# University of Florida/Egm4313/s12.team11.R1

Report 1
Intermediate Engineering Analysis Section 7566 Team 11 Due date: February 1, 2012.

## Problem R1.1: Spring-dashpot system

Solved by Francisco Arrieta

### Problem Statement

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

### Solution
#### Kinematics

    $y=y_{k}=y_{c}\!$ (1)
    Since the spring and dashpot are placed parallel to each other, they will experience the same deformation when a force $f(t)$ is applied on the system

#### Kinetics

    $f(t)-f_{k}-f_{c}=my''\!$ (2)
    Based on Newton's Second Law of Motion $\sum F=ma$ where $a$ is the second derivative of position with respect to time

#### Relations

    $f_{k}=ky_{k}\!$ (3)
    $f_{c}=cy_{c}'\!$ (4)
Using (1) in (2)&(3):
    $f_{k}=ky\!$
    $f_{c}=cy'\!$
Then (2) becomes:
Created by: Francisco Arrieta 22:05, 31 January 2012 (UTC)

## Problem R1.2: Spring-mass-dashpot system (parallel)

Solved by Kyle Gooding

### Problem Statement

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

#### Figures
### Solution
#### 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
$\sum F_{applied}+F_{m}+F_{m}+F_{c}+F_{k}=0\!$
Substitution
$0=r(t)-m*y''-c*y'-k*y\!$
And finally we find:
Created By: Kyle Gooding 01:46, 29 January 2012 (UTC)

## Problem R1.3: Spring-mass-dashpot system (series)

Solved by Luca Imponenti

### Problem Statement

For the spring-dashpot-mass system shown, draw the FBDs and derive the following equation of motion $my''+f_{I}=f(t)\!$

#### Figures
##### Free Body Diagram
### Solution

Where f$_I$ is the resistive force from both the spring and dash-pot and f(t) is our applied force.

#### Kinetics

$\sum F_{y}=ma=f(t)-f_{I}\!$
which can be re-written as
Created by: Luca Imponenti 03:01, 30 January 2012 (UTC)

## Problem R1.4: Circuit equation derivation

Solved by Gonzalo Perez

### Problem Statement

Derive (3) and (4) from (2).
Given:
$V=LC{\frac {dV_{c}^{2}}{dt^{2}}}+RC{\frac {dV_{c}}{dt}}+V_{c}$ (2)
$V'=LI''+RI'+{\frac {1}{C}}I$ (3)
$V=LQ''+RQ'+{\frac {1}{C}}Q$ (4)

### Solution

First, let's solve for (3).
Recall that:
$Q=CV_{c}=\int Idt$,
and
$I=C{\frac {dV_{c}}{dt}}={\frac {dQ}{dt}}$.
We can use this information to replace the differentiating terms accordingly.
After doing so, we get:
$V=L{I}'+RI+V_{c}\!$
but knowing that $I={\frac {dQ}{dt}}$, we can rearrange the terms to get $V_{c}={\frac {1}{C}}\int Idt$.
Using this information in the previously derived equation, we find that:
$V=LI'+RI+{\frac {1}{C}}\int Idt$.
Finally, after differentiating $V\!$ with respect to $t\!$, we get:
$V'=LI''+RI'+{\frac {1}{C}}I$.
Now, let's solve for (4).
Once again considering that $Q=CV_{c}\!$, we can solve for (4) by differentiating
twice and then plugging it into (2).
Deriving twice, we find that:
$Q=CV_{c}\!$
$Q'=C{\frac {dV_{c}}{dt}}$
$Q''=C{\frac {dV_{c}^{2}}{dt^{2}}}$
After plugging this into (2), we see that:
$V=LQ''+RQ'+V_{c}\!$.
Once rearranging $Q=CV_{c}\!$, we find that $V_{c}={\frac {Q}{C}}$.
We can plug this in to the above equation to get the solution:
Created by: Gonzalo Perez 22:07, 31 January 2012 (UTC)

## Problem R1.5: General solutions to ODEs

Solved by Jonathan Sheider
These problems are taken directly from Erwin Kreyszig's Advanced Engineering Mathematics, p.59 problems 4-5. The theory behind solving the differential equations is based on the derived mathematical approach to Solving Homogeneous Linear ODE's with Constant Coefficients (Erwin Kreyszig's Advanced Engineering Mathematics, pages 53-59).

### K 2011 p.59 pb. 4
#### Problem Statement

Find a general solution to the given ODE. Check your answer by substituting into the original equation.
Given:
${y}''+4{y}'+(\pi ^{2}+4)y=0\!$

#### Solution

The characteristic equation of this ODE is therefore:
$\lambda ^{2}+a\lambda +b=\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!$
Evaluating the discriminant:
$a^{2}-4b=4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0\!$
Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:
$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))\!$
Where: $\omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {\pi ^{2}+4-{\frac {1}{4}}(4)^{2}}}={\sqrt {\pi ^{2}}}=\pi \!$
And finally we find the general homogenous solution:
Checking:
We found that:
$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$
Differentiating $y\!$ to obtain ${y}'\!$ and ${y}''\!$ respectively:
${y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!$
${y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$
Substituting these equations into the original ODE yields:
$4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$
$+4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x)))+(\pi ^{2}+4)(e^{-2x}(Acos(\pi x)+Bsin(\pi x)))=0\!$
$4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$
$-8e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!$
$(4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x))+(-4+4)\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+(\pi ^{2}-\pi ^{2})e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!$
$0\equiv 0\!$
Therefore this solution is correct.

### K 2011 p.59 pb. 5
#### Problem Statement

Find a general solution to the given ODE. Check your answer by substituting into the original equation.
Given:
${y}''+2\pi {y}'+\pi ^{2}y=0\!$

#### Solution

The characteristic equation of this ODE is therefore:
$\lambda ^{2}+a\lambda +b=\lambda ^{2}+2\pi \lambda +\pi ^{2}=0\!$
Evaluating the discriminant:
$a^{2}-4b=(2\pi )^{2}-4(\pi ^{2})=4\pi ^{2}-4\pi ^{2}=0\!$
Therefore the equation has a real double root and a general homogenous solution of the form:
$y=(c_{1}+c_{2}x)e^{-ax/2}\!$
And finally we find the general homogenous solution:
Checking:
We found that:
$y=(c_{1}+c_{2}x)e^{-\pi x}\!$
Differentiating $y\!$ to obtain ${y}'\!$ and ${y}''\!$ respectively:
And,
${y}''=-\pi c_{2}e^{-\pi x}-\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}\!$
Substituting these equations into the original ODE yields:
$-2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+2\pi (c_{2}e^{-\pi x}-\pi (c_{1}+c_{2}x)e^{-\pi x})+\pi ^{2}((c_{1}+c_{2}x)e^{-\pi x})=0\!$
$-2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+2\pi c_{2}e^{-\pi x}-2\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}=0\!$
$(-2\pi +2\pi )c_{2}e^{-\pi x}+(\pi ^{2}-2\pi ^{2}+\pi ^{2})(c_{1}+c_{2}x)e^{-\pi x}=0\!$
$0\equiv 0\!$
Therefore this solution is correct.
Created by: Jonathan Sheider 16:27, 31 January 2012 (UTC)

## Problem R1.6: ODE linearity and superposition

Solved by Daniel Suh

### Problem Statement

For each Ordinary Differential Equations, determine the order, linearity, and show whether the principal of superposition can be applied.

#### Order of ODEs

To find the order of Ordinary Differential Equations, (ODEs) simply check the differential order of the dependent variable. If the dependent variable is differentiated once, then it is a first order ODE. If it is differentiated twice, it is a second order ODE.

#### Linearity

To find the linearity of the ODE, check the power of the dependent variable. If the dependent variable is raised to the power of anything one than 1, then it is not linear. If the dependent variable is raised to the power of 1, then it is linear.

#### Superposition Principle

   For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.
The function satisfies the superposition principle if the sum of the homogeneous solution and the particular solution is equal to ${\overline {y}}(x)\!$, with ${\overline {y}}=(y_{h}+y_{p})\!$ ${\overline {y}}(x):=y_{h}(x)+y_{p}(x)\!$

### Solutions
#### R1.6a

    $y''=g=const\!$
    Check superposition principle:
       $y''=g\!$
       $y_{h}''=0\!$
       $y_{p}''=g\!$
       $(y_{h}''+y_{p}'')\equiv {\overline {y}}''\!$

#### R1.6b

    $mv'=mg-bv^{2}\!$
    Check superposition principle:
       $mv'+bv^{2}=mg\!$
       $mv_{h}'+bv_{h}^{2}=0\!$
       $mv_{p}'+bv_{p}^{2}=mg\!$
       $m(v_{h}'+v_{p}')+b(v_{h}^{2}+v_{p}^{2})\not \equiv m{\overline {v}}'+b({\overline {v}}^{2})\!$

#### R1.6c

    $h'=-k{\sqrt {h}}\!$
    Check superposition principle:
       $h'+kh^{1/2}=0\!$
       $h_{h}'+kh_{h}^{1/2}=0\!$
       $h_{p}'+kh_{p}^{1/2}=0\!$
       $(h_{h}'+h_{p}')+k(h_{h}^{1/2}+h_{p}^{1/2})\not \equiv {\overline {h}}'+k{\overline {h}}^{1/2}\!$

#### R1.6d

    $my''+ky=0\!$
    Check superposition principle:
       $y''+{\frac {k}{m}}y=0\!$
       $y_{h}''+{\frac {k}{m}}y_{h}=0\!$
       $y_{p}''+{\frac {k}{m}}y_{p}=0\!$
       $(y_{h}''+y_{p}'')+{\frac {k}{m}}(y_{h}+y_{p})\equiv {\overline {y}}''+{\frac {k}{m}}{\overline {y}}\!$

#### R1.6e

    $y''+\omega _{o}^{2}y=\cos(\omega t)\!$
    Check superposition principle:
       $y''+\omega _{o}^{2}y=\cos(\omega t)\!$
       $y_{h}''+\omega _{o}^{2}y_{h}=0\!$
       $y_{p}''+\omega _{o}^{2}y_{p}=\cos(\omega t)\!$
       $(y_{h}''+y_{p}'')+\omega _{o}^{2}(y_{h}+y_{p})\equiv {\overline {y}}''+\omega _{o}^{2}{\overline {y}}\!$

#### R1.6f

    $LI''+RI'+{\frac {1}{C}}I=E'\!$
    Check superposition principle:
       $I''+{\frac {R}{L}}I'+{\frac {1}{LC}}I={\frac {E'}{L}}\!$
       $I_{h}''+{\frac {R}{L}}I_{h}'+{\frac {1}{LC}}I_{h}=0\!$
       $I_{p}''+{\frac {R}{L}}I_{p}'+{\frac {1}{LC}}I_{p}={\frac {E'}{L}}\!$
       $(I_{h}''+I_{p}'')+{\frac {R}{L}}(I_{h}'+I_{p}')+{\frac {1}{LC}}(I_{h}+I_{p})\equiv {\overline {I}}''+{\frac {R}{L}}{\overline {I}}'+{\frac {1}{LC}}{\overline {I}}\!$

#### R1.6g

    $EIy^{iv}=f(x)\!$
    Check superposition principle:
       $y''''-{\frac {1}{EI}}y=0\!$
       $y_{h}''''-{\frac {1}{EI}}y_{h}=0\!$
       $y_{p}''''-{\frac {1}{EI}}y_{p}=0\!$
       $(y_{h}''''+y_{p}'''')-{\frac {1}{EI}}(y_{h}+y_{p})\equiv {\overline {y}}''''-{\frac {1}{EI}}{\overline {y}}\!$

#### R1.6h

    $L\theta ''+g\sin \theta =0\!$
    Check superposition principle:
       $\theta ''+{\frac {g}{L}}\sin \theta =0\!$
       $\theta _{h}''+{\frac {g}{L}}\sin \theta _{h}=0\!$
       $\theta _{p}''+{\frac {g}{L}}\sin \theta _{p}=0\!$
       $(\theta _{h}''+\theta _{p}'')+{\frac {g}{L}}(\sin \theta _{h}+\sin \theta _{p})\not \equiv {\overline {\theta }}''+{\frac {g}{L}}\sin {\overline {\theta }}\!$

#### R1.6i

    $y_{1}^{'}=ay_{1}-by_{1}y_{2}\!$
    $y_{2}^{'}=ky_{1}y_{2}-ly_{1}\!$
Created by [Daniel Suh] 21:59, 31 January 2012 (UTC)