User:Egm4313.s12.team9/R1

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

Initial Information
From [[media:Egm4313.s12.team9.1-4.jpg|Lecture slide 1-4]] and [[media:Egm4313.s12.team9.1-5.jpg|Lecture slide 1-5]]:

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

k, c, m, f(t), y(t), y_k, y_c, f_i, f_k, f_c $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }



Methods
Kinematics:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y = y_k = y_c$$
 * $$\displaystyle Derived (Eq.1)$$
 * }

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


 * $$\displaystyle f(t) = my'' + f_i$$
 * $$\displaystyle (Eq.2)$$
 * }


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


 * $$\displaystyle f_i = f_k = f_c$$
 * $$\displaystyle (Eq.3)$$
 * }


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


 * $$\displaystyle f(t) = my'' + f_k + f_c$$
 * $$\displaystyle Derived (Eq.2,Eq.3)$$
 * }

Spring Force:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle f_k = ky_k$$
 * $$\displaystyle (Eq.4)$$
 * }

Dash-Pot Force:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle f_c = cy'_c$$
 * $$\displaystyle (Eq.5)$$
 * }

Solution
Final Equation:
 * {| style="width:100%" border="0" align="left"

$$\displaystyle f(t) = my''_k + cy'_k + ky_k$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle $$
 * }

=R1.2: Spring-mass-dashpot with applied force r(t) on the ball(Fig. 53, p.85, K2011)=

Initial Information
Variables:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle k, c, m, r(t), y_k, f_k, f_c $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

Methods
Kinematics:


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

$$
 * $$\displaystyle y=y_k $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

Kinetics:


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

$$
 * $$\displaystyle my'' + f_k + f_c = r(t) $$
 * $$\displaystyle (Eq.2)
 * $$\displaystyle (Eq.2)
 * }
 * }

Spring Force:


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

$$
 * $$\displaystyle f_k = ky_k $$
 * $$\displaystyle (Eq.3)
 * $$\displaystyle (Eq.3)
 * }
 * }

Dashpot Force:


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

$$
 * $$\displaystyle f_c = cy'_c $$
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * }
 * }

Solution

 * {| style="width:0%" border="0" align="left"

$$\displaystyle my'' + ky_k + cy'_c = r(t) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= R1.3 Spring-dashpot-mass system FBD and Derivation of Equation of Motion = This problem can be found on [[media:Egm4313.s12.team9.1-6.jpg|Lecture slide 1-6]].

Given
The spring-dashpot-mass system can be found on [[media:Egm4313.s12.team9.1-4.jpg|Lecture slide 1-4]]

Methods
Free body diagrams:



The general equilibrium expression in the y direction is:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \sum F_{y}=ma $$
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * }
 * }

From the FBD's it is apparent that:


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

$$
 * $$\displaystyle f_{I}=f_{k}=f_{c} $$
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * }
 * }

Where $$ f_{I} $$ is the internal force. Thus, substituting gives:


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

$$
 * $$\displaystyle ma=my''=f(t)-f_{I} $$
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * }
 * }

Solution
Rearranging to solve for $$f(t)$$ we get equation 2 from [[media:Egm4313.s12.team9.1-4.jpg|Lecture slide 1-4]]


 * {| style="width:0%" border="0" align="left"

$$\displaystyle my''+f_{I}=f(t) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= R1.4: RLC Circuit Modeling =

Initial Information
From [[media:Egm4313.s12.team9.2-2.jpg|Lecture slide 2-2]], a general RLC circuit Kirchhoff's Voltage Law (KVL) equation, and two alternative formulations, are given:


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

V = LC \frac{d^{2}v_{c}}{dt^{2}} + RC \frac{dv_{c}}{dt}+ v_{c}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * }
 * }


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

{V}' = L{I}'' + R{I}' + \frac{1}{C}I$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.3)
 * }
 * }


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

V = L{Q}'' + R{Q}' + \frac{1}{C}Q$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.4)
 * }
 * }

We are being asked to derive (3) and (4) from (2).

Methods
From [[media:Egm4313.s12.team9.2-2.jpg|Lecture slide 2-2]], capacitance is defined as,


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

Q = Cv_{c} \Rightarrow \int idt = Cv_{c} \Rightarrow i = C\frac{dv_{c}}{dt}$$ $$ Using this definition we can find equations (3) and (4).
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * }
 * }

Solution
Deriving (1), we get:


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

C\frac{d^{2}v_{c}}{dt^{2}} = \frac{di}{dt}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.{1}')
 * }
 * }

Also, by solving (1) for $$v_{c}$$, we obtain:


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

v_{c} = \frac{1}{C}\int idt$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.{1}'')
 * }
 * }

Substituting equations (1), (1'), and (1") into (2)


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

V = L{I}' + RI + \frac{1}{C}\int I$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.{2}')
 * }
 * }

Which is an "integro-differential equation." Therefore, to eliminate the integral we differentiate (2') with respect to t, to get:


 * {| style="width:0%" border="0" align="left"

$$\displaystyle {V}' = L{I}'' + R{I}' + \frac{1}{C}I$$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Since $$Q = \int idt$$ from (1), substituting this into (2') yields:


 * {| style="width:0%" border="0" align="left"

$$\displaystyle V = L{Q}'' + R{Q}' + \frac{1}{C}Q $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= R1.5: General Solutions of 2nd Order Linear ODEs with Constant Coefficients =

Initial Information
These problems can be found on [[media:Lecture slide 2-5.jpg|Lecture slide 2-5]].

From pg. 59 problem 4,


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

{y}''+4{y}'+({{\pi }^{2}}+4)y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.1)
 * }
 * }

And from pg. 59 problem 5,


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

{y}''+2\pi {y}'+{{\pi }^{2}}y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.2)
 * }
 * }

Find a general solution for Equations (1) and (2) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equations (1) and (2). The characteristic equation of Equation (1) is


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

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

The case of the equation can be determined by $$\displaystyle {{a}^{2}}-4b$$, where $$\displaystyle a=4$$ and $$\displaystyle b={{\pi }^{2}}+4$$ , as shown below:


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

{{a}^{2}}-4b={{4}^{2}}-4({{\pi }^{2}}+4)=16-4{{\pi }^{2}}-16=-4{{\pi }^{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since this is below 0, the equation is a Case III equation with complex conjugate roots.

Using the same method, the characteristic equation of Equation (2) is


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

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

The case of the equation can be determined by $$\displaystyle {{a}^{2}}-4b$$, where $$\displaystyle a=2\pi $$ and $$\displaystyle b={{\pi }^{2}}$$ , as shown below:
 * {| style="width:100%" border="0" align="left"

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

Since this is equal to 0, the equation is a Case II equation with a real double root.

Solution: Problem 4
The roots of the characteristic equation is given by:


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

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -4\pm \sqrt{-4{{\pi }^{2}}} \right) \\ & \lambda =-2\pm \pi i \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Therefore, in this problem,
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \omega =\pi$$
 * }
 * }
 * }

From page 57, a real general solution in Case III is


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

y={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x)$$ where A and B are arbitrary constant coefficients. In this example, the general solution is
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:0%" border="0" align="left"

$$\displaystyle y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (1).


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

& {y}'=-2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}\left( -A\pi \sin \pi x+B\pi \cos \pi x \right) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Substituting into Equation (1),


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

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

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-8{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+{{\pi }^{2}}{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


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

0=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x)$$.
 * $$\displaystyle
 * }
 * }

Solution: Problem 5
Equation (2) can be factored as shown below to determine the root.


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

& {{(\lambda +\pi )}^{2}}=0 \\ & \lambda =-\pi \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

From page 56, a basis of solutions is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {{e}^{-\pi x}}$$, $$\displaystyle x{{e}^{-\pi x}}$$
 * }
 * }
 * }

Therefore, the corresponding general solution for this equation is
 * {| style="width:0%" border="0" align="left"

$$\displaystyle y=({{c}_{1}}+{{c}_{2}}x){{e}^{-\pi x}}$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (2).
 * {| style="width:100%" border="0" align="left"

{y}'=-\pi {{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+{{c}_{2}}{{e}^{-\pi x}}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

& {y}''={{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-\pi {{c}_{2}}{{e}^{-\pi x}}-\pi {{c}_{2}}{{e}^{-\pi x}} \\ & {y}''={{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-2\pi {{c}_{2}}{{e}^{-\pi x}} \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Substituting into Equation (2),
 * {| style="width:100%" border="0" align="left"

{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-2\pi {{c}_{2}}{{e}^{-\pi x}}+2\pi \left( -\pi {{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+{{c}_{2}}{{e}^{-\pi x}} \right)+{{\pi }^{2}}\left( {{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x) \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-2{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+2\pi {{c}_{2}}{{e}^{-\pi x}}-2\pi {{c}_{2}}{{e}^{-\pi x}}=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

0=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since all terms cancel, the solution is:
 * {| style="width:100%" border="0" align="left"

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

= R1.6 Determination of ODE Order, Linearity, and Application of the Superposition Principle =

Initial Information
We are asked to determine the order, linearity and whether the principle of superposition can be applied to the following examples. The order of a differential equation is found by looking at the highest occurring derivative of the dependent variable. A differential equation is linear if the dependent variable and all of its derivatives occur linearly throughout the equation.

Background Theory

If we were to consider a function $$\bar{y}(x)$$, which is defined as the sum of a homogeneous solution $$\displaystyle{y_h(x)}$$ and a particular solution $$\displaystyle{y_p(x)} $$:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

Homogeneous Solution:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y_h''+p(x)y_h'+q(x)y_h=0$$
 * }
 * }
 * }

Particular Solution:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y_p''+p(x)y_p'+q(x)y_p=r(x)$$
 * }
 * }
 * }

Summing up both the Homogeneous Solution and the Particular Solution will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle (y_h+y_p)+p(x)(y_h'+y_p')+q(x)(y_h+y_p)=r(x)$$
 * }
 * }
 * }

Reducing the equation according to Linear Principle will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y_h+y_p=(y_h+y_p)= \bar{y}$$
 * }
 * }
 * }


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


 * $$\displaystyle y_h'+y_p'=(y_h+y_p)'= \bar{y}'$$
 * }
 * }
 * }

Now, through substitution, we obtain:


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


 * $$\displaystyle \bar{y}''+p(x)\bar{y}'+q(x)\bar{y}=r(x)$$
 * }
 * }
 * }

which is see to be the same equation as:


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


 * $$\displaystyle y''+p(x)y'+q(x)y=r(x)$$
 * }
 * }
 * }

Falling Stone <dl> <dt>Governing Equation:</dt> <dd>$$\displaystyle{y}''= g = constant$$</dd> </dl>

Order: Second Order ODE Linearity: Yes Superposition: Yes This can be proved using the previous method shown in background theory: Using:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

The homogeneous equation is set as:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y_h''}=0$$
 * }
 * }
 * }

And the particular solution is set as:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y_p''}=g$$
 * }
 * }
 * }

Now, adding both equations, we get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y_h+y_p}=g$$
 * }
 * }
 * }

Which can be simplified to:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {(y_h+y_p)''}=g$$
 * }
 * }
 * }

Which, by superposition, equals
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y}''=g$$
 * }
 * }
 * }

Parachutist</H4> <dl>

<dt>Governing Equation:</dt> <dd>$$\displaystyle{mv'}=mg-bv^2$$</dd> </dl>

Order: First Order ODE Linearity: No Superposition: No

This can be shown using the same method as the one shown in background theory. Start by rearranging the the given equation:
 * {| style="width:100%" border="0" align="left"


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

To look like:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {mv'+bv^2}=mg$$
 * }
 * }
 * }

Now, using the governing equation:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{v}(x) = v_p(x) + v_h(x$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {mv_h'+mv_h^2}=0$$
 * }
 * }
 * }

And the particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {mv_p'+mv_p^2}=mg$$
 * }
 * }
 * }

Now we add both of those equations together to get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {mv_h'+mv_h^2+mv_p'+mv_p^2}=mg$$
 * }
 * }
 * }

We can simplify that equation down to:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {m(y_h+y_p)'+m(v_h^2+v_p^2)}=mg$$
 * }
 * }
 * }

We can see that:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle (v_h^2+v_p^2)\neq \bar{v}^2$$
 * }
 * }
 * }

The method of Superposition will not hold.

Outflowing water from a tank

Govering Equation:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {h'}=-k\sqrt{h}$$
 * }
 * }
 * }

Order: First Order ODE Linearity: No Superposition: No

This can be shown using the same method as the one shown in background theory. Rearrange the governing equation to look like:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {h'}=-k\sqrt{h}$$
 * }
 * }
 * }

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


 * $$\displaystyle \bar{v}(x) = v_p(x) + v_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle mv_h'+mv_h^2 = 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle mv_p'+mv_p^2 = mg$$
 * }
 * }
 * }

Now, adding both of those equations together, we get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle mv_h'+mv_h^2+mv_p'+mv_p^2= mg$$
 * }
 * }
 * }

which will simplify to:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m(y_h+y_p)'+m(v_h^2+v_p^2) = mg$$
 * }
 * }
 * }

You can see that:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle (v_h^2+v_p^2)\neq \bar{v}^2$$
 * }
 * }
 * }

So the method of Superposition does not hold.

Vibrating mass on a spring</H4> <dl> <dt>Governing Equation:</dt> <dd>$$\displaystyle{my''+ky}=0$$</dd> </dl>

Order: Second Order ODE Linearity: Linear Superposition: Yes

This can be shown using the same method as the one shown in background theory. Using: $$\bar{y}(x) = y_p(x)+y_h(x)$$

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m{y}_h''+ky_h= 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m{y}_p''+ky_p= 0$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m{y}_h+ky_h+m{y}_p+ky_p = 0$$
 * }
 * }
 * }

You can simplify the former equation using governing equation to get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m({y}_h+{y}_p)''+k(y_h+y_p) = 0$$
 * }
 * }
 * }


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


 * $$\displaystyle m\bar{y}''+k\bar{y}= 0$$
 * }
 * }
 * }

Superposition can be applied.

Beats of a vibrating system</H4> <dl> <dt>Governing Equation:</dt> <dd>$${y}''+\omega _0^2y=\cos \omega t, \omega _0\approx \omega $$</dd> </dl>

Order: Second Order ODE Linearity: Linear Superposition: Yes

This can be shown using the same method as the one shown in background theory. Using:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y}_h''+\omega _0^2y_h= 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y}_p''+\omega _0^2y_p=\cos \omega t$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y}_h+\omega _0^2y_h+{y}_p+\omega _0^2y_p=\cos \omega t$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle ({y}_h+{y}_p)''+\omega _0^2(y_h+y_p)=\cos \omega t$$
 * }
 * }
 * }


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


 * $$\displaystyle \bar y''+\omega _0^2 \bar y=\cos \omega t$$
 * }
 * }
 * }

Superposition can be applied because the equations are the same.

Current I in an RLC Circuit <dl> <dt>Governing Equation:</dt> <dd>$$\displaystyle{V}' = L{I}'' + R{I}' + \frac{1}{C}I$$</dd> </dl>

Order: Second Order ODE Linearity: Yes Superposition: Yes

This can be shown using the same method as the one shown in background theory.

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


 * $$\displaystyle \bar{I}(x) = I_p(x) + I_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle LI_h''+RI_h'+\frac{1}{C}I_h=0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle LI_p''+RI_p'+\frac{1}{C}I_p=E'$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle LI_h+RI_h'+\frac{1}{C}I_h+LI_p+RI_p'+\frac{1}{C}I_p= E'$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L(I_h+I_p)''+R(I_h+I_p)'+\frac{1}{C}(I_h+I_p)= E'$$
 * }
 * }
 * }


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


 * $$\displaystyle L\bar I''+R\bar I'+\frac{1}{C}\bar I= E'$$
 * }
 * }
 * }

Superposition can be applied because the equations are the same.

Beam Deformation</H4> <dl> <dt>Governing Equation:</dt> <dd>$$\displaystyle{EIy^{iv}}=f(x)$$</dd> </dl>

Order: Fourth Order ODE Linearity: Linear Superposition: Yes

This can be shown using the same method as the one shown in background theory.

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


 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle EIy_h^{iv} = 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle EIy_p^{iv} = f(x)$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle EIy_h^{iv}+EIy_p^{iv} = f(x)$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle EI(y_h+y_p)^{iv} = f(x)$$
 * }
 * }
 * }


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


 * $$\displaystyle EI(\bar y)^{iv} = f(x)$$
 * }
 * }
 * }

Superposition can be applied because the equations are the same.

Pendulum</H4> <dl> <dt>Governing Equation:</dt> <dd>$$\displaystyle{L\theta''+g*sin(\theta)}=0$$</dd> </dl>

Order: Second Order ODE Linearity: No Superposition: No

This can be shown using the same method as the one shown in background theory.

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


 * $$\displaystyle \bar{\theta}(x) = \theta_p(x) + \theta_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L\theta_h'' +g\sin \theta_h =0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L\theta_p'' +g\sin \theta_p =0$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L\theta_h +g\sin \theta_h+L\theta_p +g\sin \theta_p =0$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L(\theta_h+\theta_p)''+g(\sin \theta_h +\sin \theta_p )=0$$
 * }
 * }
 * }


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


 * $$\displaystyle (\sin \theta_h +\sin \theta_p )\neq \sin \bar \theta$$
 * }
 * }
 * }

Superposition cannot be applied because the equations are not the same.

= References =

Section 2 Lecture Notes

=Contributing Team Members=