text:
User:Egm4313.s12.team6.jagolinzer.r1.4

Given:
$$V=LC\frac{d^2V_{c}}{dt^2}+RC\frac{dV_{c}}{dt}+V_{c}$$

Find:
Derive the below two equations from the given:

 * A)     $$LI''+RI'+\frac{1}{C}I=V'$$

 * and

 * B)     $$LQ''+RQ'+\frac{1}{C}Q=V$$

Part A:
From Eq. 2-2(1): $$Q=CV_{c}=\int Idt$$

$$\therefore \frac{dI}{dt}=C\frac{d^2V_{c}}{dt}$$

substituting $$Q=CV_{c}=\int Idt$$ into $$RC\frac{dV_{c}}{dt}$$ we get:   $$RI$$

substituting $$\frac{dI}{dt}=C\frac{d^2V_{c}}{dt}$$ into $$LC\frac{d^2V_{c}}{dt^2}$$ we get:    $$LI'$$

So we now have:   $$V=LI'+RI+V_{c}$$

Deriving the whole equation we get:

 * $$V'=LI''+RI'+\frac{dV_{c}}{dt}$$

by multiplying $$\frac{dV_{c}}{dt}$$ by $$/frac{C}{C}$$ we get:

 * $$V'=LI''+RI'+\frac{CdV_{c}}{Cdt}$$

we can now sub $$Q=CV_{c}=\int Idt$$ for $$\frac{CdV_{c}}{dt}$$ giving us:

Part B:
From Eq. 2-2(1): $$Q=CV_{c}=\int Idt$$

multiplying  $$V_{c}$$   by $$\frac{C}{C}$$ we get $$\frac{CV_{c}}{C}$$

Subbing $$Q=CV_{c}=\int Idt$$ into $$\frac{CV_{c}}{C}$$ we get:

 * $$V=LC\frac{d^2V_{c}}{dt^2}+RC\frac{dV_{c}}{dt}+\frac{Q}{C}$$

taking the first and second derivative of  $$Q=CV_{c}$$   we get:

 * $$Q'=C\frac{dV_{c}}{dt}$$ and $$Q''=C\frac{d^2V_{c}}{dt^2}$$

substitute $$Q'=C\frac{dV_{c}}{dt}$$ into $$RC\frac{dV_{c}}{dt}$$ we get:   $$RQ'$$

substitute $$Q=C\frac{d^2V_{c}}{dt^2}$$ into $$LC\frac{d^2V_{c}}{dt^2}$$ we get:   $$LQ$$

title: User:Egm4313.s12.team6.jagolinzer.r1.4


text:
User:Allanaxelrod

$$y''=f(t)$$

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

title: User:Allanaxelrod


text:
User:Egm4313.s12.team16.brandonw

Egm4313.s12.team16

title: User:Egm4313.s12.team16.brandonw


text:
File:FBD1.2.png

(== Summary ==

title: File:FBD1.2.png


text:
User:Egm4313.s12.team18

=Team 18=

Handy Links

 * Course Website
 * IEA S12 Lecture notes/audio/video
 * Course Wiki
 * Report Table
 * MediaWiki Formatting
 * LaTeX Formatting
 * LaTeX Equation Editor

Team Members
Alvin Beyerlein Warren Burgher Andrew Crete Michelle Gibson William Hollon Josh Roberts Alissa Windham

Reports
Report 1

Report 2

Report 3

Report 4

Report 5

Report 6

Report 7

title: User:Egm4313.s12.team18


text:
File:FBD1.3.png

(== Summary ==

title: File:FBD1.3.png


text:
User:Egm4313.s12.team13.rossmeissl

Grading Rubric

R1work

R2work

title: User:Egm4313.s12.team13.rossmeissl


text:
University of Florida/Egm4313/s12.team15.r3

< Team 15

REPORT 2

Author
Solved and typed by Reviewed By -

Author
Solved and typed by - Reviewed By -

Author
Solved and typed by - Reviewed By -

Author
Solved and typed by - Kristin Howe and Reviewed By -

Author
Solved and typed by - Reviewed By -

Author
Solved and typed by - Reviewed By -


text:
User:Egm4313.s12.team18/R1

=Report 1= Team 18

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

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

Solution
Solution:

Using methods of kinematics to note that the displacement of the dashpot is equivalent to that of the spring:
 * {| style="width:100%" border="0"

$$  \displaystyle y=y_k=y_c\! $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Using methods of kinetics, specifically Newton's 2nd Law:
 * {| style="width:100%" border="0"

$$  \displaystyle my''+f_I=f(t)\! $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle f_I=f_k+f_c\! $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Constitutive Relations of forces for spring and dashpot:
 * {| style="width:100%" border="0"

$$  \displaystyle f_k=ky_k=ky\! $$     (4)
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 * style="width:95%" |
 *
 * }

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

$$  \displaystyle f_c=cv=cy_c'\!=cy' $$     (5)
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 * style="width:95%" |
 *
 * }

From (3), (4), and (5):
 * {| style="width:100%" border="0"

$$  \displaystyle f_I=ky+cy' $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Inserting (2) into (6):
 * {| style="width:100%" border="2"

$$  \displaystyle my''+cy'+ky=f(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Author
Solved and typed by - Egm4313.s12.team18.crete 16:25, 29 January 2012 (UTC)

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

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

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

$$  \displaystyle {F_c = cy'} $$     (1)
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 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {F_k = ky} $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Using Newton's Second Law:
 * {| style="width:100%" border="0"

$$  \displaystyle {\Sigma\vec{F}=m\vec{a}} $$
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 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {\vec{a}=y''} $$
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 * style="width:95%" |
 *
 * }

From the Free Body Diagram above:

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

$$  \displaystyle -ma=F_c+F_k-r(t) $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Or

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

$$  \displaystyle r(t)=F_c+F_k+ma $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Plug Equation 1 and Equation 2 into Equation 3.

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

$$  \displaystyle {r(t)=cy'+ky+my} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Or

 * {| style="width:100%" border="2"

$$  \displaystyle {r(t)=my''+cy'+ky} $$
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 * style="width:95%" |
 * }

Author
Solved and typed by - Egm4313.s12.team18.gibson.mg 21:49, 30 January 2012 (UTC)

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

Spring-Mass-Dashpot System:

Solution
Spring FBD: Dashpot FBD: Mass FBD: Given:
 * {| style="width:100%" border="0"

$$ \displaystyle f_k=f_c=f_I $$     (1) Using Newton's Second Law:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \sum \mathbf{F} = m\frac{\partial V}{\partial t} $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$ \displaystyle \frac{\partial V}{\partial t}=a(t)=y{}'' $$     (3) Therefore
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \sum \mathbf{F} = my{}'' $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Analyzing Mass FBD with Equation (1):
 * {| style="width:100%" border="0"

$$ \displaystyle \sum \mathbf{F} = f(t)-f_I $$     (5) Combining Equation (4) & (5):
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle f(t)-f_I= my{}''
 * style="width:95%" |
 * style="width:95%" |

$$     (6) Therefore:
 *

 * }
 * {| style="width:100%" border="1"

$$ \displaystyle f(t)= my{}''+f_I $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Author
Solved and typed by - Egm4313.s12.team18.alvinb 17:13, 29 January 2012 (UTC)

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

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

$$  \displaystyle Q=Cv_C \Rightarrow \int i\,dt=Cv_c \Rightarrow i = C \frac{dv_C}{dt} $$     (1a, 1b, 1c)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle V=LC\frac{d^2v_C}{dt^2}+RC\frac{dv_C}{dt}+v_C $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle LI''+RI'+\frac{1}{C}I=V' $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Deriving (3) from (2)
In order to derive first alternate version, (3), of the circuit equation, (2), we begin by taking the derivative of (2) with respect to time, the result of which can be seen below in (5)

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

$$  \displaystyle V'=LC \frac{d^3 v_C}{dt^3} + RC \frac{d^2 v_C}{dt^2} + \frac{d v_C}{dt} $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Next, by rearranging the right hand equation from (1) we obtain the following relation:

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$$  \displaystyle \frac{I}{C}=\frac{d v_C}{dt} $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And by taking subsequent derivatives, we also get:
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{I'}{C}=\frac{d^2 v_C}{dt^2} \, \, \& \, \, \frac{I''}{C}=\frac{d^3 v_C}{dt^3} $$     (7) & (8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By plugging the relations from (6), (7), and (8) into equation (2) and reducing the fractions we obtain the variant seen in (3)
 * {| style="width:100%" border="2"

$$  \displaystyle LI'' + RI' + \frac{1}{C}I=V' $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Deriving (4) from (2)
We can derive the second alternate equation, (4), from circuit equation (2) using a method similar to the one used to derive (3) as shown above.

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

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

$$  \displaystyle Q=Cv_C \, \, \& \, \, Q'=C \frac{dv_C}{dt} \, \, \& \, \, Q''=C \frac{d^2 v_C}{dt^2} $$     (9a, 9b, 9c)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Once again, plugging (9a, 9b, 9c) into (2) and reducing the results we can obtain the second variation of the circuit equation: (4)
 * {| style="width:100%" border="2"

$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and typed by - Egm4313.s12.team18.hollon 19:22, 28 January 2012 (UTC)

Problem
K 2011 p.59 pb 4.

Find a general solution. Check your answer by substitution.

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

$$  \displaystyle y''+4y'+(\pi^2+4)y=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
Note This is a homogenous linear 2nd order ordinary differential equation (ODE) with constant coefficients. Trial Solution:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{\lambda x} $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y'=\lambda e^{\lambda x} $$ (3)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y''=\lambda^2 e^{\lambda x} $$ (4) Plugging (2),(3), and (4) into (1)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle (\lambda^2 + 4\lambda + (\pi^2+4))e^{\lambda x} = 0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The characteristic equation is
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2 + 4\lambda + (\pi^2+4) = 0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Note The coefficients of this quadratic equation are
 * {| style="width:100%" border="0"

$$  \displaystyle a=1 $$
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 * style="width:95%" |

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

$$  \displaystyle b=4 $$
 * style="width:95%" |
 * style="width:95%" |

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

$$  \displaystyle c=\pi^2+4 $$
 * style="width:95%" |
 * style="width:95%" |

 * }

The discriminant can be found as follows
 * {| style="width:100%" border="0"

$$  \displaystyle \Delta = b^2 - 4ac $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Delta = 4^2 - 4(1)(\pi^2+4) $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Note The discriminant is negative. Therefore, Eq (6) has complex roots of the form

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

$$  \displaystyle \lambda = -1/2a \pm i\omega $$     (8) where
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \omega=\sqrt{c- b^2/4} $$     (9)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y(x)=e^{-bx/2} (Acos\omega x + Bsin\omega x) $$ (10) Plugging known quantities into (9) and (10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(x)=e^{-2x} (Acos\sqrt{(\pi^2+4)-4}\,\,x + Bsin\sqrt{(\pi^2+4)-4}\,\,x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Now The general solution to the ODE is
 * {| style="width:100%" border="1"

$$  \displaystyle y(x)=e^{-2x} (Acos\pi\,x + Bsin\pi\,x) $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Problem
K 2011 p.59 pb 5. Find a general solution. Check your answer by substation.

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

$$  \displaystyle y'' + 2{\pi}y' +{\pi ^2}y = 0
 * style="width:95%" |
 * style="width:95%" |

$$     (1)
 *

 * }

Solution
This is a homogeneous linear ODE with constant coefficients.
 * {| style="width:100%" border="0"

$$  \displaystyle a=2\pi $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b=\pi ^2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle a^2 -4b $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitution equations 2 and 3 into equation 4 we see that
 * {| style="width:100%" border="0"

$$  \displaystyle a^2 -4b = (2 \pi) ^2 - 4( \pi ^2)= 4 \pi ^2 -4 \pi ^2 = 0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Because equation 5 is equal to 0, the solution is a real double root. Therefore the solution will be in the form
 * {| style="width:100%" border="0"

$$  \displaystyle y = (c_1 + c_2 x)e^ {\lambda x}
 * style="width:95%" |
 * style="width:95%" |

$$     (6)
 *

 * }

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

$$  \displaystyle \lambda = -\frac{a}{2}= -\frac{2 \pi}{2} = -\pi
 * style="width:95%" |
 * style="width:95%" |

$$     (7)
 *

 * }

By combining equations 6 and 7 we see that the general solution is
 * {| style="width:100%" border="0"

$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
 * style="width:95%" |
 * style="width:95%" |

$$     (8)
 *

 * }

In order to check the solution, we take the first and second derivatives of equation 8

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

$$  \displaystyle y' = c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}
 * style="width:95%" |
 * style="width:95%" |

$$     (9)
 *

 * }

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

$$  \displaystyle y'' = -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x}
 * style="width:95%" |
 * style="width:95%" |

$$     (10)
 *

 * }

After taking the derivates, we substitute equations 8, 9, and 10 back into equation 1

 * {| style="width:100%" border="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} - \pi (c_1 + c_2 x) e^{-\pi x}) + {\pi} ^2 ( (c_1 + c_2 x)e^ {\lambda x}) = 0
 * style="width:95%" |
 * style="width:95%" |

$$     (11)
 *

 * }

Equation 11 is true. Therefore,
 * {| style="width:100%" border="1"

$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Is a general solution to equation 1.

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

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

Tower of Pisa

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

$$ \displaystyle { y{}''=g } $$ (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order: Second Linear Principle of Superposition applies:

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

$$ \displaystyle { y_{h}{}''=0} $$ (1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle { y_{p}{}''=g} $$ (1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(y_{h}+y_{p})=\bar{y}} $$ (1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle { (y_{h}+y_{p}){}''=g} $$ (1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{\bar{ y}{}''=g}} $$ Is the same as $$ \displaystyle { \mathbf{y{}''=g}} $$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Parachutist

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

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

 * }
 * }

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

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

$$ \displaystyle { mv_{h}{}'+bv_{h}^{2}=0} $$ (2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle { mv_{p}{}'+bv_{p}^{2}=mg} $$ (2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(v_{h}+v_{p})=\bar{v}} $$ (2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle { m(v_{h}+v_{p}){}'+b(v_{h}^{2}+v_{p}^{2})=mg} $$ (2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{ m\bar{v}{}'+b(v_{h}^{2}+v_{p}^{2})=mg} } $$ Is not the same as $$ \displaystyle { \mathbf{mv{}'+bv^{2}=mg}} $$ (2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Out-flowing Water

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

$$ \displaystyle {h{}'=-k\sqrt{h}} $$ (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

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

$$ \displaystyle { h_{h}{}'+k\sqrt{h_{h}}=0} $$ (3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle { h_{p}{}'+k\sqrt{h_{p}}=0} $$ (3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(h_{h}+h_{p})=\bar{h}} $$ (3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle { (h_{h}+h_{p}){}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0} $$ (3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{\bar{h}{}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0 }} $$ Is not the same as $$ \displaystyle {\mathbf{ h{}'+k\sqrt{h}=0}} $$ (3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Vibrating Mass on a String

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

$$ \displaystyle {my{}''+ky=0} $$ (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order:2nd Linear Principle of Superposition applies:

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

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

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

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$$ \displaystyle {y_{h}{}''+\omega _{0}^{2}y_{h}=0} $$ (5.1)
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$$ \displaystyle {y_{p}{}''+\omega _{0}^{2}y_{p}=\cos( \omega t)} $$ (5.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (5.3)
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$$ \displaystyle {(y_{p}+y_{h}){}''+\omega _{0}^{2}(y_{p}+y_{h})=\cos( \omega t)} $$ (5.4)
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$$ \displaystyle {\mathbf{\bar{y}{}''+\omega _{0}^{2}\bar{y}=\cos( \omega t)}} $$ Is the same as $$ \displaystyle {\mathbf{ y{}''+\omega _{0}^{2}y=\cos( \omega t)}} $$ (5.5)
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Current I in RLC Circuit

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$$ \displaystyle {LI{}''+RI{}'+\frac{1}{c}I=E{}'} $$ (6)
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Order:2nd Linear Principle of Superposition applies:

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$$ \displaystyle {LI_{h}{}''+RI_{h}{}'+\frac{1}{c}I_{h}=0} $$ (6.1)
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$$ \displaystyle {LI_{p}{}''+RI_{p}{}'+\frac{1}{c}I_{p}=E{}'} $$ (6.2)
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$$ \displaystyle {(I_{p}+I_{h})=\bar{I}} $$ (6.3)
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$$ \displaystyle {L(I_{p}+I_{h}){}''+R(I_{p}+I_{h}){}'+\frac{1}{c}(I_{p}+I_{h})=E{}'} $$ (6.4)
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$$ \displaystyle {\mathbf{L\bar{I}{}''+R\bar{I}{}'+\frac{1}{c}\bar{I}=E{}' }} $$ is the same as $$ \displaystyle {\mathbf{ LI{}''+RI{}'+\frac{1}{c}I=E{}'}} $$ (6.5)
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Deformation of a Beam

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

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$$ \displaystyle {EIy_{h}^{iv}=0} $$ (7.1)
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$$ \displaystyle {EIy_{p}^{iv}=f(x)} $$ (7.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (7.3)
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$$ \displaystyle {EI(y_{h}+y_{p})^{iv}=f(x)} $$ (7.4)
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$$ \displaystyle {\mathbf{EI\bar{y}^{iv}=f(x)} } $$ Is the same as $$ \displaystyle {\mathbf{ EIy^{iv}=f(x)}} $$ (7.5)
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Pendulum

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$$ \displaystyle {L\Theta {}''+gsin(\Theta )=0} $$ (8)
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Order:Second Non-Linear Principle of Superposition does not apply:

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$$ \displaystyle {L\Theta_{h} {}''+gsin(\Theta_{h} )=0} $$ (8.1)
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$$ \displaystyle {L\Theta_{p} {}''+gsin(\Theta_{p} )=0} $$ (8.2)
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$$ \displaystyle {(\Theta_{p}+\Theta_{h})=\bar{\Theta}} $$ (8.3)
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$$ \displaystyle {L(\Theta_{p}+\Theta_{h}){}''+gsin(\Theta_{p}+\Theta_{h})=0} $$ (8.4)
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$$ \displaystyle {\mathbf{L\bar{\Theta} {}+gsin(\Theta_{p}+\Theta_{h} )=0}} $$ Is not the same as $$ \mathbf{L\Theta {}+gsin(\Theta )=0} $$
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Author
Solved and typed by: Egm4313.s12.team18.windham 18:36, 30 January 2012 (UCT)

title: User:Egm4313.s12.team18/R1


text:
User:Egm4313.s12.team16.mcdaniel

team16

title: User:Egm4313.s12.team16.mcdaniel


text:
University of Florida/Egm4313/s12.team15.r2

<Team 15

REPORT 2

Given
$$\lambda _{1}=-2\!$$

$$\lambda _{2}=+5\!$$

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

$$y{}'(0)=0\!$$

$$r(x)=0\!$$

Find
Find the non-homogeneous linear 2nd order ordinary differential equation with constant coefficients in standard form and the solution in terms of the initial conditions and the general excitation $$r(x)$$.

Then plot the solution.

Solution
Some of the following standard equations were referenced from the textbook Advanced Engineering Mechanics 10th Ed. Kreyszig 2011.

Standard Form: $$\lambda ^{2}+a\lambda+b=0\!$$

$$(\lambda-\lambda_{1})(\lambda-\lambda_{2})=0\!$$

$$\lambda^{2}-\lambda\lambda_{2}-\lambda\lambda_{1}+\lambda_{1}\lambda_{2}=0\!$$

$$\lambda^{2}-5\lambda+2\lambda-10=0\!$$

$$\lambda^{2}-3\lambda-10=0\!$$

Therfore, $$a=-3\!$$

$$b=-10\!$$

Standard Form: $$y{}''+ay{}'+by=r(x)=0\!$$

$$y{}''-3y{}'-10y=0\!$$

Standard Form: $$y(x)=y_{h}(x)+y_{p}(x)\!$$

$$y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$$

$$y_{p}(x)=0\!$$

Therefore, $$y_{p}{}'(x)=0\!$$

$$y(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$$

$$y{}'(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}\!$$

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

then $$1=c_{1}e^{-2(0)}+c_{2}e^{5(0)}\!$$

$$1=c_{1}+c_{2}\!$$

If $$y{}'(0)=0\!$$

then $$0=c_{1}e^{-2(0)}+c_{2}e^{5(0)}\!$$

$$0=c_{1}+c_{2}\!$$

$$2c_{1}=5c_{2}\!$$

$$c_{1}=\frac{5}{2}c_{2}\!$$

$$1=\frac{5}{2}c_{2}+c_{2}\!$$

$$1=\frac{7}{2}c_{2}\!$$

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

$$1=c_{1}+c_{2}\!$$

$$c_{1}=1-c_{2}=1-\frac{2}{7}=\frac{5}{7}\!$$

Final Solution:

$$y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x}\!$$

Plot of Solution:

Author
Solved and typed by Kristin Howe Reviewed By -

Author
Solved and typed by - Reviewed By -

R2.3
Problems referenced from Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.3-4

Given
3a) $$ y{}''+6{}y'+8.96y=0\!$$

3b) $$ y'' + 4y' + (\pi^2 + 4)y = 0\!$$

Find
Find the general solution of the ODEs.

Solution
3a)

$$ y{}''+6{}y'+8.96y=0\!$$

$$ a=6\!$$

$$ b=8.96\!$$

$$\lambda ^2+a\lambda +b=0 \! $$

$$\lambda ^2+6\lambda +8.96=0 \! $$

Using the quadratic formula:

$$ \lambda_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2}\,\!$$ where $$\lambda_{1/2},\!$$ are the 2 different roots generated using quadratic formula.

Plug-in and solve using known variables (b=6, a=1, c=8.96)

$$ \lambda_{1/2} = \frac{-6 \pm \sqrt{0.16}}{2}\,\!$$ $$ = \frac{-6 \pm 0.4}{2}\,\!$$ $$ = -3 \pm 0.2\,\!$$

Therefore,

$$ \lambda_{1} = -2.8 \,\!$$ $$ \lambda_{2} = -3.2 \,\!$$

The General Solution formula for Distinct Real roots is $$y(x)=C_1 e^{\lambda_1 x}+C_2 e^{\lambda_2 x}$$

Now substitute the known roots into the General Solution formula: $$ y(x)= C_1e^{-2.8x}+C_2e^{-3.2x} \! $$

Check using substitution. First, find the 1st and 2nd derivative of general solution formula:

$$ {y}' = -3.2C_{1}e^{-3.2x} - 2.8C_{2}e^{-2.8x} \!$$

$$ {y}'' = 10.24C_{1}e^{-3.2x} + 7.84C_{2}e^{-2.8x} \!$$

Then, plug in the known variables into the original equation:

$$ y{}''+6{}y'+8.96y=0\!$$ (Original Equation)

$$ 10.24c_{1}e^{-3.2x} + 7.84c_{2}e^{-2.8x} - 19.2c_{1}e^{-3.2x} - 16.8c_{2}e^{-2.8x} + 8.96c_{1}e^{-3.2x} + 8.96c_{2}e^{-2.8x} = 0 \!$$

By inspection, all of the terms on the left side of the equation cancel out to 0, making the expression correct. Therefore, the general solution is correct.

3b)

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

$$ a=4\!$$

$$ b=(\pi^2 + 4)\!$$

$$\lambda ^2+a\lambda +b=0 \! $$

$$\lambda ^2+4\lambda +(\pi^2 + 4)=0 \! $$

Using the quadratic formula:

$$ \lambda_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2}\,\!$$ where $$\lambda_{1/2},\!$$ are the 2 different roots generated using quadratic formula.

Plug-in and solve using known variables (b=4, a=1, c=[$$\pi^2$$ + 4])

$$ \lambda_{1/2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$

$$ \lambda_{2} = -2 - \pi i \,\!$$

The General Solution formula for Complex Conjugate roots is $$y(x)=e^{-\frac{\alpha x}{2}}(C_1\cos \beta x+C_2\sin \beta x) \!$$

Now substitute the known roots into the General Solution formula: $$y(x)=e^{-2x}(C_1\cos \pi x+C_2\sin \pi x) \!$$

Check using substitution. First, find the 1st and 2nd derivative of general solution formula:

$$ {y}' = e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x))-2e^{-2x}(C_1cos(\pi x)+C_2sin(\pi x)) \!$$

$$ {y}''=e^{-2x}(-C_1\pi ^{2}cos(\pi x)-C_2\pi ^{2}sin(\pi x)+2C_1\pi sin(\pi x)-2C_2\pi sin(\pi x))-2e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x)-2C_1cos(\pi x)-2C_2sin(\pi x)) \!$$

Then, plug in the known variables into the original equation:

$$ y'' + 4y' + (\pi^2 + 4)y = 0\!$$ (Original Equation)

$$ e^{-2x}(-C_1\pi ^{2}cos(\pi x)-C_2\pi ^{2}sin(\pi x)+2C_1\pi sin(\pi x)-2C_2\pi sin(\pi x))-2e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x)-2C_1cos(\pi x)-2C_2sin(\pi x)) \!$$

$$ + 4(e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x))-2e^{-2x}(C_1cos(\pi x)+C_2sin(\pi x))) \!$$

$$+ (\pi^2 + 4)(e^{-2x}(C_1\cos \pi x+C_2\sin \pi x)) = 0\!$$

By inspection, all of the terms on the left side of the equation cancel out to 0, making the expression correct. Therefore, the general solution is correct.

Author
Solved and typed by - Tim Pham 19:36, 07 February 2012 (UTC) Reviewed By -

R2.4
The following problems are referenced from Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59.

For additional information and/or practice reference Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.53-59 and the lecture notes Lecture 5.

 * Note the lecture notes and Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 textbook were used to solve the following.

Given
4a) $$ y{}''+2\pi{}y'+\pi^{2}y=0\!$$

4b) $$ 10y{}''-32{}y'+25.6{}y=0\!$$

Find
Find the general solution. Check answer by substitution.

Solution
4a)

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

a=2\pi\!

$$ b=\pi^{2}\!$$

$$\lambda ^2+a\lambda +b=0 \! $$

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

Finding the roots:

$$\lambda_1=\frac{1}{2}(-2\pi+\sqrt{(-2\pi)^2-4(\pi^{2})})=-\pi\!$$

$$\lambda_2=\frac{1}{2}(-2\pi-\sqrt{(-2\pi)^2-4(\pi^{2})})=-\pi\!$$

Both solutions above show the double root characteristic so the following is applied:

$$y_1=e^{-\pi{x}}\!$$

$$y_2=uy_1\!$$

$$ u=c_1x+c_2\!$$         This is due to double integration.

So, $$y_2=(c_1x+c_2)y_1\!$$

Choose $$ c_1= 1 c_2=0\!$$

$$y_2= (x+0)y_1\!$$

$$y_2=xy_1\!$$

$$ y_2=xe^{-\pi{x}}\!$$

$$ y=(c_1+c_2x)e^{-\pi{x}}\!$$ General Solution

Check by substitution

$$ y=(c_1+c_2x)e^{-\pi{x}}\!$$

$$ y'=-\pi{}c_1e^{-\pi{x}}+c_2e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}\!$$

$$ y''=\pi^{2}c_1e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+\pi^{2}c_2xe^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}\!$$

$$\pi^{2}c_1e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+\pi^{2}c_2xe^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+ 2\pi{}(-\pi{}c_1e^{-\pi{x}}+c_2e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}})+ \pi^{2}(e^{-\pi{x}}c_1+e^{-\pi{x}}xc_2)=0 \!$$

Since all terms cancel out then by substitution the general solution $$ y=(c_1+c_2x)e^{-\pi{x}}\!$$ does hold true.

4b) $$ 10y{}''-32{}y'+25.6{}y=0\!$$

Divide by 10 to simplify equation :

$$ y{}''-3.2{}y'+2.56{}y=0\!$$

$$ a=-3.2\!$$

$$ b=2.56\!$$

$$ \lambda ^2+a\lambda +b=0 \! $$

$$ \lambda ^2-3.2\lambda +2.56=0 \! $$

Solving for the roots:

$$ \lambda_1=\frac{1}{2}(-(-3.2)+\sqrt{(-3.2)^2-4(2.56)})=1.6\!$$

$$ \lambda_2=\frac{1}{2}(-(-3.2)-\sqrt{(-3.2)^2-4(2.56)})=1.6\!$$

$$\lambda_1=\lambda_2\!$$ This shows the double root characteristic.

So, the following is applied

$$ y_1=e^{1.6x}\!$$

$$y_2=uy_1\!$$

$$ u=c_1x+c_2\!$$  This is due to double integration

So

$$y_2=(c_1x+c_2)y_1\!$$

Choose $$ c_1= 1 c_2=0\!$$

$$y_2= (x+0)y_1\!$$

$$y_2=xy_1\!$$

$$ y_2=xe^{1.6x}\!$$

$$ y=(c_1+c_2x)e^{1.6x}\!$$  General Solution

 Check by Substitution

$$ y''=2.56c_1e^{1.6x}+1.6c_2e^{1.6x}+1.6c_2xe^{1.6x}+2.56c_2xe^{1.6x}\!$$

$$y'=1.6c_1e^{1.6x}+c_2e^{1.6x}+ 1.6c_2xe^{1.6x} \!$$

$$y=c_1e^{1.6x}+c_2xe^{1.6x} \!$$

$$10(2.56c_1e^{1.6x}+1.6c_2e^{1.6x}+1.6c_2xe^{1.6x}+2.56c_2xe^{1.6x})-32(1.6c_1e^{1.6x}+c_2e^{1.6x}+ 1.6c_2xe^{1.6x})+25.6(c_1e^{1.6x}+c_2xe^{1.6x})=0\!$$

Since all terms cancel out then through substitution it can be confirmed that the general solution

$$ y=(c_1+c_2x)e^{1.6x}\!$$ does hold true.

Author
Solved and typed by - Cynthia Hernandez Reviewed By -

Given
Problem 16, P 59 Kreyszig:

$$ e^{2.6x}, e^{-4.3x}\!$$

Problem 17, P 59 Kreyszig:

$$e^{-\sqrt{5}x},xe^{-\sqrt{5}x}\!$$

Find
Problem 16, P 59 Kreyszig:

ODE in the form:

$${y}''+a{y}'+by=0\!$$

Problem 17, P 59 Kreyszig:

ODE in the form:

$${y}''+a{y}'+by=0\!$$

Solution
Problem 16, P 59 Kreyszig:

For two distinct real-roots:

$$y=c_{1}e^{\lambda_{1}x}+c_{2}e^{\lambda_{2}x}\!$$

Where:

$$\lambda_{1}=2.6\!$$ and $$\lambda_{2}=-4.3\!$$

The characteristic equation is:

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

Multiplying the roots of the equation together:

$$(\lambda-2.6)(\lambda+4.3)=\lambda^{2}+1.7\lambda-11.18\!$$

Therefore the ODE is:

$$y''+1.7y'-11.18y=0\!$$

Problem 17, P 59 Kreyszig:

For real double root:

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

Where:

$$\lambda=-\sqrt{5}\!$$

The characteristic equation is:

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

Multiplying the roots of the equation together:

$$(\lambda+\sqrt{5})^{2}=\lambda^{2}+2\sqrt{5}\lambda+5\!$$

Therefore the ODE is:

$$y''+2\sqrt{5}y'+5y=0\!$$

Author
Solved and typed by - Neil Tidwell 1:55, 06 February 2012 (UTC) Reviewed By -

Given
Spring-Dashpot-Mass System from Sec 1 lecture notes.

Has double real root $$\lambda=-3\!$$

Find
$$k,c,m\!$$

Solution
For spring-mass-dashpot system:

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

This simplifies to:

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

Multiply roots together to get characteristic equation:

$$(\lambda+3)^2=\lambda^2+6\lambda+9\!$$

Set coefficients of simplified spring-mass-dashpot equation equal to those of characteristic equation:

$$\frac{k}{mc}=6\!$$

$$\frac{k}{m}=9\!$$

Have one degree of freedom, so arbitrarily choose $$m=1\!$$

Solve system:

$$m=1\!$$ $$k=9\!$$ $$c=1.5\!$$

Author
Solved and typed by - Neil Tidwell 2:14, 06 February 2012 (UTC) Reviewed By -

Given
Develop the MacLaurin Series (Taylor Series at $$ t = 0\! $$) for $$ e^t,\cos t,\sin t\! $$ $$ L[y] and  L[w]  then L[y+w] \! $$ $$ L[y+w] = L[y]+ L[w] \! $$ $$ L[cy] = cL[y] $$ $$ L[kw] = kL[w] \! $$

Author
Solved and typed by - Jenny Schulze Reviewed By -

Given
Find a general solution to the equations, and check answers by substitution. Problem 8

$$ y''+y'+3.25y=0 \! $$  (8.0)

Problem 15

$$ y''+0.54y'+(0.0729+\pi )y=0 \!$$  (8.1)

Find
Factor as in the text and solve.

Author
Solved and typed by - Jenny Schulze Reviewed By -

Given
Initial Conditions:

$$ y(0)=1, y'(0)=0\ $$

No excitation: $$ r(x)=0\ $$

Find
Find and plot the solution for the L2-ODE-CC corresponding to:

$$\lambda ^{2}+4\lambda+13=0\! $$

In another Fig., superimpose 3 Figs.: (a) this Fig., (b) the Fig. in R2.6 p.5-6, (c) the Fig. in R2.1 p.3-7.

Solution
The corresponding ODE in standard form:

$$ y''+4y'+13y=r(x) \!$$

To find the roots of the ODE, convert ODE to this form:

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

which in this case is:

$$ \lambda^{2}+4\lambda+13=0\! $$

use the formula to find the roots:

$$\lambda_{1,2}=\frac{-b\pm i\sqrt{b^{2}-4ac}}{2a}=-\alpha\pm i\omega $$

$$\lambda_{1,2}=\frac{-4\pm i\sqrt{4^{2}-4*1*13}}{2*1}=-2\pm 3i $$

because the roots are complex conjugates, the general solution can be given as:

$$ y(t)=e^{-\alpha t}(Acos\omega t+bsin\omega t)=e^{-2 t}(Acos3t+Bsin3t) \!$$

to find A and B, we use the initial conditions and take the first derivative of y(t):

$$ y(0)=1=e^{-2*0}(Acos(3*0)+Bsin(3*0)=1*A=A \!$$

$$ A=1 \!$$

$$ y'(t)=e^{-2t}((-2A+3B) cos(3t)-(3A+2B)sin(3t)) \!$$

$$ y'(0)=0=e^{-2*0}((-2*1+3B) cos(3*0)-(3*1+2B)sin(3*0))=1*(-2+3B)=-2+3B \!$$

$$ B=\frac{2}{3} \!$$

thus:

$$ y(t)=e^{-2 t}(cos3t+\frac{2}{3}sin3t) \!$$

The graph of the solved equation is shown below in Fig. 1.

The three graphed equations are shown together below in Fig. 2

Equation 1 (R2.9):

$$ y_1(t) = e^{-2 t}(cos3t+\frac{2}{3}sin3t) \!$$

Equation 2 (R2.6):

$$ y_2(t) = e^{-3t}+3te^{-3t} \!$$

Equation 3 (R2.1):

$$ y_3(t) = \frac{5}{7}e^{-2t}+\frac{2}{7}e^{5t} \!$$

Author
Solved and typed by - --James Moncrief 01:42, 8 February 2012 (UTC) Reviewed By -


text:
User:Egm4313.s12.team2.plunkett.jp

=hi=

title: User:Egm4313.s12.team2.plunkett.jp


text:
User:Egm4313.s12.team18/R2

=Report 2=

Problem
Given the two roots and the initial conditions:
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$$  \displaystyle \lambda_1=-2\qquad\qquad\lambda_2 = +5 $$
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 * }
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$$  \displaystyle y(0)=1\qquad\qquad y'(0)=0 $$ Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x)=0$$.
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 * }

Consider no excitation:
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$$  \displaystyle r(x)=0 $$
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Plot the solution.

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values above as the 2 roots of the corresponding characteristic equation.

Solution

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$$  \displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) $$     (2.1.1)
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 *
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Now Insert the given roots into (1)
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$$  \displaystyle (\lambda-(-2))(\lambda-5) $$     (2.1.2) By the FOIL method,  (2) gives the characteristic equation for the unknown L2-ODE-CC.
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 *
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$$  \displaystyle \lambda^2-3\lambda-10=0 $$     (2.1.3)
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 *
 * }

From this characteristic equation, (3), we can find the non-homogeneous L2-ODE-CC.
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$$  \displaystyle y''-3y'-10=r(x) $$     (2.1.4)
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 *
 * }

Now It is known from the given roots that the homogeneous L2-ODE-CC has the solution
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$$  \displaystyle y_H(x)=C_1e^{-2x} + C_2e^{5x} $$     (2.1.5)
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 *
 * }

Therefore, the overall solution is
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$$  \displaystyle y(x)=y_H(x)+y_P(x) $$     (2.1.6)
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 *
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$$  \displaystyle y(x)=C_1e^{-2x}+C_2e^{5x}+y_P(x) $$     (2.1.7) where $$y_P(x)$$ is the particular solution.
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 *
 * }

Now Using the given initial conditions
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$$  \displaystyle y(0)=1\qquad y'(0)=0 $$
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$$  \displaystyle y(0)=C_1+C_2+y_P(0)=1 $$     (2.1.8) Differentiating $$y(x)$$ and then inserting the initial condition.
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 *
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$$  \displaystyle y'(x)=-2C_1e^{-2x}+5C_2e^{5x}+y_P'(x) $$     (2.1.9)
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 *
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$$  \displaystyle y'(0)=-2C_1+5C_2+y_P'(0)=0 $$     (2.1.10)  Now Multiplying (2.1.8) by 2 and adding it to (2.1.10)
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 *
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$$  \displaystyle (2C_1+2_C2+2y_P(0)=2) + (-2C_1+5C_2+y_P'(0)=0) $$
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$$  \displaystyle C_2=\frac{2}{7}-\frac{2}{7}y_P(0)-\frac{1}{7}y_P'(0) $$     (2.1.11) By inserting (2.1.11) into (2.1.8) and combining like terms $$C_1$$ is also found.
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 *
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$$  \displaystyle C_1=\frac{5}{7}-\frac{5}{7}y_P(0)+\frac{1}{7}y_P'(0) $$     (2.1.12)  Now The overall solution $$y(x)$$ is given in terms of the initial conditions.
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 *
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$$  \displaystyle y(x)=\frac{1}{7}[(5-5y_P(0)+y_P'(0))e^{-2x}+(2-2y_P(0)-y_P'(0))e^{5x}] $$     (2.1.13)
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 *
 * }

Now Considering no excitation

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$$  \displaystyle r(x)=0 $$ Which causes
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$$  \displaystyle y_P(x)=0\,\,,\,\,y_P'(x)=0 $$ So
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$$  \displaystyle C_1=\frac{5}{7} $$
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 *  (2.1.14)
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$$  \displaystyle C_2=\frac{2}{7} $$ The homogeneous solution to the L2-ODE-CC is therefore
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 *  (2.1.15)
 * }
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$$  \displaystyle y_H(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$
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 *  (2.1.16)
 * }

This solution will now be plotted using MATLAB.

Now Generate 3 non-standard and non-homogeneous L2-ODE-CC that satisfy the roots given in the problem statement.
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$$  \displaystyle 2y''-6y'-20=r(x) $$
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 *
 (2.1.17)
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$$  \displaystyle \frac{3}{2}y''-\frac{9}{2}y'-15y=r(x) $$
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 *
 (2.1.18)
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$$  \displaystyle 3y''-9y'-30y=r(x) $$
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 *
 (2.1.19)
 * }

Author
Solved and typed by - Egm4313.s12.team18.roberts.jr 18:28, 7 February 2012 (UTC)

Problem
Initial Conditions: $$y(0)=1 \, \ y'(0)=0$$ No excitation: $$r(x)=0$$ Find and plot the solution for L2-ODE-CC (4) p.5-5.:
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$$ \displaystyle y''-10y'+25y=r(x) $$
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 * }

Solution
Replacing y with $$\lambda$$:
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$$ \displaystyle \lambda^{2}-10\lambda+25=0 $$     (1)
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 *

 * }

This can be factored to:

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$$ \displaystyle (\lambda-5)^{2}=0 $$     (2)
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 *

 * }

Meaning there is a real double-root of
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$$ \displaystyle \lambda=5 $$     (3)
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 *

 * }

Therefore the equations can be set up as:

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$$ \displaystyle y_{g}=C_{1}e^{\lambda x}+C_{2}xe^{\lambda x} $$ (4)
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 *

 * }

Using the double-root value for lambda:

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$$ \displaystyle y_{g}=C_{1}e^{5x}+C_{2}xe^{5x} $$     (5) Using the initial conditions given, solve for C1 and C2: y(0)=1 Using values: y=1, x=0
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$$ \displaystyle 1=C_{1}e^{5(0)}+C_{2}(0)e^{5(0)} $$     (6)
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$$ \displaystyle 1=C_{1} $$     (7)
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 *

 * }

Using the derivative of y'(0)=0: y'=0, x=0, C1=1

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$$ \displaystyle y_{g}=C_{1}e^{5x}+C_{2}xe^{5x} $$     (8)
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 *

 * }
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$$ \displaystyle {y_{g}}'=5C_{1}e^{5x}+C_{2}e^{5x}(5X+1) $$     (9) Inputting values:
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$$ \displaystyle 0=5(1)e^{5(0)}+C_{2}e^{5(0)}(1) $$     (10)
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 *

 * }

Therefore:
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$$ \displaystyle 0=5+C_{2} $$     (11)
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$$ \displaystyle C_{2}=-5 $$     (12) Final Solution:
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$$ \displaystyle y_{g}=e^{5x}-5xe^{5x} $$     (13)
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 *

 * }

Author
Solved and typed by - Egm4313.s12.team18.alvinb 13:53, 6 February 2012 (UTC)

Problem 2.3
Solve problems 3 and 4 on page 59 in the textbook, Kreyszig 2011. "Find a general solution for the ODE and check by substitution."

3.$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$

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

Solution
Background information:

For homogeneous linear ODE's with constant coefficients, a general equation can be described as, $$ \displaystyle {y{}''+ay{}'+by=0.} $$

The function $$ \displaystyle {y=e^{\lambda x}} $$ may be tried as a solution for the general equation above.

After substituting this equation and its derivatives: $$ \displaystyle {y{}'=\lambda e^{\lambda x}} $$ and $$ \displaystyle {y{}''=\lambda ^{2}e^{\lambda x}} $$ into the general equation,the result is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)e^{\lambda x}=0} $$

and the characteristic equation is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)=0} $$ which means that if $$ \displaystyle {\lambda} $$ is a solution of this characteristic equation, then the exponential function is a solution of the general equation. If we solve this characteristic equation for $$ \displaystyle {\lambda} $$, the result is$$ \displaystyle {\lambda _{1,2}=\frac{1}{2}(-a\pm \sqrt{a^{^{2}}-4b})} $$

Therefore there are 3 cases when it comes to finding the solution to Homogeneous Linear ODE's with CC:

Case 1:    Two real roots if  $$ \displaystyle {a^{^{2}}-4b>0} $$

Case 2:    A real double root if $$\displaystyle {a^{^{2}}-4b=0} $$

Case 3: Complex conjugate roots if $$\displaystyle {a^{^{2}}-4b=0} $$

-Problem 3 page 59 :
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$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$ (1) The discriminant for this equation (1) is:
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 *

 * }
 * }
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$$ \displaystyle { \sqrt{6^{^{2}}-4(8.96)}=.16} $$ which is >0 and therefore a case 1 (2)
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 *

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 * }

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$$ \displaystyle {\lambda _{1}=-2.8, \lambda _{2}=-3.2} $$ and $$ \displaystyle {y _{1}=e^{-2.8x}, y _{2}=e^{-3.2x}} $$ (3)
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 *

 * }
 * }

The general solution to (1) is

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$$ \displaystyle {y=c_{1}e^{-2.8x}+c_{2}e^{^{-3.2x}}} $$ (4) Check this solution by plugging in
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 *

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$$ \displaystyle {y _{1}=e^{-2.8x}} $$ and $$ \displaystyle {y _{2}=e^{-3.2x}} $$ (5) into (1). The results for plugging in the above equations(5) and their derivatives:
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 *

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$$ \displaystyle {y_{1}=e^{-2.8x},y_{1}{}'=-2.8e^{-2.8x},y_{1}{}''=7.84e^{-2.8x}} $$ and $$ \displaystyle {y_{2}=e^{-3.2x},y_{2}{}'=-3.2e^{-3.2x},y_{2}{}''=10.24e^{-3.2x}} $$ (6) into equation (1) respectively are:
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$$ \displaystyle {e^{-2.8x}(7.84-(6)(2.8)+8.96)=0} $$ and $$ \displaystyle {e^{-3.2x}(10.24-(6)(3.2)+8.96)=0} $$
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(7)
 *

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 * }

Both of these reduce to
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$$ \displaystyle {0=0} $$ (8)
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 *

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 * }

Problem 4 page 59:
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$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$ (9)
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 *

 * }
 * }

The discriminant for this equation is:

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$$ \displaystyle {4^{2}-4(\pi ^{2}+4)=-4\pi^{2}} $$ (10) which is a case 3 and solutions to the characteristic equation are:
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 *

 * }
 * }
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$$ \displaystyle {\lambda _{1}=-\frac{1}{2}a+i\omega, \lambda _{2}=-\frac{1}{2}a-i\omega} $$ (11) and our particular solutions for characteristic equation for this problem are:
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 *

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 * }
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$$ \displaystyle {\lambda _{1}=-\frac{1}{2}(4)+i\pi, \lambda _{2}=-\frac{1}{2}4-i\pi} $$ (12)
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 *

 * }
 * }

The general solution to (9) is

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$$ \displaystyle {y=e^{-\frac{4}{2}x}(A\cos \pi x+Bsin\pi x)} $$ (13) Check this solution by plugging in
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 *

 * }
 * }
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$$ \displaystyle {y _{1}=e^{-\frac{4}{2}x}\cos \pi x} $$ and $$ \displaystyle {y _{2}=e^{-\frac{4}{2}x}\sin \pi x} $$ (14) into (1). The results for plugging in the equations (14) and their derivatives:
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 *

 * }
 * }

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$$ \displaystyle {y_{1}=e^{-2x}\cos \pi x, y_{1}{}'=-\pi e^{-2x}-2e^{-2x}\cos \pi x, y_{1}{}''=e^{-2x}((4-\pi ^{2})\cos \pi x+4\pi \sin \pi x)} $$ (15)
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 *

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and
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$$ \displaystyle {y_{2}=e^{-2x}\sin \pi x, y_{2}{}'=e^{-2x}(\pi \cos \pi x-2\sin \pi x), y_{2}{}''=e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)} $$ (16)  into equation (1) respectively are:
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 *

 * }
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$$ \displaystyle {e^{-2x}((4-\pi ^{2})\cos \pi x+4\sin \pi x+4(-\pi \sin \pi x-2\cos \pi x)+(\pi ^{2}+4\cos \pi x)=0} $$ (17)
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 *

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$$ \displaystyle {e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)+4e^{-2x}(\pi \cos \pi x-2\sin \pi x)+(\pi ^{2}+4)e^{-2x}\sin \pi x=0} $$ (18) Both of these reduce to
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$$ \displaystyle {0=0} $$ (19) -
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Author
Solved and typed by - Egm4313.s12.team18.windham 22:41, 7 February 2012 (UTC)

Problem
K 2011 p.59 pb 5. Find a general solution. Check your answer by substation.

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$$  \displaystyle y'' + 2{\pi}y' +{\pi ^2}y = 0
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$$     (1)
 *

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Solution
This is a homogeneous linear ODE with constant coefficients.
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$$  \displaystyle a=2\pi $$     (2)
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$$  \displaystyle b=\pi ^2 $$     (3)
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$$  \displaystyle a^2 -4b $$     (4)
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Substitution equations 2 and 3 into equation 4 we see that
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$$  \displaystyle a^2 -4b = (2 \pi) ^2 - 4( \pi ^2)= 4 \pi ^2 -4 \pi ^2 = 0 $$     (5)
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 *

 * }

Because equation 5 is equal to 0, the solution is a real double root. Therefore the solution will be in the form
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$$  \displaystyle y = (c_1 + c_2 x)e^ {\lambda x}
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$$     (6)
 *

 * }

Where
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$$  \displaystyle \lambda = -\frac{a}{2}= -\frac{2 \pi}{2} = -\pi
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$$     (7)
 *

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By combining equations 6 and 7 we see that the general solution is
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$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
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$$     (8)
 *

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In order to check the solution, we take the first and second derivatives of equation 8

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$$  \displaystyle y' = c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}
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$$     (9)
 *

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$$  \displaystyle y'' = -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x}
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$$     (10)
 *

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After taking the derivates, we substitute equations 8, 9, and 10 back into equation 1

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$$  \displaystyle -2\pi c_2 e^ {-\pi x} + {\pi ^2} (c_1 + c_2 x) e^{-\pi x} + 2\pi (c_2 e^ {-\pi x} - \pi (c_1 + c_2 x) e^{-\pi x}) + {\pi} ^2 ( (c_1 + c_2 x)e^ {\lambda x}) = 0
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$$     (11)
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Equation 11 is true. Therefore,
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$$  \displaystyle y = (c_1 + c_2 x)e^ {-\pi x}
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$$
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 * }

Is a general solution to equation 1.

Solved and typed by - Egm4313.s12.team18.burgher.wb 14:28, 7 February 2012

Problem
K 2011 p.59 pb 6. Find a general solution. Check your answer by substation.

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$$  \displaystyle 10y'' - 32y' +25.6y = 0
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$$     (1)
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Solution
This is a homogeneous linear ODE with constant coefficients.
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$$  \displaystyle a=10 $$     (2)
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$$  \displaystyle b=-32 $$     (3)
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To find the value of the discriminant we use the equation
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$$  \displaystyle a^2 -4b $$     (4)
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By substitution equations 2 and 3 into equation 4 we see that
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$$  \displaystyle a^2 -4b = 10^2 - 4(-32)= 100 +128 = 228 $$     (5)
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 *

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Because equation 5 is greater than 0, the solution is has two real roots. Therefore the solution will be in the form
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$$  \displaystyle y = c_1 e ^ {\lambda_1 x} + c_2 e ^ {\lambda_2 x}
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$$     (6)
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Where
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$$  \displaystyle \lambda_1 = \frac{1}{2} (-a + \sqrt{a^2 -4b})
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$$
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 * }

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$$  \displaystyle \lambda_1 = \frac{1}{2} (-10 + \sqrt{10^2 -4(-32})
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$$
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 * }

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$$  \displaystyle \lambda_1 = \frac{1}{2} (-10 + \sqrt{100 +128})
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$$
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$$  \displaystyle \lambda_1 = \frac{1}{2} (-10 + 15.0997) = 2.55
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$$
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And

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$$  \displaystyle \lambda_2 = \frac{1}{2} (-a - \sqrt{a^2 -4b})
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$$
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$$  \displaystyle \lambda_2 = \frac{1}{2} (-10 - \sqrt{10^2 -4(-32})
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$$
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$$  \displaystyle \lambda_2 = \frac{1}{2} (-10 - \sqrt{100 +128})
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$$
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$$  \displaystyle \lambda_2 = \frac{1}{2} (-10 - 15.0997) = -12.55
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$$
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 * }

By substitution \lambda_1 and \lambda_2 into equation 6 we see that the general solution is
 * {| style="width:100%" border="0"

$$  \displaystyle y = c_1 e ^ {2.55 x} + c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$     (7)
 *

 * }

In order to check the solution, we take the first and second derivatives of equation 7

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

$$  \displaystyle y' = 2.55 c_1 e ^ {2.55 x} - 12.55 c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$     (8)
 *

 * }

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

$$  \displaystyle y'' = 6.5025 c_1 e ^ {2.55 x} + 157.5025 c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$     (9)
 *

 * }

After taking the derivates, we substitute equations 7, 8, and 9 back into equation 1

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

$$  \displaystyle 10 (6.5025 c_1 e ^ {2.55 x} + 157.5025 c_2 e ^ {-12.55 x}) - 32 (2.55 c_1 e ^ {2.55 x} - 12.55 c_2 e ^ {-12.55 x}) + 25.6 (c_1 e ^ {2.55 x} + c_2 e ^ {-12.55 x}) = 0
 * style="width:95%" |
 * style="width:95%" |

$$     (10)
 *

 * }

Equation 10 is true. Therefore,
 * {| style="width:100%" border="1"

$$  \displaystyle y = c_1 e ^ {2.55 x} + c_2 e ^ {-12.55 x}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }
 * }

Is a general solution to equation 1.

Solved and typed by - Egm4313.s12.team18.burgher.wb 14:28, 7 February 2012

Problem
K 2011 p.59 problems 16 and 17.

Solution
Find an ODA of form $$y''+ay'+by=0$$ for the given basis: #16
 * {| style="width:100%" border="0"

Given: $$ \displaystyle e^{2.6x}, e^{-4.3x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

From the given, we can see that this is a scenario of two distinct roots of form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b})=2.6 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b})=-4.3 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Introducing a variable $$c$$ to simplify calculations:

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

$$  \displaystyle c=\sqrt{a^2-4b} $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting (3) into (1) and (2):
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=\frac{1}{2}(-a+c)=2.6 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_2=\frac{1}{2}(-a-c)=-4.3 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From (4) and (5) we have two equations and two unknowns. Begin by solving for $$a$$ via adding (4) and (5):

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

$$  \displaystyle -a=-1.7 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \rightarrow a=1.7 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting the value from (6) into (4) and solving for $$c$$:

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

$$  \displaystyle \frac{1}{2}(-a+c)=2.6 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \rightarrow c=6.9 $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting the value from (7) into (3) and solving for $$b$$:

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

$$  \displaystyle 6.9=\sqrt{1.7^2-4b} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle b=\frac{1.7^2-6.9^2}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \rightarrow b=-11.18 $$     (8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting the values from (6) and (8) into the general form yields:

 * {| style="width:100%" border="2"

$$  \displaystyle y''+1.6y'-11.18y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

#17
 * {| style="width:100%" border="0"

Given: $$ \displaystyle e^{-\sqrt{5}x}, xe^{-\sqrt{5}x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

From the given we can see that this is a scenario of a real double root:

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

$$  \displaystyle \lambda=-\frac{1}{2}a $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle a^2-4b=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From the given basis:

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

$$  \displaystyle e^{\frac{-ax}{2}} $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Setting (3) equal to (1) and solving for $$a$$:

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

$$  \displaystyle e^{\frac{-ax}{2}}=e^{-\sqrt{5}x} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \frac{a}{2}=\sqrt{5} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \rightarrow a=2\sqrt{5} $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving for $$b$$ by substituting (4) into (2):

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

$$  \displaystyle {2\sqrt{5}}^2-4b=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle b=\frac{{2\sqrt{5}}^2}{4} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \rightarrow b=\sqrt{5} $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting the results from (4) and (5) into the general form yields:

 * {| style="width:100%" border="2"

$$  \displaystyle y''+2\sqrt{5}y'+5y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Problem
Realize spring-dashpot-mass systems in series as shown in Fig. p.1-4 with the similar characteristic as in (3) pp.5-5, but with double real root $$\lambda=-3$$, i.e., find the values for the parameters $$k,c,m$$.

Equation for Spring-Dashpot-Mass System:
 * {| style="width:100%" border="0"

$$ \displaystyle My_{k}''+\frac{Mk}{c}y_{k}'+ky=f(t) $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
Given the real double-root$$\lambda=-3$$:
 * {| style="width:100%" border="0"

$$ \displaystyle (\lambda+3)^{2}=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \lambda^{2}+6\lambda+9=0 $$     (3) Therefore:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle M=1 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \frac{Mk}{c}=6 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle k=9 $$     (6) Solving for all unknowns: M=1 k=9 c=1.5
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and typed by - Egm4313.s12.team18.alvinb 14:20, 6 February 2012 (UTC)

Problem
Develop the MacLaurin series (Taylor series at T=0) for: $$e^t \, \ \cos t \ , \ \sin t $$

Problem
K 2011 p. 60 problems 8 and 15. These problems give you an equation and ask for the general solution. Once the general solution has been found, use substitution to check the answer.

Problem 8) $$\displaystyle y^{''}+y^{'}+3.25y=0 $$

Problem 15) $$\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi)y=0 $$

Solution
Problem 8)
 * $$ \displaystyle y^{''}+y^{'}+3.25y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+\lambda+3.25 =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:

 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12 $$

Or
 * $$\displaystyle a^{2}-4b=-12 $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:

 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{3.25-\frac{1}{4}(1)^{2}}) \rightarrow \omega=\sqrt{3} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{3}$$
 * $$ \displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

Checking the solution by using substitution:

 * $$\displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

 * $$\displaystyle y^{'}=\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{-2}+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{'}=-\frac{1}{2}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$

 * $$\displaystyle y^{''}=-\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{4}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{''}=-\frac{1}{4}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$

 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$

 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-x/2}Bsin(\sqrt{3}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

Problem 15)
 * $$ \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+0.54\lambda+(0.0729+\pi ) =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:

 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi)=-4\pi $$

Or
 * $$\displaystyle a^{2}-4b=-4\pi $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:

 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{(0.0729-\pi)-\frac{1}{4}(0.54)^{2}}) =\sqrt{\pi} \rightarrow \omega=\sqrt{\pi} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{\pi}$$ $$ \displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

Checking the solution by using substitution:

 * $$\displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

 * $$\displaystyle y^{'}=-0.27{e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))}+e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)) $$

 * $$\displaystyle y^{''}=0.0729e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$

 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$

 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-0.27}Bsin(\sqrt{\pi}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

Author
Solved and typed by - Egm4313.s12.team18.gibson.mg 18:22, 7 February 2012 (UTC)

Problem
Initial conditions: $$y(0)=1 \, \ y'(0)=0$$ No excitation:$$r(x)=0$$ Find and plot the solution for L2-ODE-CC corresponding to (1). In another Fig. superpose 3 Figs.: (a) this Fig., (b) the Fig in R2.6 p..5-6, (c) the Fig. in R2.5 p.3-6.

title: User:Egm4313.s12.team18/R2


text:
User:Egm4313.s12.team19

Group members

1. Gabriel Arab

2. Adam Burton

3. Joshua Campbell

4. Charles Chiamchittrong

5. Santiago Marin

6. Grant (Kyle) Uppercue

7. Vladimir Horwitz

Reports

R1

R2

Anything below this line is only for team members and is not to be graded.

Code test page 2

title: User:Egm4313.s12.team19


text:
User:Hopeton87

R1

title: User:Hopeton87


text: File:IEA 1.jpg

title: File:IEA 1.jpg


text: File:IEA.PNG

title: File:IEA.PNG


text: File:Spring.JPG

title: File:Spring.JPG


title: File:Iea.s12.sec6.djvu


text:
User:Alex8937

=R1.1: Spring-dash pot system in parallel with a mass and applied force f(t)=

Initial Information
From lecture slide 1-4

 * {| style="width:50%" border="-5" align="center"

Variables:$$k,$$ $$c,$$  $$m,$$  $$f(t),$$  $$y(t),$$  $$y_k,$$  $$y_c,$$  $$f_i,$$  $$f_k,$$  $$f_c$$
 * style="width:100%; padding:10px; border:2px solid #00ff00" |
 * style="width:100%; padding:10px; border:2px solid #00ff00" |
 * style= |
 * }
 * }

title: User:Alex8937


text:
University of Florida/Egm4313/s12.team15.r4

< Team 15

REPORT 4

Author
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Author
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Author
Solved and typed by - Kristin Howe and Reviewed By -

Author
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Author
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text:
User:Egm4313.s12.team17.deaver.md/Report 1 Problem 1 4

= Report 1 =


text:
User:Egm4313.s12.team17.deaver.md/Report 1

= Problem 4 - Modeling: Electric Circuits =

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

$$  \displaystyle Q = \int I dt = C\nu_C $$     (Eq.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle V = LC\frac{d^2\nu_C}{dt^2} + RC\frac{d\nu_C}{dt} + \nu_C $$     (Eq.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle V^\prime = LI^{\prime\prime} + RI^\prime + \frac{1}{C}I $$     (Eq.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle V = LQ^{\prime\prime} + RQ^\prime + \frac{1}{C}Q $$     (Eq.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Problem Statement
Use Eq. 2 to derive the alternative forms of the circuit equation 3 and 4.[2]

Solution
Solve for \nu_C \, in terms of I for Eq. 1.
 * {| style="width:100%" border="0"

$$  \displaystyle \nu_C = \frac{1}{C}\int I dt = \frac{1}{C}Q $$     (Eq.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substitute Eq. 5 into Eq. 2.
 * {| style="width:100%" border="0"

$$  \displaystyle V = LC\frac{d^2\nu_C}{dt^2} + RC\frac{d\nu_C}{dt} + \frac{1}{C}\int I dt $$ (Eq.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Take the differential of Eq. 1 twice with respect to t \, in terms of Q and \nu_C \,.
 * {| style="width:100%" border="0"

$$  \displaystyle Q^\prime = C\frac{d\nu_C}{dt} $$     (Eq.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle Q^{\prime\prime} = C\frac{d^2\nu_C}{dt^2} $$     (Eq.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substitute Eq. 5, 7, and 8 into Eq. 6.
 * {| style="width:100%" border="0"

$$  \displaystyle V = LQ^{\prime\prime} + RQ^\prime + \frac{1}{C}Q $$     (Eq.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Take the differential of Eq. 9 with respect to t \,.
 * {| style="width:100%" border="0"

$$  \displaystyle V^\prime = LQ^{\prime\prime\prime} + RQ^{\prime\prime} + \frac{1}{C}Q^\prime $$     (Eq.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Take the differential of Eq. 1 three time with respect to t \, in terms of Q and I.
 * {| style="width:100%" border="0"

$$  \displaystyle Q^\prime = I $$ (Eq.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle Q^{\prime\prime} = I^\prime $$     (Eq.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle Q^{\prime\prime\prime} = I^{\prime\prime} $$     (Eq.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substitute Eq. 11, 12, and 13 into Eq. 10.
 * {| style="width:100%" border="0"

$$  \displaystyle V^\prime = LI^{\prime\prime} + RI^\prime + \frac{1}{C}I $$     (Eq.14)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *
 * }


text:
University of Florida/Egm4313/s12.team15.r5

< Team 15

REPORT 5

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text:
University of Florida/Egm4313/s12.team15.r6

< Team 15

REPORT 6

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text:
User:Egm4313.s12.team19.Burton

My contribution to Report 1

For Report 1, I did problem 1.2, based on lecture notes page (Page number).

title: User:Egm4313.s12.team19.Burton


text:
User:Egm4313.s12.team19.Campbell

My contribution to Report 1

For Report 1, I did problem 1.3.

My contribution to Report 2

I did problem 2.4

and proofread

problem 2.5

title: User:Egm4313.s12.team19.Campbell


text:
Draft:Original research/Early telescopes

The early telescopes as manufactured in the Netherlands are refracting. The first known practical telescopes were invented in the Netherlands at the beginning of the 17th century, using glass lenses.

The earliest recorded working telescopes were the refracting telescopes that appeared in the Netherlands in 1608. Their development is credited to three individuals: Hans Lippershey and Zacharias Janssen, who were spectacle makers in Middelburg, and Jacob Metius of Alkmaar. "The Hague discussed the patent applications first of Hans Lipperhey of Middelburg, and then of Jacob Metius of Alkmaar... another citizen of Middelburg, Sacharias Janssen had a telescope at about the same time but was at the Frankfurt Fair where he tried to sell it".

"Galileo always insisted that the ancients had telescopes."

"The Russians have found crystal lenses, perfectly spherical and of great precision, in ancient Egypt, during the African-dominated period."

"It is also only within recent years that the discovery of a seminal black kingdom in the Nile Valley, predating the Egyptian dynasties, has settled the question, once and for all, of the roots of classical Egyptian culture and technology (see Journal of African Civilizations, vol. 4, no. 2, November 1982)."

Telescopes and lenses
“For we can so shape transparent bodies, and arrange them in such a way with respect to our sight and objects of vision, that the rays will be reflected and bent in any direction we desire, and under any angle we wish, we may see the object near or at a distance … So we might also cause the sun, moon and stars in appearance to descend here below.”

Minerals
Quartz is the second-most-abundant mineral in the Earth's continental crust, after feldspar. Pure quartz, traditionally called rock crystal (sometimes called clear quartz), is colorless and transparent or translucent.

Iceland spar, formerly known as Iceland crystal, is a transparent variety of calcite, or crystallized calcium carbonate. "It has been speculated that the sunstone (a different mineral than the gem-quality sunstone) mentioned in medieval Icelandic texts was Iceland spar and that Vikings used its light-polarizing property to tell the direction of the sun on cloudy days, for navigational purposes.

A chemically pure and structurally perfect diamond is perfectly transparent with no hue, or color. However, in reality almost no gem-sized natural diamonds are absolutely perfect.

"c. 3500 B.C. Phoenicians cooking on sand discover glass."

"Natural glass has existed since the beginnings of time, formed when certain types of rocks melt as a result of high-temperature phenomena such as volcanic eruptions, lightning strikes or the impact of meteorites, and then cool and solidify rapidly. Stone-age man is believed to have used cutting tools made of obsidian (a natural glass of volcanic origin also known as hyalopsite, Iceland agate, or mountain mahogany) and tektites (naturally-formed glasses of extraterrestrial or other origin, also referred to as obsidianites)."

"According to the ancient-Roman historian Pliny (AD 23-79), Phoenician merchants transporting stone actually discovered glass (or rather became aware of its existence accidentally) in the region of Syria around 5000 BC [7,000 b2k]. Pliny tells how the merchants, after landing, rested cooking pots on blocks of nitrate placed by their fire. With the intense heat of the fire, the blocks eventually melted and mixed with the sand of the beach to form an opaque liquid."

"The earliest man-made glass objects, mainly non-transparent glass beads, are thought to date back to around 3500 BC, with finds in Egypt and Eastern Mesopotamia. In the third millennium, in central Mesopotamia, the basic raw materials of glass were being used principally to produce glazes on pots and vases. The discovery may have been coincidental, with calciferous sand finding its way into an overheated kiln and combining with soda to form a coloured glaze on the ceramics. It was then, above all, Phoenician merchants and sailors who spread this new art along the coasts of the Mediterranean."

Theoretical telescopes
Def. "[a] monocular optical instrument possessing magnification for observing distant objects" is called a telescope.

Def. "any instrument ... for observing distant objects" is called a telescope.

"From Latin telescopium, from Greek τηλεσκόπος (tēleskopos), far-seeing, from τῆλε (tēle, "afar") + σκοπέω (skopeō, "I look at)."

Keplerian telescopes
The Keplerian Telescope, invented by Johannes Kepler in 1611, is an improvement on Galileo's design. It uses a [plano]convex lens as the eyepiece instead of Galileo's [double] concave one. The advantage of this arrangement is [that] the rays of light emerging from the eyepiece are converging. This allows for a much wider field of view and greater eye relief but the image for the viewer is inverted. Considerably higher magnifications can be reached with this design but to overcome aberrations the simple objective lens needs to have a very high [focal ratio] f-ratio.

All refracting telescopes use the same principles. The combination of an objective lens 1 and some type of eyepiece 2 is used to gather more light than the human eye could collect on its own, focus it 5, and present the viewer with a brighter, clearer, and magnified virtual image 6.

Air telescopes
At right is a woodcut illustration of a 45 m focal length Keplerian air, or aerial telescope built by Johannes Hevelius, and even longer tubeless "aerial telescopes" were constructed.

An aerial telescope is a type of very-long-focal-length refracting telescope built in the second half of the 17th century that did not use a tube. "After about 1675, therefore, astronomers did away with the telescope tube. The objective was mounted on a building or pole by means of a ball-joint and aimed by means of a string..." Instead, the objective was mounted on a pole, tree, tower, building or other structure on a swivel ball-joint. The observer stood on the ground and held the eyepiece, which was connected to the objective by a string or connecting rod. By holding the string tight and maneuvering the eyepiece, the observer could aim the telescope at objects in the sky. The idea for this type of telescope may have originated in the late 17th century with the Dutch mathematician, astronomer and physicist Christiaan Huygens and his brother Constantijn Huygens, Jr., though it is not clear that they invented it.

An even simpler air telescope is shown in the second image at left. To have only an air telescope, an ancient needs lenses. The ones in the schematic are called planoconvex. Convex on one side towards the target and planar on the viewing side.

Possible telescopes
The image at right is a painting by artist Giorgio Vasari (1511–1574). Behind and above the main focus on Cronus (Saturn) castrating Uranus (the Greek sky god before Zeus) is what appears to be "some kind of Armillary sphere". Specifically, the "[w]hole thing is some kind of Armillary sphere".

Even more than that is an object angled slightly upward (right to left) that appears to be pointed at or above the large metallic ring. This object seems to be painted as if it extends through the top hole of the armillary sphere. At the upper left end of the object is an embellished end piece that may have a lens (objective lens) or piece of material the diameter of which matches that of the tube behind the embellishment. The midsection appears to taper to a smaller diameter near the right end. The right end has an even smaller piece attached or inserted that looks like an eyepiece lens holder. This tubular object seems to be receiving hands in the air praise from two people to its left. The armillary sphere may be tilted ~23 1/2°.

The tubular object bares a striking resemblance to a telescope. The image taken as a whole suggests an interaction between Cronus and Uranus as observed near the plane of the ecliptic by someone through a telescope.

Another suggestion: "It would be the sceptre, the symbolic attribute of the ruler of the deities and of the universe. It was an attribute of Uranus and is about to pass to Cronus. The nature of a sceptre as a phallic symbol seems especially obvious in the context of this painting. The ruling god and owner of the sceptre during a given mythical era is also the main progenitor of that era. (Thus mirroring an alpha male in primate and some other animal societies.)". Bold added.

The Castration of Uranus is apparently a fresco by Vasari & Cristofano Gherardi (c. 1560, Sala di Cosimo I, Palazzo Vecchio, Florence).

"This painting is new to me, I must admit, and I have no idea what the object is that you refer to. It looks as though it might be [a] spindle of some kind."

"I don't think it really looks like a telescope, apart from anything else it would hardly fit into the iconography of the mutilation of Uranus by Saturn. Having looked at other paintings relating to this myth, I think your so-called spindle is more likely a torch that has gone out in the chaos of the scene depicted - Maybe a metaphor for upsetting the light of the Sun or similar, who knows."

The frontispiece from Johannes Kepler's Rudolphine Tables appears to have a telescope drawn in on the upper roof extension on the left side. While this would be no surprise if the drawing dates from after 1609, as suspected, it would be a surprise if this drawing dates from the 1500's.

"The central piece of information uncovered by de Waard was an entry in the unpublished journal of Isaac Beeckman, the rector of the Latin school in Dordrecht, and a friend of Descartes.37 Beeckman learned how to grind lenses for telescopes in an effort to obtain better instruments. In the early 1630's he took lessons from a spectacle-maker in Middelburg named Johannes Sachariassen, the son of Sacharias Janssen. Beeckman recorded in his journal that during one of these lessons Johannes Sachariassen told him that the first telescope in the Netherlands had been made in 1604 by his father, after the model of an instrument in the possession of an Italian. This instrument bore the date of 1590.38"

As many Italians were in the Netherlands fighting the Spanish, some with experience in grinding lenses for optical uses, it is likely a few carried with them optical transits such as the one dated to 1590 made in Italy which were so common as to be taken for granted by historians, modifiable into telescopes.

Transits
The first occurrence of the word "theodolite" is found in the surveying textbook A geometric practice named Pantometria (1571) by Leonard Digges, which was published posthumously by his son, Thomas Digges.

The first instrument more like a true theodolite was likely the one built by Joshua Habermel in Germany in 1576, complete with compass and tripod.

Armillary spheres
Although lost to Europe since the terminus of the Greco-Roman era, Gerbert reintroduced the astronomical armillary sphere to Latin Europe via Al-Andalus in the late 10th century.

Richer wrote about another of Gerbert's last armillary sphere, which had sighting tubes fixed on the axis of the hollow sphere that could observe the constellations, the forms of which he hung on iron and copper wires. This armillary sphere was also described by Gerbert in a letter to his colleague Constantine. Gerbert instructed Constantine that, if doubtful of the position of the pole star, he should fix the sighting tube of the armillary sphere into position to view the star he suspected was it, and if the star did not move out of sight, it was thus the pole star. Furthermore, Gerbert instructed Constantine that the north pole could be measured with the upper and lower sighting tubes, the Arctic Circle through another tube, the Tropic of Cancer through another tube, the equator through another tube, and the Tropic of Capricorn through another tube.

The first two images at the right dating from 1480 and 1547 show the sphere without sighting tubes or any device for observing astronomical objects.

The first two images at the left show spheres dating to after 1608 and the Dutch invention of the telescope.

It appears to be the case that armillary spheres are often displayed without any attachment for viewing such as a sighting tube or telescope.

Of the some 458 images of armillary spheres in commons, only the Vasari painting shows a sighting tube or telescope of any kind within or attached to the sphere.

Sighting tubes
On the use of the sighting tube to fix the position of the pole star, Shen Kuo wrote:

"Before Han times it was believed that the pole star was in the center of the sky, so it was called Jixing (Summit star). Zi Geng(-zhi) found out with the help of the sighting tube that the point in the sky which really does not move was a little more than 1 degree away from the summit star. In the Xining reign-period (1068-1077) I accepted the order of the emperor to take charge of the Bureau of the Calendar. I then tried to find the true pole by means of the tube. On the very first night I noticed that the star which could be seen through the tube moved after a while outside the field of view. I realized, therefore, that the tube was too small, so I increased the size of the tube by stages. After three months' trials I adjusted it so that the star would go round and round within the field of view without disappearing. In this way I found that the pole star was distant from the true pole somewhat more than 3 degrees. We used to make the diagrams of the field, plotting the positions of the star from the time when it entered the field of view, observing after nightfall, at midnight, and early in the morning before dawn. Two hundred of such diagrams showed that the 'pole star' was really a circumpolar star. And this I stated in my detailed report to the emperor. "

On the left is an image showing a small telescope. Note the cylindrical pieces needed to accommodate lenses and tube movement. In the drawing on the right from the 13th century an astronomer holds a sighting tube. The tube is only one continuous cylinder.

The second image down on the left shows a tapered metal observation tube with end pieces similar to lens holders.

Dioptras
A dioptra [the image at the right] is a classical astronomical and surveying instrument, dating from the 3rd century BCE. The dioptra was a sighting tube or, alternatively, a rod with a sight at both ends, attached to a stand. If fitted with protractors, it could be used to measure angles.

Greek astronomers used the dioptra to measure the positions of stars; both Euclid and Geminus refer to the dioptra in their astronomical works.

Camera obscura
Girolamo Cardano (1501-1576) made "a camera obscura with a diverting spectacle ... appears ... to have initiated the use of a convex lens in the aperture."

Tubes
"There are indeed ancient tablets that mention astronomers' lenses supported by a golden tube to enlarge the pupil, and in Nineveh a rock crystal lens was found (Pettinato 1998). Maybe one day a new archaeological excavation will find a Babylonian telescope for the first time."

Sceptres
The sceptre imaged at the right is a piece of the Portuguese crown jewels. Note that the image is vertical as is the sceptre.

The image at the left shows a sceptre dated to before 1380.

A key point is that the sceptres both display an object setting above and removed from the shaft of the septre which is either tapered or cylindrical.

The object in the Uranus/Saturn painting from about 1560 does not have an object displayed on top of the sceptre but shows an object closely resembling an objective lens inside its housing.

The second image on the right shows that earlier scepters, or perhaps walking sticks were lengthy, approximately two-thirds a person's height. The ornamentation at the scepters upper end varied greatly as the sculpture at the second left shows.

More recent scepters such as the one shown at third right and third left are sword-like.

The fourth image at the right demonstrates that scepters may be much shorter. The scepter is in the statue's right hand.

Spindles
In machine tools, a spindle is a rotating axis of the machine, which often has a shaft at its heart. The shaft itself is called a spindle, but also, in shop-floor practice, the word often is used metonymically to refer to the entire rotary unit, including not only the shaft itself, but its bearings and anything attached to it (chuck, etc.).

Torches
On the right is a photograph of an apparently 16th century artwork which shows a blazing torch immediately behind the Pope.

The torch painted in second image down the right is much longer and narrower than the one higher or lower on the left.

The second image down on the left shows only the torch from Michele Rocca's "Angelica and Medoro" dated to "between circa 1720 and circa 1750".

Eyes
"Old Kingdom Egyptian statues from the first to the sixth dynasties had lenses placed in their eyes." "The thickness of the lenses was not intended to duplicate the thickness of the real human eye, but rather to create an effect so that the eye follows the person passing by when looked from other angles." "[T]he illusion of the following eye technique."

The Archaic or Early Dynastic Period of Egypt immediately follows the unification of Lower and Upper Egypt c. 3100 BC. It [is] generally taken to include the First and Second Dynasties, lasting from the Protodynastic Period of Egypt until about 2686 BC, or the beginning of the Old Kingdom.

The eyes of these statues contain rock-crystals. "These are probably the oldest portrait statues in the world. These people who sit before us side by colored to the life, fresh and glowing as the day they gave the artist his last sitting lived at a time when the great pyramids were not yet built and at a date which is variously calculated as from about 4,000 to 6,300 years from the present day. The princess wears her hair precisely as it is still worn in Nubia and her necklace is of a pattern still favored. The eyes of both statues are inserted. The eyeball which is set in an eyelid of bronze, is made of opaque white quartz with an iris of rock-crystal enclosing a pupil of some kind of brilliant metal. This treatment gives to the eyes a look of intelligence which is almost appalling." Amelia B. Edwards "These incomparable statues are most expressive and stand in vitality to the works of any later age in Egypt. They were found in the tomb chamber- Ra-Ho-tep is entitled a royal son [probably of Seneferu]- The signs carved in these tombs are among the earliest known. Instead of full-length burial with coffins, head rests, vases, and provision for a future life, the more usual method of burial at Medum is lying on the left side with the knees drawn up facing the east and without vases or other objects, showing a diversity of beliefs and probably of races." W.M. Flinders Petrie.

"The composition of these eyes is a lens of polished rock crystal (either alpha silica or fused silica, formerly known as cystalline quartz and fused quartz which had a convex front surface and a near hemispherical concave ground pupil surface in a flat iris plane (normally covered with resin) at the rear of the lens."

The Prehistory of Egypt spans the period of earliest human settlement to the beginning of the Early Dynastic Period of Egypt in ca. 3100 BC, starting with King Menes/Narmer.

The Predynastic Period is traditionally equivalent to the Neolithic period, beginning ca. 6000 BC and including the Protodynastic Period (Naqada III).

Early lenses
"Lens-shaped crystals have long been known from Bronze Age contexts". These are "usually recognized as short-focus magnifying lenses."

The slightly oval lens [40 x 35 mm] has been roughly ground and has a focal point about 110 millimetres (4.5 in) from the flat side.

"There are now 23 ancient lenses on display in the Archaeological Museum at Herakleion and many more are in storage there. They are also made of rock crystal and are of optical quality, with generated plano-convex surfaces."

There is more about lenses more recently from Visby, Gotland. "What intrigues the researchers is that the lenses are of such high quality that they could have been used to make a telescope some 500 years before the first known crude telescopes were constructed in Europe in the last few years of the 16th century." "Made from rock-crystal, the lenses have an accurate shape that betrays the work of a master craftsman. The best example of the lenses measures 50 mm (2 inches) in diameter and 30 mm (1 inch) thick at its centre." "The [Visby] Gotland crystals provide the first evidence that sophisticated lens-making techniques were being used by craftsmen over a 1,000 years ago."

"If one Italian scientist is correct then the telescope was not invented sometime in the 16th century by Dutch spectacle makers, but by ancient Assyrian astronomers nearly three thousand years earlier. According to Professor Giovanni Pettinato of the University of Rome, a rock crystal lens, currently on show in the British museum, could rewrite the history of science. He believes that it could explain why the ancient Assyrians knew so much about astronomy."

Hypotheses

 * 1) The ancients may have used air or aerial telescopes.


text:
User:Egm4313.s12.team19.Chiamchittrong

My contribution to Report 1

For Report 1, I did problem 1.4, based on lecture notes page (Page number).

title: User:Egm4313.s12.team19.Chiamchittrong


text:
User:Egm4313.s12.team19.Marin

My contribution to Report 1

For Report 1, I did problem 1.5.

title: User:Egm4313.s12.team19.Marin


text:
User:Egm4313.s12.team19.Uppercue

My contribution to Report 1

For Report 1, I did Problem R1.6, based on lecture notes Sec 2(d). Here's the link to the notes:

My contribution to Report 2 For Report 2, I did Problem R2.7, based on lecture notes Sec 6(b). Here's the link to the notes: [[media:iea.s12.sec6.djvu|Sec 6 (b)]]

My contribution to Report 3 For Report3, I did Problems R3.1 and R3.8, based on lecture notes sec7 (b).

title: User:Egm4313.s12.team19.Uppercue


text:
User:Egm4313.s12.team19.Arab

My contribution to Report 1

For Report 1, I did problem 1.1.

Furthermore, I made the Facebook page for our group, which we used to talk outside of class.

title: User:Egm4313.s12.team19.Arab


text: File:Parallel Spring Dashpot.jpg

title: File:Parallel Spring Dashpot.jpg


text:
University of Florida/Egm4313/s12.team11.imponenti/R1.3

Problem Statement
For the spring-dashpot-mass system shown, draw the FBDs and derive the following equation of motion $$my''+f_I=f(t)$$

Free Body Diagram

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

Kinetics
Newton's second law of motion in the y-direction

Σ$$F_y=ma=f(t)-f_I$$ where $$a=y''$$

which can be re-written as

$$my''+f_I=f(t)$$

Egm4313.s12.team11.imponenti 03:01, 30 January 2012 (UTC)