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:

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

From Figure 1 -
Spring:

Mass:

Damp:

Forcing Funtion:

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):

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

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)

${\displaystyle f=my''\!}$ 

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

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

Author: Ryan McDaniel
Reviewed by: Erin Standish

Given:

Solution:
Deriving equation (3):

Taking the derivative of (Eq. 2) gives:

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):

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

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:

(1)

and we have the relation

(2)

Its first two derivatives are

(3)

and

(4)

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

Factoring out the exponential term,

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

Recognizing the relation, (7) can be factored to

to reveal the roots of the characteristic equation to be

and

so the standard form of the equation can be written

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

and we have the relation

its first two derivatives are

and

Incorporating these equations it can be shown that

Dividing out,

we have

and since

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

Factoring we have,

Which places a double root at

So to describe the system in standard form,

Author: Brandon Wright
Reviewed by: Matt Hilsenrath

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:

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

Therefore superposition IS valid.

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:

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

Therefore, superposition is NOT valid.

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:

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

Therefore superposition is NOT valid.

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:

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

Therefore superposition IS valid.

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:

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

Therefore superposition IS valid.

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:

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

Therefore superposition IS valid.

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:

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

Therefore superposition is NOT valid.

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:

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

Therefore superposition is NOT valid.

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