = f(x) $$.

Thus, the Superposition Principle can be applied.

Pendulum (Sec. 4.5)
Order: 2nd Linearity: Nonlinear Superposition: No

The Homogeneous Solution $$ \displaystyle L \theta_h''+g \sin \theta_h = 0 $$

plus the Particular Solution $$ \displaystyle L \theta_p''+g \sin \theta_p = 0 $$

yields $$ \displaystyle L ( \theta_h+ \theta_p)+g{( \sin \theta_h+ \sin \theta_p)} = L ( \theta_h+ \theta_p)''+g{( \sin \theta_h+ \sin \theta_p)} = 0 $$

But, $$ g{( \sin \theta_h+ \sin \theta_p)} \neq g \sin ( \theta_h + \theta_p) $$

Therefore, the Superposition Principle CANNOT be applied.

title: User:Egm4313.s12.team8.colocar/R1.6


title: File:PEA1.F09.Mtg24.pg3.fig1.new.svg


text:
Talk:Radiation astronomy/Gamma rays

Plagiarism and copyright violation check
Pass. WMF copyvio tool using TurnItIn. Short phrases or sentences such as "For X-rays, the index of refraction is defined by Rayleigh scattering,..." were similarly used in external pages, quoted and referenced here, but are not regarded as plagiarism or copyright violation. --Marshallsumter (discuss • contribs) 14:58, 19 January 2018 (UTC)

Popularity
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title: Talk:Radiation astronomy/Gamma rays


text:
User talk:Marshallsumter/Keynote lectures (draft)/Theoretical radiation astronomy

Dead link
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--Mu301Bot 08:39, 29 January 2012 (UTC)

 * Link works fine. No clue why bot's having trouble. Marshallsumter 17:08, 29 January 2012 (UTC)

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 * Link works fine. No clue why bot's having trouble. Marshallsumter 17:42, 19 February 2012 (UTC)

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 * Link works fine. No clue why bot's having trouble. Marshallsumter 17:46, 19 February 2012 (UTC)

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 * In Theoretical radiation astronomy on 2012-02-19 08:41:20, HTTP Redirect Loop: ->  ->  ->  ->

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Dispute the speedy deletion request
Custodians or Curators considering Speedy deletion of this resource, please see this same section title on Talk:Craters by radiation. --Marshallsumter (discuss • contribs) 03:56, 4 August 2018 (UTC)

 * Let's discuss this before we move it out of draft space.--Guy vandegrift (discuss • contribs) 20:12, 5 August 2018 (UTC)

 * FYI: Craters by radiation, Neutron radiation astronomy, Cosmic radiation astronomy, Theoretical radiation astronomy, First astronomical sources, Satellites for radiation astronomy, and Telescopes for radiation astronomy were never in Draft ns. See Draft talk:Craters by radiation. --Marshallsumter (discuss • contribs) 21:39, 5 August 2018 (UTC)