User:Egm4313.s12.team16.r1

Report 1:
By: Team 16

R1.1


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 $$ \dot{y} \! $$ and $$ \ddot{y} \! $$ respectively.

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

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

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

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

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

where $$ c \! $$ is the coefficient of viscous friction

Equating the kinematics to the kinetics we preliminarily have

$$ m\ddot{y}+ky+c\ddot{y}=\ddot{F_{y}} \! $$ and simplifying with: $$ y=y_{k}=y_{c} \! $$ $$ \dot{y}=\dot{y_{c}}=\dot{y_{k}} \! $$

Final Equation:

Author: Nyal Jennings

Last Edit: 01/28/2012 - 11:01 PM

Reviewed By: Martin Dolgiej

R1.2
From Figure 1 - Spring:

Mass:

Damp:

Forcing Funtion:

Balancing the forces in the y direction: $$ \sum F_y = F_m - F_r = F_k - F_c \! $$

Substituting equations (1),(2),(3), and (4): $$ my'' - r(t) = -ky - cy' \! $$

Author: Martin Dolgiej

Last Edit: 01/28/2012 - 11:32 PM

Reviewed By: Matt Hilsenrath

R1.3
FBD of Spring



$$f_s = k*y_s \! $$

FBD of Dashpot



$$ Cy_D' = f_c \! $$

FBD of Mass



(Newton's 2nd Law)

$$ f = my'' \! $$

$$\sum F(t) = my'' + f_s - f_c \! $$

$$ f_1 = f_s - F_c \! $$

Author: Ryan McDaniel

Reviewed by: Erin Standish

R1.4
Given:

Solution:

Deriving equation (3):

Taking the derivative of (Eq. 2) gives:

By definition:

$$ i=C\frac{dv_C}{dt} $$

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

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

Plugging theses definitions into (Eq. A):

Deriving equation (4):

By definition:

$$ Q=Cv_c \! $$

$$ Q'=C\frac{dv_C}{dt} $$

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

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

Authors: Erin Standish, James Welch

Reviewed by: Brandon Wright Ryan McDaniel

R1.5
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

Part A:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{y}(x) = y_p(x) + y_h(x) \! $$ From Figure 1 - $$ y'' = g \! $$ Splitting differential equation into particular and homogeneous: $$ y_p'' = g \! $$ $$ y_h'' = 0 \! $$ Summing above equations together: $$ y_p + y_h = g \! $$ Therefore superposition IS valid.

Part B:
Order:     First Order ODE Linearity: Non-linear Superposition: $$ \bar{v}(x) = v_p(x) + v_h(x) \! $$ From Figure 2 - $$ mv' = mg - bv^2 \! $$ Splitting differential equation into particular and homogeneous: $$ mv_p' + bv_p^2 = mg \! $$ $$ mv_h' + bv_h^2 = 0 \! $$ Summing above equations together: $$ mv_p' + mv_h' + bv_p^2 + bv_h^2 = mg \! $$ Therefore, superposition is NOT valid.

Part C:
Order:     First Order ODE Linearity: Non-linear Superposition: $$ \bar{h}(x) = h_p(x) + h_h(x) \! $$ From Figure 3 - $$ h' = -kh^{1/2} \! $$ Splitting differential equation into particular and homogeneous: $$ h_h' = -kh_h^{1/2} \! $$ $$ h_p' = -kh_p^{1/2} \! $$ Summing above equations together: $$ h_h' + h_p' = -kh_h^{1/2} - kh_p^{1/2} \! $$ Therefore superposition is NOT valid.

Part D:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{y}(x) = y_p(x) + y_h(x) \! $$ From Figure 4 - $$ my'' + ky = 0 \! $$ Splitting differential equation into particular and homogeneous: $$ my_h'' + ky_h = 0 \! $$ $$ my_p'' + ky_p = 0 \! $$ Summing above equations together: $$ my_h + my_p + ky_h + ky_p = 0 \! $$ Therefore superposition IS valid.

Part E:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{y}(x) = y_p(x) + y_h(x) \! $$ From Figure 5 - $$ y'' + w_0^2y = \cos{wt} \! $$ Splitting differential equation into particular and homogeneous: $$ y_h'' + w_0^2y_h = 0 \! $$ $$ y_p'' + w_0^2y_p = \cos{wt} \! $$ Summing above equations together: $$ y_p + y_h + w_0^2y_p + w_0^2y_h = \cos{wt} \! $$ Therefore superposition IS valid.

Part F:
Order:     Second Order ODE Linearity: Linear Superposition: $$ \bar{I}(x) = I_p + I_h \! $$ From Figure 6 - $$ LI'' + RI' + \frac{1}{C}I = E' \! $$ Splitting differential equation into particular and homogeneous: $$ LI_p'' + RI_p' + \frac{1}{C}I_p = E' \! $$ $$ LI_h'' + RI_h' + \frac{1}{C}I_h = 0 \! $$ Summing above equations together: $$ LI_p + LI_h + RI_p' + RI_h' + \frac{1}{C}I_p + \frac{1}{C}I_h = E' \! $$ Therefore superposition IS valid.

Part G:
Order:     iv Order ODE Linearity: Non-linear Superposition: $$ \bar{y}(x) = y_p + y_h \! $$ From Figure 7 - $$ EIy^{iv} = f(x) \! $$ Splitting differential equation into particular and homogeneous: $$ EIy_p^{iv} = f(x) \! $$ $$ EIy_h^{iv} = 0 \! $$ Summing above equations together: $$ EIy_p^{iv} + EIy_h^{iv} = f(x) \! $$ Therefore superposition is NOT valid.

Part H:
Order:     Second Order ODE Linearity: Non-linear Superposition: $$ \bar{\theta} = \theta_p + \theta_h \! $$ From Figure 8 - $$ L\theta'' + g\sin{\theta} = 0 \! $$ Splitting differential equation into particular and homogeneous: $$ L\theta_h'' + g\sin{\theta_h} = 0 \! $$ $$ L\theta_p'' + g\sin{\theta_p} = 0 \! $$ Summing above equations together: $$ L\theta_h + L\theta_p + g\sin{\theta_h} + g\sin{\theta_p} = 0 \! $$ Therefore superposition is NOT valid.

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