text:
User:Egm4313.s12.team3/Report1

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/Report1


text:
User:Egm4313.s12.team20.stevens

R1.2

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

title: User:Egm4313.s12.team20.stevens


text:
User:Egm4313.s12.team20.stevens.t

R1.2

title: User:Egm4313.s12.team20.stevens.t


text:
User:Egm4313.s12.team20.stevens.t/Report 1

Given
Derive the equation of motion of a spring-dashpot in Fig.53, in K 2011 p.85, with an applied force r(t) on the ball. The figure is shown below.

Figure 2

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

$$ \displaystyle y=y_k=y_c$$
 * style="width:95%" |
 * (2.0)
 * }

Kinetics:

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

$$ \displaystyle r(t)=-my''-f_k-f_c$$
 * style="width:95%" |
 * (2.1)
 * }

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

$$ \displaystyle f_k=ky_k$$
 * style="width:95%" |
 * (2.2)
 * }

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

$$ \displaystyle f_c=cy_c'$$
 * style="width:95%" |
 * (2.3)
 * }

After substituting in for $$\displaystyle y $$ with (2.1):


text:
User:Aburton426

My contribution to Report 1

For Report 1, I did problem 1.2

title: User:Aburton426


text:
User:Egm4313.s12.team3.katz/Brian Katz Report 2 Questions

Report 2
R2.1 (sec3 page 7) Find solution for 2 systems and plot R2.2 (sec5 page 7) L2 ODE no excitation R2.3 (sec5 page 7) similar to probs in previous report R2.4 (sec5 page 7) L2 find general solution R2.5 (sec5 page 7) Given solution find the ODE R2.6 (sec5 page 7) S-D-M system in seris find values

Problem Statement
R2.6 (sec5 page 7) S-D-M system in series find values

Problem Solution
Figure of a spring-dashpot-mass system in series Using the final equation describing the spring-dashpot-mass system:
 * {| style="width:100%" border="0"

$$\displaystyle f(t)=m({y_k}''+\frac{k}{c}{y_k}')+ky_k \;\;\; becomes $$ $$\displaystyle f(t)=m{y_k}''+m\frac{k}{c}{y_k}'+ky_k $$ Now using the homogeneous equation:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle (\lambda-(-3)^2) = {\lambda}^2-6\lambda+9=0 $$ So replacing $$\lambda$$ with y':
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle f(t)={y''}^2+6y'+9=0 $$ So now
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:40%" border="1"

$$  \displaystyle m=1 $$ $$ \displaystyle m\frac{k}{c}=6 \;\;\; \rightarrow \;\;\; \frac{(1)(9)}{c}=6 \;\;\; \rightarrow  \;\;\; 9=6c \;\;\; \rightarrow \;\;\; c=1.5 $$ $$ \displaystyle k=9 $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Problem Statement
R2.7 (sec6 page 7) MacLaurin

Part A
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
 * {| style="width:100%" border="0"

$$\displaystyle f(t) = e^t \,\,\,\,\,\,\,, \,\,\,\,\,\, f(0) = e^0 = 1 $$ $$\displaystyle f'(t) = e^t \,\,\,\,\,, \,\,\,\,\, f'(0) = e^0 = 1 $$ $$\displaystyle f(t) = e^t \,\,\,\,, \,\,\,\, f(0) = e^0 = 1 $$ $$\displaystyle f(t) = e^t \,\,\,, \,\,\, f(0) = e^0 = 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

The general form of Taylor series is:
 * {| style="width:100%" border="0"

$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$ For this problem the general form becomes:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle P_n(t)= f(0)+\frac {f'(0)}{1!} (t-0)+ \frac{f''(0)}{2!} (t-0)^2+\frac{f^{(3)}(0)}{3!}(t-0)^3+ \cdots $$ $$\displaystyle P_n(t)= 1+t+\frac {1}{2} (t)^2+ \frac{1}{6} (t)^3+ \cdots $$ In Sigma notation:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle e^t=\sum_{n=0}^{\infty }\frac{t^n}{n!} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Part B
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
 * {| style="width:100%" border="0"

$$\displaystyle f(t) = cos(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\,\, f(0) = cos(0) = 1 $$ $$\displaystyle f'(t) = -sin(t) \,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\, f'(0) = -sin(0) = 0 $$ $$\displaystyle f(t) = -cos(t) \,\,\,\,\,\,\,, \,\,\,\,\,\,\, f(0) = -cos(0) = -1 $$ $$\displaystyle f(t) = sin(t) \,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\, f(0) = sin(0) = 0 $$ $$\displaystyle f^{(4)}(t) = cos(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(4)}(0) = cos(0) = 1 $$ $$\displaystyle f^{(5)}(t) = -sin(t) \,\,\,\,, \,\,\,\, f^{(5)}(0) = -sin(0) = 0 $$ $$\displaystyle f^{(6)}(t) = -cos(t) \,\,\,\,, \,\,\,\, f^{(6)}(0) = -cos(0) = -1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

The general form of Taylor series is:
 * {| style="width:100%" border="0"

$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$ For this problem the general form becomes:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle P_n(t)= f(0)+ \frac{f''(0)}{2!} (t-0)^2+\frac{f^{(4)}(0)}{4!}(t-0)^4+\frac{f^{(6)}(0)}{6!}(t-0)^6\cdots $$ $$\displaystyle P_n(t)= 1-\frac{1}{2}(t)^2+ \frac{1}{24} (t)^4- \frac{1}{720} (t)^6 +\cdots $$ In Sigma notation:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle cos(t)=\sum_{n=0}^{\infty }\frac{(-1)^kt^{2k}}{(2k)!} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Part C
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
 * {| style="width:100%" border="0"

$$\displaystyle f(t) = sin(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\, f'''(0) = sin(0) = 0 $$ $$\displaystyle f'(t) = cos(t) \,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\, f'(0) = cos(0) = 1 $$ $$\displaystyle f(t) = -sin(t) \,\,\,\,\,\,\,, \,\,\,\,\,\,\, f(0) = -sin(0) = 0 $$ $$\displaystyle f(t) = -cos(t) \,\,\,\,\,\,, \,\,\,\,\,\, f(0) = -cos(0) = -1 $$ $$\displaystyle f^{(4)}(t) = sin(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(4)}(0) = sin(0) = 0 $$ $$\displaystyle f^{(5)}(t) = cos(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(5)}(0) = cos(0) = 1 $$ $$\displaystyle f^{(6)}(t) = -sin(t) \,\,\,\,, \,\,\,\, f^{(6)}(0) = -sin(0) = 0 $$ $$\displaystyle f^{(7)}(t) = -cos(t) \,\,\,\,, \,\,\,\, f^{(7)}(0) = -cos(0) = -1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

The general form of Taylor series is:
 * {| style="width:100%" border="0"

$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$
 * style="width:95%" |
 * style="width:95%" |
 * }

For this problem the general form becomes:
 * {| style="width:100%" border="0"

$$\displaystyle P_n(t)= \frac {f'(0)}{1!} (t-0)+\frac{f^{(3)}(0)}{3!}(t-0)^3+ \frac{f^{(5)}(0)}{5!}(t-0)^5+ \frac{f^{(7)}(0)}{7!}(t-0)^7 +\cdots $$ $$\displaystyle P_n(t)= t-\frac{1}{6}(t)^3+ \frac{1}{120} (t)^5- \frac{1}{5040} (t)^7 +\cdots $$ In Sigma notation:
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle sin(t)=\sum_{n=0}^{\infty }\frac{(-1)^kt^{2k+1}}{(2k+1)!} $$
 * style="width:95%" |
 * style="width:95%" |
 * }


text:
User:Egm4313.sp12.team4.diroma

test http://en.wikiversity.org/wiki/User:Egm4313.sp12.team4.diroma/report1

$$ function \!$$

=Problem 1=

Given the two roots:

$$\displaystyle \lambda_1 = -2 \! $$

$$\displaystyle \lambda_2 = 5 \! $$

and initial conditions:

$$\displaystyle y(0) = 1 \!$$

$$\displaystyle y'(0) = 0 \!$$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$ r(x) \! $$

 * Consider $$\displaystyle r(x) = 0 \! $$

 * Plot the solution

Solution
Characteristic Equation:

$$(\lambda-\lambda_1)(\lambda-\lambda_2) \!$$

Using the Given values for $$ \lambda_1 \! $$ and $$ \lambda_2 \! $$ we can write:

$$ (\lambda+2)(\lambda-5) = 0  \rightarrow   \lambda^2-3\lambda-10 = 0  $$

Non-homogeneous L2-ODE-CC:

$$ y''-3y'-10y = r(x) \! $$

The homogeneous solution is given by:

$$ y_h = c_1 e^{-2x} + c_2 e^{5x} \! $$

The overall solution is given by:

$$ y(x) = y_h + y_p \! $$;           where $$ y_h $$ is the homogeneous solution and $$ y_p $$ is the particular solution.

When the excitation factor $$ r(x) \! $$ is zero (as in this case), $$ y_p $$ is zero, so we can use the given initial conditions to find constants $$ c_1 \!$$ and $$ c_2 \!$$.

To do this we must first take the derivative $$ \frac{dy}{dx} $$ of $$ y_h \!$$.

Doing so, we find $$ y'_h = 0 = -2c_1+5c_2 \!$$

Multiplying $$ y_h \!$$ by 2 and adding to $$ y'_h \!$$ we can solve for $$ c_2 \!$$

$$ c_2 = \frac{2}{7} \!$$

Then multiplying $$ y_h $$ by -5 and adding to $$ y'_h $$ we can solve for $$ c_1 $$

$$ c_1 = \frac{5}{7} \!$$

Substituting in our new values for $$ c_1 \!$$ and $$ c_2 \!$$ we find the final solution to be:

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

 * style="width:95%" |
 * style="width:95%" |
 * }

To obtain three non-standard (non homogeneous) L2-ODE-CC that correspond to the same roots of our characteristic equation, we can simply multiply our characteristic equation by any non-zero constant. So choosing 2,3 and 4 as our constants we obtain:

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

 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

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 *
 * }

These will all yield the same roots; $$ \lambda_1 \! $$ and $$ \lambda_2 \! $$

Author
Solved and typed by - Egm4313.sp12.team4.diroma 04:23, 8 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.sp12.team4.diroma 04:23, 8 February 2012 (UTC)

title: User:Egm4313.sp12.team4.diroma


text:
User:Erkan Yilmaz/observations/2012 February

observations from Erkan Yilmaz for 2012 February

You are invited to add your comments (e.g. but my perception of this issue is different) - please sign anything you add (howto Signature).

PLE (again)
Is the question too simple? too dangerous? not relevant? "I don't need a special term like: PLE to help me be happy here in WV"? Who knows... Personal learning environment (PLE) question: retry after 4 months

contacting more people

thoughts about Wikiversity
The below points show some of my thoughts about what WV (could be|is)
 * I've noticed over time that concepts like learning, interaction, ... are most of the times blurry for others
 * some example: a user comes into chat and states: I don't understand what WV is for
 * discussing + user states: that sounds different than what I read/saw at WV
 * So? what does this mean?
 * for the moment I seem to focus on: products and/or/vs interaction
 * Why products? Because people (tend to) see these first
 * products is a bad term, better would be: learning resources, but that again blurs probably?

I started this from being unhappy or a more negative feeling after seeing some things yesterday + today

Where might be pointers?

 * inside
 * mental models, definitions
 * outside
 * structures people can not change + "obey" those then

changing "inside" is hard or at least not fast possible, but changing form, state is perhaps easier to achieve

"blah blah blah"...
The sections here may not always be related, so jump to the next one if you want...

 * Wiki inherits (OOP) problems due to its structure
 * an example: on my own wiki there are no problems. Why? I limit access. So: interaction or thoughts with others lead to (un)happiness?
 * actually it's the (spam) bots I'm unhappy with
 * needs Mediawiki can't fulfill
 * Where should people go instead of WV? other projects (WP, WB, ...) have their goal limitations...
 * and since they decide to be at home here: People try to form things after their (mental) models
 * There's nothing wrong with that, everyone has that right to voice their ideas + execute them
 * see harmony section below
 * Is the learning definition (see 2nd part) again which limits people?
 * "It is about the P (P stands for people and not: pages or process)": people still see "only products" and NOT people (interaction)
 * but isn't it again influenced due to software, structure wiki?
 * products are faster to see, realize than interaction
 * it's not only them: I noticed this for me also for a certain aspect AFTER adding the pic here
 * a wiki can just die, it's a normal, natural process: things start and somehow end, what stays from that? Memories? Remains in stone or other less destroyable things?
 * criticizing mental things (models, ... whatever) probably never brings a happy face - short term speaking
 * critizing where? public, private?
 * Who are the people trying to "fix" mental models? (could go into direction like: the one-eyed is King in between blind people)
 * Or: errors done in the past (by whom? teachers, parents, society ...) are to be fixed by WV participants? Even possible?

interaction

 * So, are there no successful examples of the so-called interaction after so many years of WV history?
 * There are, but people do not see them
 * So? better marketing?
 * "do not see them" or: "do not WANT TO see them" ? (should be seen less as enemy picture or so)
 * this counts for many sides:
 * admitting it may work: does that mean being wrong? But being wrong is nothing bad: no one if perfect or has all info since humans are limited (a machine/software may have perhaps facts but misses some abilities of humans. So? A mix of human + machine/software as a way out?)
 * does one have the time to see the interactions? Everything costs time, so: if I do not care about i. then I also do not want to devote time + shut down immediately

 * How could interaction be measured?
 * output: products
 * perhaps good interaction leads to more output?
 * e.g. as stimulator, awakening of creativity?
 * not necessarily: good interaction leads first to thoughts?
 * and these lead to other things, but how can one measure things separately? all connected
 * products reflect only a certain point in time + are already old when I save them
 * readers normally associate the state of a product to the current time frame, but: there has time passed + interaction progressed
 * the past influencing present/future
 * individual feedback
 * like: "I learned something valuable today"
 * but it requires someone to give feedback
 * e.g. from the past courses can be seen teachers/classes are not so willing to give feedback, see here
 * open culture, no blaming for errors or current state of a person
 * must not be direct adressed, could also be in a reflecting way, e.g. in a blog form
 * interactions can not exist alone
 * Chicken or the egg
 * so: trying to achieve harmony?
 * are humans made for harmony?
 * (perhaps in a separate section)
 * in the "good ol' days" many active in WV
 * explosion + many left
 * What do they do now in real life?
 * Their experiences (from explosion) shut down those ideals? Or are working on fulfilling them: elsewhere? Rebooting Wikiversity
 * This is also a result + one of the things the learning platform Wikiversity is for (?)
 * possible dangers: majority <-> minority decisions, manipulation, socket accounts, ...
 * so: additional effort, but then again back to aspect: who gives me the time to achieve WV-wide harmony?
 * A few people do this by spending valuable time
 * Did they get a "Thank you" for that?
 * or more negative things? (when seeing only now as good or bad)
 * Which technology is best for interaction?
 * for each individual different
 * I had to think that chat/IRC is a nice way for me
 * Why? because of its fast/timely feedback possibility?
 * Which solution could be outside of concepts like products, interaction?
 * Is there a solution?
 * it does not end or stop at a point, or?

intro
These are some options, but: they all require someone doing it. So: who might do it? and: why? Do people see this as important?

from chat:
 *  I think people have given up hope in WV
 *  people being also the internal WV people
 *  fighting alone, for oneself
 * other reasons: general decline in participation Wikimedia wide; "headquarters" not using WV in their new activities, ...
 * people maturing, analogy perhaps to federation:
 * 1. back then many used identi.ca (because not so "evil" as twitter)
 * 2. since it became more present people also use the software on their own servers
 * so like: people run now their own wikis

or... people take all for granted?
 * it's free so I do not have to care about it
 * let's misuse it? tragedy of the commons?
 * à propos misusing: people might see a blog or this blog also as misusing. But still it's possible (right now) in WV to do this (one of the freedoms)
 * I could also write it somewhere else, e.g. in my own wiki, bu then it feels like it's not belonging to WV anymore (OK, it could be imported again), but it gives additional burdens, e.g. when someone wants to add a comment, must register account...

steps

 * showing the good parts of WV
 * no one collects feedback from external institutions using WV
 * e.g. we created this page: Brick and mortar collaboration
 * when was it created? when people wanted to shut down WV
 * see would now more focus on brick and mortar institutions help WV ?
 * motivate others to talk more about "their learning" (aspects, definitions, ...)
 * could be done by asking on talk page + collecting
 * showing on front page as: user xyz thinks: Learning is/involves...
 * in meetings: Wikiversity:Meetings/Learning on Wikiversity (June 2008, summary for meeting)
 * there are no efforts anymore to do things over the various WVs :-(
 * including all of the WVs for a global vision, task, ...
 * I mean there are few contributors + WV participants might understand the situation better + more willingly help out, give feedback...
 * Perhaps initiating again Wikiversity:Reports might be easier
 * a propos vison: Vision
 * compare the length in 2009 + 2010
 * see here: blogging, using language that encourages, summarizing task force, ...

2012 February 3: reconstruction of the past is...

 * ...hard:
 * I mean I've problems in my life time to find out why these peaks happened:
 * blog
 * PLE
 * So I decided to not believe any archeologist who reconstructs something from a few thousand years back ;-)

 * ...attached with emotions:
 * Had to think of the past - wonder where all the guys are now? made me feel sad (I just saw that the below chapter also includes the word sad)

 * But some hours ago something positive:
 * had some nice grins when reading the BaB page - WV pages as time machine is just wonderful (see about time conservation: 2011-10-24 in the PLE)

 * I wonder if the humour is still alive?

2012 February 2 - it is sad when a wiki dies/disappears
Netknowledge.org is not accessible anymore (http 403) since about 1 month I think. I've emailed people for another link. If all fails, hopefully someone can give me a dump so I can host it in my wiki with same licence.
 * 2012-02-13:
 * no feedback yet
 * just reading here and can't find cached version (google or archive.org (The machine that serves this file is down. We're working on it.)) of H.'s Research:User editing patterns and collaboration

 * Hi Erkan! Unfortunately no.  I have been taking part in the Stanford online course experiment, the first of which is here.  I am coming to realise that we may need to go beyond wikimedia to have a functioning wikiversity, since, for example, the guys will never turn on the youtube extension.  Hillgentleman | //\\ |Talk 16:04, 14 February 2012 (UTC)
 * Yeah, the extension freedom is a problem. I could imagine with some fancy features, people could turn their "boring" courses into nice things (e.g. wikieducator is more courageous in using extensions).
 * On the other hand: NK did not work out + is actually gone now, wikademia has no active contributions
 * I guess the answer is: looking at all of the similar efforts and pick the best things and start "something" with those features
 * Or: just join the best initiative, since building something up alone needs more time :-( Erkan Yilmaz uses the Chat + Identi.ca 16:33, 14 February 2012 (UTC)

 * 2012-09-08:
 * I'm starting to recover that wiki's content here, though I think a major part is lost forever :-(
 * after 90 pages there come thoughts like this:
 * people who look at it might think it was an amateurish thing, but remember: you see pages only from cache from its beginning of NetKnowledge :-(
 * which reminds me how any work in its early stages (be it on paper, virtual, e.g. wiki) must look to others in its baby steps -> praise the learning process one does over time ! :-)

 * 2014-11-18:
 * 24% of the wikis I participated in the past are gone by now :-(

2012 February 1: Sandbox server - where are thou?
Sandbox server: status (not possible to set up for >1 year?) (mailing list)


text: File:R 1.2.png

title: File:R 1.2.png


text: User talk:Addihockey10 (automated)

title: User talk:Addihockey10 (automated)


text:
User:EGM4313.s12.team4.jones/R1.3

=Problem 3=

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

Solution
FBD: Force balance equation of spring-dashpot-mass system:

$$\sum F_x=f(t)-f_I=ma$$ where $$a={y}''\,\!$$

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

$$m{y}''+f_I=f(t)\,\!$$ Internal force:
 * $$\displaystyle (Eq. 2) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$f_I=f_k=f_c\,\!$$
 * $$\displaystyle (Eq. 3) $$
 * }
 * }

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

$$f_k=ky_k\,\!$$
 * $$\displaystyle (Eq. 4) $$
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$f_c={y_c}''\,\!$$ where:
 * $$\displaystyle (Eq. 5) $$
 * }
 * }

c=dashpot constant

k=spring constant

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

$$y=y_k+y_c\,\!$$ -total displacement is displacement from the spring + displacement from the dashpot
 * $$\displaystyle (Eq. 1) $$
 * }
 * }

Taking the double integral of $$y$$ you get:
 * {| style="width:100%" border="0"

$${y}={y_k}+{y_c}''\,\!$$
 * $$\displaystyle (Eq. 6) $$
 * }
 * }

Get $$y_c$$ in terms of $$y_k$$: $$f_k=f_c\,\!$$

$$ky_k=c{y_c}'\,\!$$

$${y_c}''=\frac{k}{c}y_k\,\!$$

Plug back into $${y}''\,\!$$: $${y}={y_k}+{({y_c}')}'\,\!$$

$${({y_c}')}'=\frac{k}{c}{y_k}'\,\!$$

$$\rightarrow {y}={y_k}+\frac{k}{c}{y_k}'\,\!$$

Now get $$f_I$$ in terms of $$y_k$$: $$f_I=f_k\,\!$$

$$f_k=ky_k\,\!$$

$$\rightarrow f_I=ky_k\,\!$$

When plugging in all variables into original equation you get:

$$m({y_k}''+\frac{k}{c}{y_k}')+ky_k=f(t)\,\!$$

title: User:EGM4313.s12.team4.jones/R1.3