# 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 ${\displaystyle f(t)}$.

### Solution

#### Kinematics

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

#### Kinetics

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

#### Relations

${\displaystyle f_{k}=ky_{k}\!}$ (3)
${\displaystyle f_{c}=cy_{c}'\!}$ (4)

Using (1) in (2)&(3):

${\displaystyle f_{k}=ky\!}$
${\displaystyle f_{c}=cy'\!}$

Then (2) becomes:

                                                     ${\displaystyle f(t)-ky-cy'=my''\!}$


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 ${\displaystyle r(t)\!}$ on the ball.

### Solution

#### Kinematics

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

#### Kinetics

Spring

${\displaystyle F_{k}=-k*y\!}$

Dashpot

${\displaystyle F_{c}=-c*y'\!}$

Mass

${\displaystyle F_{m}=m*y''\!}$

Forcing Function

${\displaystyle F_{applied}=r(t)\!}$

Sum of Forces

${\displaystyle \sum F_{applied}+F_{m}+F_{m}+F_{c}+F_{k}=0\!}$

Substitution

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

And finally we find:

                                                 ${\displaystyle r(t)=m*y''+c*y'+k*y\!}$


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
${\displaystyle my''+f_{I}=f(t)\!}$

### Solution

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

#### Kinetics

${\displaystyle \sum F_{y}=ma=f(t)-f_{I}\!}$

which can be re-written as

                                                        ${\displaystyle my''+f_{I}=f(t)\!}$


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:

${\displaystyle V=LC{\frac {dV_{c}^{2}}{dt^{2}}}+RC{\frac {dV_{c}}{dt}}+V_{c}}$ (2)

${\displaystyle V'=LI''+RI'+{\frac {1}{C}}I}$ (3)

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

### Solution

First, let's solve for (3).

Recall that:

${\displaystyle Q=CV_{c}=\int Idt}$,

and

${\displaystyle I=C{\frac {dV_{c}}{dt}}={\frac {dQ}{dt}}}$.

We can use this information to replace the differentiating terms accordingly.

After doing so, we get:

${\displaystyle V=L{I}'+RI+V_{c}\!}$

but knowing that ${\displaystyle I={\frac {dQ}{dt}}}$, we can rearrange the terms to get ${\displaystyle V_{c}={\frac {1}{C}}\int Idt}$.

Using this information in the previously derived equation, we find that:

${\displaystyle V=LI'+RI+{\frac {1}{C}}\int Idt}$.

Finally, after differentiating ${\displaystyle V\!}$ with respect to ${\displaystyle t\!}$, we get:

${\displaystyle V'=LI''+RI'+{\frac {1}{C}}I}$.

Now, let's solve for (4).

Once again considering that ${\displaystyle Q=CV_{c}\!}$, we can solve for (4) by differentiating

twice and then plugging it into (2).

Deriving twice, we find that:

${\displaystyle Q=CV_{c}\!}$

${\displaystyle Q'=C{\frac {dV_{c}}{dt}}}$

${\displaystyle Q''=C{\frac {dV_{c}^{2}}{dt^{2}}}}$

After plugging this into (2), we see that:

${\displaystyle V=LQ''+RQ'+V_{c}\!}$.

Once rearranging ${\displaystyle Q=CV_{c}\!}$, we find that ${\displaystyle V_{c}={\frac {Q}{C}}}$.

We can plug this in to the above equation to get the solution:

                                                       ${\displaystyle V=LQ''+RQ'+{\frac {1}{C}}Q}$


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:

${\displaystyle {y}''+4{y}'+(\pi ^{2}+4)y=0\!}$

#### Solution

The characteristic equation of this ODE is therefore:

${\displaystyle \lambda ^{2}+a\lambda +b=\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!}$

Evaluating the discriminant:

${\displaystyle a^{2}-4b=4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0\!}$

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

${\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))\!}$ [1]

Where: ${\displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {\pi ^{2}+4-{\frac {1}{4}}(4)^{2}}}={\sqrt {\pi ^{2}}}=\pi \!}$

And finally we find the general homogenous solution:

                                                  ${\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$


Checking:

We found that:

${\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

Differentiating ${\displaystyle y\!}$ to obtain ${\displaystyle {y}'\!}$ and ${\displaystyle {y}''\!}$ respectively:

${\displaystyle {y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!}$

                                   ${\displaystyle {y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))\!}$


${\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

                 ${\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$


Substituting these equations into the original ODE yields:

${\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$
${\displaystyle +4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x)))+(\pi ^{2}+4)(e^{-2x}(Acos(\pi x)+Bsin(\pi x)))=0\!}$

${\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$
${\displaystyle -8e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}$

${\displaystyle (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x))+(-4+4)\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+(\pi ^{2}-\pi ^{2})e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}$

${\displaystyle 0\equiv 0\!}$

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:

${\displaystyle {y}''+2\pi {y}'+\pi ^{2}y=0\!}$

#### Solution

The characteristic equation of this ODE is therefore:

${\displaystyle \lambda ^{2}+a\lambda +b=\lambda ^{2}+2\pi \lambda +\pi ^{2}=0\!}$

Evaluating the discriminant:

${\displaystyle a^{2}-4b=(2\pi )^{2}-4(\pi ^{2})=4\pi ^{2}-4\pi ^{2}=0\!}$

Therefore the equation has a real double root and a general homogenous solution of the form:

${\displaystyle y=(c_{1}+c_{2}x)e^{-ax/2}\!}$ [2]

And finally we find the general homogenous solution:

                                                       ${\displaystyle y=(c_{1}+c_{2}x)e^{-\pi x}\!}$


Checking:

We found that:

${\displaystyle y=(c_{1}+c_{2}x)e^{-\pi x}\!}$

Differentiating ${\displaystyle y\!}$ to obtain ${\displaystyle {y}'\!}$ and ${\displaystyle {y}''\!}$ respectively:

                                                  ${\displaystyle {y}'=c_{2}e^{-\pi x}-\pi (c_{1}+c_{2}x)e^{-\pi x}\!}$


And,

${\displaystyle {y}''=-\pi c_{2}e^{-\pi x}-\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}\!}$

                                                 ${\displaystyle {y}''=-2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}\!}$


Substituting these equations into the original ODE yields:

${\displaystyle -2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+2\pi (c_{2}e^{-\pi x}-\pi (c_{1}+c_{2}x)e^{-\pi x})+\pi ^{2}((c_{1}+c_{2}x)e^{-\pi x})=0\!}$

${\displaystyle -2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+2\pi c_{2}e^{-\pi x}-2\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}=0\!}$

${\displaystyle (-2\pi +2\pi )c_{2}e^{-\pi x}+(\pi ^{2}-2\pi ^{2}+\pi ^{2})(c_{1}+c_{2}x)e^{-\pi x}=0\!}$

${\displaystyle 0\equiv 0\!}$

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.[3]

The function satisfies the superposition principle if the sum of the homogeneous solution and the particular solution is equal to ${\displaystyle {\overline {y}}(x)\!}$, with ${\displaystyle {\overline {y}}=(y_{h}+y_{p})\!}$
${\displaystyle {\overline {y}}(x):=y_{h}(x)+y_{p}(x)\!}$

### Solutions

#### R1.6a

${\displaystyle y''=g=const\!}$
Check superposition principle:
${\displaystyle y''=g\!}$
${\displaystyle y_{h}''=0\!}$
${\displaystyle y_{p}''=g\!}$
${\displaystyle (y_{h}''+y_{p}'')\equiv {\overline {y}}''\!}$
Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

#### R1.6b

${\displaystyle mv'=mg-bv^{2}\!}$
Check superposition principle:
${\displaystyle mv'+bv^{2}=mg\!}$
${\displaystyle mv_{h}'+bv_{h}^{2}=0\!}$
${\displaystyle mv_{p}'+bv_{p}^{2}=mg\!}$
${\displaystyle m(v_{h}'+v_{p}')+b(v_{h}^{2}+v_{p}^{2})\not \equiv m{\overline {v}}'+b({\overline {v}}^{2})\!}$
Order of ODE: 1st Order
Linearity: Non-linear
Superposition Principle: No


#### R1.6c

${\displaystyle h'=-k{\sqrt {h}}\!}$
Check superposition principle:
${\displaystyle h'+kh^{1/2}=0\!}$
${\displaystyle h_{h}'+kh_{h}^{1/2}=0\!}$
${\displaystyle h_{p}'+kh_{p}^{1/2}=0\!}$
${\displaystyle (h_{h}'+h_{p}')+k(h_{h}^{1/2}+h_{p}^{1/2})\not \equiv {\overline {h}}'+k{\overline {h}}^{1/2}\!}$
Order of ODE: 1st Order
Linearity: Non-linear
Superposition Principle: No

#### R1.6d

${\displaystyle my''+ky=0\!}$
Check superposition principle:
${\displaystyle y''+{\frac {k}{m}}y=0\!}$
${\displaystyle y_{h}''+{\frac {k}{m}}y_{h}=0\!}$
${\displaystyle y_{p}''+{\frac {k}{m}}y_{p}=0\!}$
${\displaystyle (y_{h}''+y_{p}'')+{\frac {k}{m}}(y_{h}+y_{p})\equiv {\overline {y}}''+{\frac {k}{m}}{\overline {y}}\!}$
Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

#### R1.6e

${\displaystyle y''+\omega _{o}^{2}y=\cos(\omega t)\!}$
Check superposition principle:
${\displaystyle y''+\omega _{o}^{2}y=\cos(\omega t)\!}$
${\displaystyle y_{h}''+\omega _{o}^{2}y_{h}=0\!}$
${\displaystyle y_{p}''+\omega _{o}^{2}y_{p}=\cos(\omega t)\!}$
${\displaystyle (y_{h}''+y_{p}'')+\omega _{o}^{2}(y_{h}+y_{p})\equiv {\overline {y}}''+\omega _{o}^{2}{\overline {y}}\!}$
Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

#### R1.6f

${\displaystyle LI''+RI'+{\frac {1}{C}}I=E'\!}$
Check superposition principle:
${\displaystyle I''+{\frac {R}{L}}I'+{\frac {1}{LC}}I={\frac {E'}{L}}\!}$
${\displaystyle I_{h}''+{\frac {R}{L}}I_{h}'+{\frac {1}{LC}}I_{h}=0\!}$
${\displaystyle I_{p}''+{\frac {R}{L}}I_{p}'+{\frac {1}{LC}}I_{p}={\frac {E'}{L}}\!}$
${\displaystyle (I_{h}''+I_{p}'')+{\frac {R}{L}}(I_{h}'+I_{p}')+{\frac {1}{LC}}(I_{h}+I_{p})\equiv {\overline {I}}''+{\frac {R}{L}}{\overline {I}}'+{\frac {1}{LC}}{\overline {I}}\!}$
Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

#### R1.6g

${\displaystyle EIy^{iv}=f(x)\!}$
Check superposition principle:
${\displaystyle y''''-{\frac {1}{EI}}y=0\!}$
${\displaystyle y_{h}''''-{\frac {1}{EI}}y_{h}=0\!}$
${\displaystyle y_{p}''''-{\frac {1}{EI}}y_{p}=0\!}$
${\displaystyle (y_{h}''''+y_{p}'''')-{\frac {1}{EI}}(y_{h}+y_{p})\equiv {\overline {y}}''''-{\frac {1}{EI}}{\overline {y}}\!}$
Order of ODE: 4th Order
Linearity: Linear
Superposition Principle: Yes

#### R1.6h

${\displaystyle L\theta ''+g\sin \theta =0\!}$
Check superposition principle:
${\displaystyle \theta ''+{\frac {g}{L}}\sin \theta =0\!}$
${\displaystyle \theta _{h}''+{\frac {g}{L}}\sin \theta _{h}=0\!}$
${\displaystyle \theta _{p}''+{\frac {g}{L}}\sin \theta _{p}=0\!}$
${\displaystyle (\theta _{h}''+\theta _{p}'')+{\frac {g}{L}}(\sin \theta _{h}+\sin \theta _{p})\not \equiv {\overline {\theta }}''+{\frac {g}{L}}\sin {\overline {\theta }}\!}$
Order of ODE: 2nd Order
Linearity: Non-linear
Superposition Principle: No

#### R1.6i

${\displaystyle y_{1}^{'}=ay_{1}-by_{1}y_{2}\!}$
${\displaystyle y_{2}^{'}=ky_{1}y_{2}-ly_{1}\!}$
Order of ODE: 1st Order
Linearity: Non-linear
Superposition Principle: N/A

Created by [Daniel Suh] 21:59, 31 January 2012 (UTC)

## References

Kreyszig, Erwin; Herbert Kreyszig, Edward J. Norminton (2011). Advanced Engineering Mathematics. John Wiley & Sons, Inc. ISBN 978-0-470-45836-5.

1. Kreyszig 2011, p.58-59.
2. Kreyszig 2011, p.54-57.
3. Kreyszig 2011, p.48.

## Table of contributions

Name Responsibilities Link to lecture Checked by Signature
Francisco Arrieta Problem R1.1 Sec 1 (d) Andrea Vargas --[Andrea Vargas] 16:39, 1 February 2012 (UTC)
Kyle Gooding Problem R1.2 Sec 1 (d) Andrea Vargas --[Andrea Vargas] 16:39, 1 February 2012 (UTC)
Luca Imponenti Problem R1.3 Sec 1 (d) Andrea Vargas --[Andrea Vargas] 16:39, 1 February 2012 (UTC)
Gonzalo Perez Problem R1.4 Sec 2 (d) Andrea Vargas --[Andrea Vargas] 16:39, 1 February 2012 (UTC)
Jonathan Sheider Problem R1.5 Sec 2 (d) Andrea Vargas --[Andrea Vargas] 16:39, 1 February 2012 (UTC)
Daniel Suh Problem R1.6 Sec 2 (d) Andrea Vargas --[Andrea Vargas] 16:39, 1 February 2012 (UTC)
Andrea Vargas Report typing and formatting N/A Team 11 -- [Team 11] 16:57, 1 February 2012 (UTC)