User:Egm4313.s12.team16.r1

By: Team 16

File:R1.1.jpg

From Figure: Mass, Spring, Damper system with spring and damper in parallel

First, we take the + y direction to be downward and z to be out of the page. Within this coordinate system, the displacement is described as a vector quantity y with the velocity and acceleration of the system denoted by ${\displaystyle {\dot {y}}\!}$ and ${\displaystyle {\ddot {y}}\!}$ respectively.

Multiplying by the mass denoted ${\displaystyle m\!}$ we derive the kinematics of the system to be

${\displaystyle m{\ddot {y}}+F1=F_{y}\!}$

${\displaystyle F1=F_{k}+F_{c}\!}$

To develop the kinetics of the system, we consider the two forces:

For the force limiting displacement in the y direction, we look at Hooke's law

${\displaystyle F_{k}=-ky\!}$

For the force opposing the motion, the damping is related to the velocity of the system by

${\displaystyle F_{c}=-c{\dot {y}}\!}$

where ${\displaystyle c\!}$ is the coefficient of viscous friction

Equating the kinematics to the kinetics we preliminarily have

${\displaystyle m{\ddot {y}}+ky+c{\ddot {y}}={\ddot {F_{y}}}\!}$

and simplifying with:

${\displaystyle y=y_{k}=y_{c}\!}$

${\displaystyle {\dot {y}}={\dot {y_{c}}}={\dot {y_{k}}}\!}$

Final Equation:

${\displaystyle F(t)=m{\ddot {y}}+c{\dot {y}}+ky\!}$

Author: Nyal Jennings
Last Edit: 01/28/2012 - 11:01 PM
Reviewed By: Martin Dolgiej

Example alt text

From Figure 1 -
Spring:

${\displaystyle \displaystyle -ky=F_{k}}$

${\displaystyle \displaystyle (Eq.1)}$

Mass:

${\displaystyle \displaystyle my''=F_{m}}$

${\displaystyle \displaystyle (Eq.2)}$

Damp:

${\displaystyle \displaystyle cy'=F_{c}}$

${\displaystyle \displaystyle (Eq.3)}$

Forcing Funtion:

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

${\displaystyle \displaystyle (Eq.4)}$

Balancing the forces in the y direction:

${\displaystyle \sum F_{y}=F_{m}-F_{r}=F_{k}-F_{c}\!}$

Substituting equations (1),(2),(3), and (4):

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

${\displaystyle my''+cy'+ky=r(t)\!}$

Author: Martin Dolgiej
Last Edit: 01/28/2012 - 11:32 PM
Reviewed By: Matt Hilsenrath

FBD of Spring

${\displaystyle f_{s}=k*y_{s}\!}$

FBD of Dashpot

${\displaystyle Cy_{D}'=f_{c}\!}$

FBD of Mass

(Newton's 2^nd Law)

${\displaystyle f=my''\!}$

${\displaystyle \sum F(t)=my''+f_{s}-f_{c}\!}$

${\displaystyle f_{1}=f_{s}-F_{c}\!}$

${\displaystyle F(t)=my''+f_{1}\!}$

Author: Ryan McDaniel
Reviewed by: Erin Standish

Given:

${\displaystyle \displaystyle V=LC{\frac {d^{2}v_{C}}{dt^{2}}}+RC{\frac {dv_{C}}{dt}}+v_{C}}$

${\displaystyle \displaystyle (Eq.2)}$

Solution:
Deriving equation (3):

Taking the derivative of (Eq. 2) gives:

${\displaystyle \displaystyle V'=LC{\frac {d^{3}v_{C}}{dt^{3}}}+RC{\frac {d^{2}v_{c}}{dt^{2}}}+{\frac {dv_{C}}{dt}}}$

${\displaystyle \displaystyle (Eq.A)}$

By definition:
${\displaystyle i=C{\frac {dv_{C}}{dt}}}$

${\displaystyle i'=C{\frac {d^{2}v_{C}}{dt^{2}}}}$

${\displaystyle i''=C{\frac {d^{3}v_{C}}{dt^{3}}}}$

Plugging theses definitions into (Eq. A):

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

${\displaystyle \displaystyle (Eq.3)}$

Deriving equation (4):
By definition:

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

${\displaystyle Q'=C{\frac {dv_{C}}{dt}}}$

${\displaystyle Q''=C{\frac {d^{2}v_{C}}{dt^{2}}}}$

Using Q and its time derivatives, (2) can be written as

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

${\displaystyle \displaystyle (Eq.4)}$

Authors: Erin Standish, James Welch
Reviewed by: Brandon Wright Ryan McDaniel

For question 1.5a we have to find the general solution to the equation:

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

${\displaystyle \displaystyle }$

(1)

and we have the relation

${\displaystyle \displaystyle y=e^{\lambda x}}$

${\displaystyle \displaystyle }$

(2)

Its first two derivatives are

${\displaystyle \displaystyle y'=\lambda e^{\lambda x}}$

${\displaystyle \displaystyle }$

(3)

and

${\displaystyle \displaystyle y''=\lambda ^{2}e^{\lambda x}}$

${\displaystyle \displaystyle }$

(4)

Incorporating (2), (3), and (4) into equation (1), it can be shown that

${\displaystyle \displaystyle \lambda ^{2}e^{\lambda x}+4\lambda e^{\lambda x}+(\pi ^{2}+4)e^{\lambda x}=0}$

${\displaystyle \displaystyle }$

Factoring out the exponential term,

${\displaystyle \displaystyle e^{\lambda x}(\lambda ^{2}+4\lambda +\pi ^{2}+4)=0}$

${\displaystyle \displaystyle }$

We recognize that an exponential term cannot equal zero, so it can be divided out. Now we are left with the characteristic equation

${\displaystyle \displaystyle \lambda ^{2}+4\lambda +\pi ^{2}+4=0}$

${\displaystyle \displaystyle }$

Recognizing the relation, (7) can be factored to

${\displaystyle \displaystyle (\lambda +2)^{2}+\pi ^{2}=0}$

${\displaystyle \displaystyle }$

to reveal the roots of the characteristic equation to be

${\displaystyle \displaystyle \lambda =-2+\pi i}$

${\displaystyle \displaystyle }$

and

${\displaystyle \displaystyle \lambda =-2-\pi i}$

${\displaystyle \displaystyle }$

so the standard form of the equation can be written

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

${\displaystyle \displaystyle }$

For question 1.5b we have to find the general solution to the equation:

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

${\displaystyle \displaystyle }$

and we have the relation

${\displaystyle \displaystyle y=e^{\lambda x}}$

${\displaystyle \displaystyle }$

its first two derivatives are

${\displaystyle \displaystyle y'=\lambda e^{\lambda x}}$

${\displaystyle \displaystyle }$

and

${\displaystyle \displaystyle y''=\lambda ^{2}e^{\lambda x}}$

${\displaystyle \displaystyle }$

Incorporating these equations it can be shown that

${\displaystyle \displaystyle \lambda ^{2}e^{\lambda x}+2\pi \lambda e^{\lambda x}+\pi ^{2}e^{\lambda x}=0}$

${\displaystyle \displaystyle }$

Dividing out,

${\displaystyle \displaystyle e^{\lambda x}}$

${\displaystyle \displaystyle }$

we have

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

${\displaystyle \displaystyle }$

and since

${\displaystyle \displaystyle e^{\lambda x}}$

${\displaystyle \displaystyle }$

can't equal 0, we must evaluate the roots of the equation.

${\displaystyle \displaystyle (\lambda ^{2}+2\pi \lambda +\pi ^{2})=0}$

${\displaystyle \displaystyle }$

Factoring we have,

${\displaystyle \displaystyle (\lambda +\pi )^{2}=0}$

${\displaystyle \displaystyle }$

Which places a double root at

${\displaystyle \displaystyle x=-\pi }$

${\displaystyle \displaystyle }$

So to describe the system in standard form,

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

${\displaystyle \displaystyle }$

Author: Brandon Wright
Reviewed by: Matt Hilsenrath

Example alt text

Order: Second Order ODE

Linearity: Linear

Superposition:

${\displaystyle {\bar {y}}(x)=y_{p}(x)+y_{h}(x)\!}$

From Figure 1 -

${\displaystyle y''=g\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle y_{p}''=g\!}$

${\displaystyle y_{h}''=0\!}$

Summing above equations together:

${\displaystyle y_{p}''+y_{h}''=g\!}$

${\displaystyle (y_{p}+y_{h})''=g={\bar {y}}(x)''\!}$

Therefore superposition IS valid.

Example alt text

Order: First Order ODE

Linearity: Non-linear

Superposition:

${\displaystyle {\bar {v}}(x)=v_{p}(x)+v_{h}(x)\!}$

From Figure 2 -

${\displaystyle mv'=mg-bv^{2}\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle mv_{p}'+bv_{p}^{2}=mg\!}$

${\displaystyle mv_{h}'+bv_{h}^{2}=0\!}$

Summing above equations together:

${\displaystyle mv_{p}'+mv_{h}'+bv_{p}^{2}+bv_{h}^{2}=mg\!}$

${\displaystyle m(v_{p}+v_{h})'+b(v_{p}^{2}+bv_{h}^{2})=mg\not =m(v_{p}+v_{h})'+b(v_{p}+bv_{h})^{2}\!}$

Therefore, superposition is NOT valid.

Example alt text

Order: First Order ODE

Linearity: Non-linear

Superposition:

${\displaystyle {\bar {h}}(x)=h_{p}(x)+h_{h}(x)\!}$

From Figure 3 -

${\displaystyle h'=-kh^{1/2}\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle h_{h}'=-kh_{h}^{1/2}\!}$

${\displaystyle h_{p}'=-kh_{p}^{1/2}\!}$

Summing above equations together:

${\displaystyle h_{h}'+h_{p}'=-kh_{h}^{1/2}-kh_{p}^{1/2}\!}$

${\displaystyle (h_{h}+h_{p})'=-kh_{h}^{1/2}-kh_{p}^{1/2}\not =-k(h_{h}+h_{p})^{1/2}\!}$

Therefore superposition is NOT valid.

Example alt text

Order: Second Order ODE

Linearity: Linear

Superposition:

${\displaystyle {\bar {y}}(x)=y_{p}(x)+y_{h}(x)\!}$

From Figure 4 -

${\displaystyle my''+ky=0\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle my_{h}''+ky_{h}=0\!}$

${\displaystyle my_{p}''+ky_{p}=0\!}$

Summing above equations together:

${\displaystyle my_{h}''+my_{p}''+ky_{h}+ky_{p}=0\!}$

${\displaystyle m(y_{h}+y_{p})''+k(y_{h}+y_{p})=0=m{\bar {y}}(x)''+k{\bar {y}}(x)\!}$

Therefore superposition IS valid.

Example alt text

Order: Second Order ODE

Linearity: Linear

Superposition:

${\displaystyle {\bar {y}}(x)=y_{p}(x)+y_{h}(x)\!}$

From Figure 5 -

${\displaystyle y''+w_{0}^{2}y=\cos {wt}\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle y_{h}''+w_{0}^{2}y_{h}=0\!}$

${\displaystyle y_{p}''+w_{0}^{2}y_{p}=\cos {wt}\!}$

Summing above equations together:

${\displaystyle y_{p}''+y_{h}''+w_{0}^{2}y_{p}+w_{0}^{2}y_{h}=\cos {wt}\!}$

${\displaystyle (y_{p}+y_{h})''+w_{0}^{2}(y_{p}+y_{h})=\cos {wt}={\bar {y}}(x)''+w_{0}^{2}{\bar {y}}(x)\!}$

Therefore superposition IS valid.

Example alt text

Order: Second Order ODE

Linearity: Linear

Superposition:

${\displaystyle {\bar {I}}(x)=I_{p}+I_{h}\!}$

From Figure 6 -

${\displaystyle LI''+RI'+{\frac {1}{C}}I=E'\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle LI_{p}''+RI_{p}'+{\frac {1}{C}}I_{p}=E'\!}$

${\displaystyle LI_{h}''+RI_{h}'+{\frac {1}{C}}I_{h}=0\!}$

Summing above equations together:

${\displaystyle LI_{p}''+LI_{h}''+RI_{p}'+RI_{h}'+{\frac {1}{C}}I_{p}+{\frac {1}{C}}I_{h}=E'\!}$

${\displaystyle L(I_{p}+I_{h})''+R(I_{p}+I_{h})'+{\frac {1}{C}}(I_{p}+I_{h})=E'=L{\bar {I}}''+R{\bar {I}}'+{\frac {1}{C}}{\bar {I}}\!}$

Therefore superposition IS valid.

Example alt text

Order: iv Order ODE

Linearity: Non-linear

Superposition:

${\displaystyle {\bar {y}}(x)=y_{p}+y_{h}\!}$

From Figure 7 -

${\displaystyle EIy^{iv}=f(x)\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle EIy_{p}^{iv}=f(x)\!}$

${\displaystyle EIy_{h}^{iv}=0\!}$

Summing above equations together:

${\displaystyle EIy_{p}^{iv}+EIy_{h}^{iv}=f(x)\!}$

${\displaystyle EI(y_{p}^{iv}+y_{h}^{iv})=f(x)\not =EI{\bar {y}}^{iv}=EI(y_{p}+y_{h})^{iv}\!}$

Therefore superposition is NOT valid.

Example alt text

Order: Second Order ODE

Linearity: Non-linear

Superposition:

${\displaystyle {\bar {\theta }}=\theta _{p}+\theta _{h}\!}$

From Figure 8 -

${\displaystyle L\theta ''+g\sin {\theta }=0\!}$

Splitting differential equation into particular and homogeneous:

${\displaystyle L\theta _{h}''+g\sin {\theta _{h}}=0\!}$

${\displaystyle L\theta _{p}''+g\sin {\theta _{p}}=0\!}$

Summing above equations together:

${\displaystyle L\theta _{h}''+L\theta _{p}''+g\sin {\theta _{h}}+g\sin {\theta _{p}}=0\!}$

${\displaystyle L(\theta _{h}+\theta _{p})''+g(\sin {\theta _{h}}+\sin {\theta _{p}})=0\not =L{\bar {\theta }}''+g\sin {\bar {\theta }}\!}$

Therefore superposition is NOT valid.

Author: Martin Dolgiej
Last Edited: 01/28/2011 - 4:10 PM
Reviewed by:James Welch, Matt Hilsenrath

Team Contribution Table
Problem || Solved/Typed By || Proofread By
R1.1 in Sec 1 p. 1-5	Andrew	Martin
R1.2 in Sec 1 p. 1-4	Martin	Matt
R1.3 in Sec 1 p. 1-5	Ryan	Erin
R1.4 in Sec 2 p. 2-2	Erin, James	Brandon, Ryan
R1.5 in Sec 2 p. 2-5	Brandon	Matt
R1.6 in Sec 2 p. 2-7	Martin	James, Matt