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University of Florida/Egm4313/s12.team7/Report1

==R.1.1: Equation of Motion for a Spring-dashpot Parallel System in Series with Mass and Applied Force ==

Problem Statement
Given a spring-dashpot system in parallel with an applied force, find the equation of motion.

Background Theory
For this problem, Newton's second law is used,

and applied to the mass at the end of a spring and damper in parallel.

Solution
Assuming no rotation of the mass,

Therefore, the spring and damper forces can be written as, respectively,

and

And the resultant force on the mass can be written as

Now, from equations ($$) and ($$) all of the forces can be substituted into Newton's second law, and since each distance is equal,

Substituting these exact force equations,

A little algebraic manipulation yields

Finally, dividing by the mass to put the equation in standard form gives the final equation of motion for the mass:

Problem Statement
For this problem, the task was to derive the equation of motion of the spring-mass-dashpot in Fig. 53, in K 2011 p.85 with an applied force r(t).

Background Theory
To solve this problem, note that the ODE of a damped mass-spring system is

When there is a external formal added to the model r(t) on the right. This then gives up the equation

Solution
In this problem referencing back to equation ($$)

The resultant force for the system can be described as stated in equation ($$)

Since r(t) is the external force,

The model of a mass-spring system ODE with external force on the right is modeled as

The internal force is equal to the force of the spring making this equation

When a dashpot is added, the force of the dashpot cy’ is added to the equation making it

For the characteristic equation m is divided throughout the whole equation making it

Problem Statement
For this problem we were to draw the FBDs and derive the equation of motion for the following system.

Background Theory
To start solving this problem, we have to first look at how the spring, dashpot, and mass interact with each other by analyzing their respective free-body diagrams. Next we take a look at what we already know.

We know from kinematics that:

We also know from kinetics that:

From the constitutive relations we know:

We also know:

by solving for $$ y_{c}' $$ we get

Solution
To derive the equation of motion we have to manipulate and combine a few formulas that we know. We first take what we found in ($$) and substitute in

for $$y''$$. From equations ($$), ($$), and ($$) we can substitute $$ f{_{i}}$$ for After the substitutions into equation ($$) we have

From the equation ($$) we know We can then substitute this into equation ($$) getting

To get the answer in its final form we divide both sides of the equation by m, and our final answer is

Problem Overview
For this problem, the goal is to derive two different equations from the circuit equation

Each separate derivation is presented in Voltage-Charge Derivation and Voltage-Current Derivation

Background Theory
To solve these derivation, it is wise to note that capacitance is

It can also be noted that

 * $$ Q=\int idt $$

This means that
 * $$ \int idt = Cv_{C} $$

Completing the integration results in

It should also be noted that ($$) can be written in the form

It should also be noted that ($$) can be written in the form

Problem Statement
For this problem, the Voltage-Charge equation

are derived from ($$)

Solution
Taking the derivative of ($$) are taken with respect to time

Substituting ($$), ($$), and ($$) into ($$) results in ($$) such that

Problem Statement
For this problem, the following equation

are derived from ($$)

Solution
The first step is to take the derivative of ($$) resulting in

Taking the derivative of ($$) are taken with respect to time

Substituting ($$), ($$), and ($$) into ($$) results in ($$) such that

R.1.5: Solutions of General 2nd Order ODEs
===Background Theory ===

Consider a second-order homogeneous linear ODE with constant coefficients a and b.

To solve this problem, note that the solution to a first-order linear ODE of the form:

is an exponential function, yielding a solution of the form:

Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing $$y'$$ and $$y''$$, and so we will take the derivatives (with respect to x) of ($$).

We will now substitute ($$), ($$), and ($$) into ($$) to obtain the relationship:

 * Simplifying, we have:

Since ($$) follows the same form as the quadratic equation, we can solve for $$\lambda _{1}$$ and $$\lambda _{2}$$ as follows:

Referring back to algebra, we know that the solution to these two equations can be one of three cases:

Case 1
Two real roots if...

These two roots give us two solutions:

The corresponding general solution then takes the form of the following:
 * $$y_{1} = e^{\lambda _{1} x}$$ || || and || || $$y_{2} = e^{\lambda _{2} x}$$
 * }
 * }

Case 2
A real double root if...

This yields only one solution:
 * $$y_{1} = e^{\lambda x}$$

In order to form a basis, a set of two linearly independent solutions, we must find another solution. To do so, we will use the method of reduction of order, where:
 * $$y_{2} = uy_{1}$$
 * $$y'_{2} = u'y_{1} + uy'_{1}$$
 * and
 * $$y_{2} = uy_{1} + 2u'y'_{1} + uy''_{1}$$
 * yeilding
 * $$ (uy_{1} + 2u'y'_{1} + uy_{1}) + a(u'y_{1} + uy'_{1}) + buy_{1} = 0 $$
 * $$ uy_{1} + u'(2y'_{1} + ay_{1}) + u(y_{1} + a'y_{1} + by_{1}) = 0 $$

Since $$y_{1}$$ is a solution of (5-1), the last set of parentheses is zero. Similarly, the first parentheses is zero as well, because
 * $$2y'_{1} = -ae^{ax/2} = -ay_{1}$$

From integration, we get the solution:
 * $$ u = c_{1}x + c_{2} $$

If we set $$c_{1} = 1$$ and $$c_{2} = 0$$
 * $$y_{2} = xy_{1}$$

Thus, the general solution is:

Case 3
Complex conjugate roots if...

In this case, the roots of ($$) are complex:
 * $$ y_{1} = e^{-ax/2} cos( \omega x) $$
 * and
 * $$ y_{2} = e^{-ax/2} sin( \omega x) $$

Thus, the corresponding general solution is of the form:

Problem Statement
For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:

Solution
Following the process that yields ($$), we find the equation:

To find the form of the general solution, we must solve for the values of $$\lambda$$ using the quadratic formula:

Since the solutions to $$\lambda$$ are complex numbers, our solution takes the form of ($$):

A and B are constant coefficients of unknown value. Had we been given initial values of $$y(x)$$ and $$y'(x)$$, we would solve for those values as well.

Confirmation of Solution
To confirm that the solution is true, we will substitute $$y$$, $$y'$$, and $$y''$$ into ($$). If the final result is a tautology then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.

Substituting ($$), ($$), and ($$) into ($$), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

Final Solution
Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:
 * $$y = e^{-2x}(Acos(\pi x)+ Bsin(\pi x)) $$

Problem Statement
For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:

Solution
Following the process that yields ($$), we find the equation:

To find the form of the general solution, we must solve for the values of $$\lambda$$. Instead of going through the quadratic formula, we can analyze the discriminant.

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of ($$), and we have the following:

where $$c_{1}$$ and $$ c_{2}$$ are unknown constant coefficients. If this were an initial value problem with values of $$y(x)$$ and $$y'(x)$$, we would solve for those values as well.

Confirmation of Solution
To confirm that the solution is true, we will substitute $$y$$, $$y'$$, and $$y''$$ into ($$) and check to see if the final result is a tautology or a contradiction.

Substituting ($$), ($$), and ($$) into ($$), we are left with:

After combining similar terms, we have the form:

Final Solution
Since the above result is a tautology for all values of $$c_{1}$$ and $$ c_{2}$$, ($$) is confirmed and the general solution to ($$) is:

Problem Statement
Given the eight Ordinary Differential Equations shown below: (1) Determine the order (2) Determine the linearity (or lack of) (3) Show whether the principle of superposition can be applied for each equation

=== Background Theory === Consider a function $$\bar{y}(x) $$, defined as the sum of the homogeneous solution, $$ y_h(x) $$, and the particular solution, $$ y_p(x) $$:

Example:

Consider the Differential Equation in standard form:

The homogeneous solution is then:

and the particular solution is:

Now we sum up equations ($$) and ($$) to get:

If the Equation is Linear then some reductions can take place:

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

and

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

Now we can make some substitutions into ($$)using ($$)and ($$):

Which is the same as equation ($$):

$$y''+p(x)y'+q(x)y=r(x)$$

Part A
Order:     $$ 2^{nd}$$ Order ODE Linearity:  Linear Superposition: Yes

This can be shown using the method from the background theory.

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Since ($$) and ($$) are the same then we can apply Superposition.

Part B
Order:     $$ 1^{st}$$ Order ODE Linearity:  Non-Linear Superposition: No

This can be shown using the method from the background theory. Before we start it is easier to do the problem if the equation is rearranged as such:

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Note that:
 * $$(v_h^2+v_p^2)\neq \bar{v}^2$$

So we cannot apply Super position.

Part C
Order:     $$ 1^{st}$$ Order ODE Linearity:  Non-Linear Superposition: No

This can be shown using the method from the background theory. Before we start it is easier to do the problem if the equation is rearranged like this:

Recall:
 * $$\bar{h}(x) = h_h(x) + h_p(x)$$

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Note that:
 * $$(\sqrt{h_h}+\sqrt{h_p})\neq \sqrt\bar{h}$$

So we cannot apply Super position.

Part D
Order:     $$ 2^{nd}$$ Order ODE Linearity:  Linear Superposition: Yes

This can be shown using the method from the background theory.

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Since ($$) and ($$) are the same then we can apply Superposition.

Part E
Order:     $$ 2^{nd}$$ Order ODE Linearity:  Linear Superposition: Yes

This can be shown using the method from the background theory.

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Since ($$) and ($$) are the same then we can apply Superposition.

Part F
Order:     $$ 2^{nd}$$ Order ODE Linearity:  Linear Superposition: Yes

This can be shown using the method from the background theory.

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Since ($$) and ($$) are the same then we can apply Superposition.

Part G
Order:     $$ 4^{th}$$ Order ODE Linearity:  Linear Superposition: Yes

This can be shown using the method from the background theory.

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Since ($$) and ($$) are the same then we can apply Superposition.

Part H
Order:     $$ 2^{nd}$$ Order ODE Linearity:  Non-Linear Superposition: No

This can be shown using the method from the background theory.

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

The homogeneous solution is:

The particular solution is:

Now we add Equations ($$) and ($$):

We can use ($$) to simplify ($$):

Note that:
 * $$(\sin \theta_h +\sin \theta_p )\neq \sin \bar \theta$$

So we cannot apply Super position.


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

title: File:R1-1FBD.png


text:
User:Egm4313.s12.team15.cynthiah

R1.6.1

Order: 2

Linearity: Linear Superposition : Yes

$$ y''= g = const  \!$$   (1) $$ y''_h=0 \!$$  homogeneous solution (2) $$ y''_p=g \!$$  particular solution (3) By solving we get the following: $$ \bar{y}=y_h + y_p = (y_h + y_p) = 0 +g = g \!$$       (4) $$ \bar{y''}=g \!$$       Since (4) = (1) Superposition is applied (5)

R1.6.2

Order: 1 Linearity: Non-linear

Superposition: No $$ mv'= mg - bv^2  \!$$  (Eq.0) $$ mv' + bv^2 = mg  \!$$  (Eq.0 Rearranged ) $$ mv_h'+b(v_h)^2=0 \!$$   homogeneous solution (Eq. 1 ) $$ mv_p'+b(v_p)^2=mg \!$$  Particular solution (Eq. 2 )

By solving we get the following :

$$\bar{v}^2=v_h+v_p\!$$

$$ m(v_h'+v_p')+ b[(v_h)^2 + (v_p)^2]=mg \!$$

$$m(v_h +v_p)' +b[(v_h)^2 +(v_p)^2]=mg\!$$ $$\bar{v'} + b(v_h+v_p)^2=mg\!$$

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

Therefore, superposition cannot be applied.

R1.6.3 Order: 1 Linearity:Non- linear

Superposition: No

$$ h'=-k\sqrt{h}\!$$  (Eq.0) $$ h'+k\sqrt{h}=0\!$$  (Eq.0 Rearranged) $${h_h}'+k\sqrt{h_h}= 0 \!$$ Homogeneous solution (Eq. 1 ) $${h_p}'+k\sqrt{h_p}= 0\!$$ Particular solution (Eq. 2)

By solving we get the following:

$$\bar{h}=h_h+h_p\!$$

$${h_h}'+k\sqrt{h_h}+{h_p}'+k\sqrt{h_p}=0\!$$ $$({h_h}'+{h_p}')+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$({h_h}+{h_p}) '+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$(\sqrt{h_h}+\sqrt{h_p})\neq \sqrt\bar{h}\!$$

So superposition cannot be applied.

R1.6.4

Order:2

Linearity: Linear

Superposition: Yes

$$ my''+ky = 0 \!$$  (Eq.0) $$m{y}_h''+ky_h= 0\!$$  Homogeneous solution (Eq. 1 ) $$m{y}_p''+ky_p= 0\!$$   Particular  solution (Eq. 2)

By solving we get the following:

$$\bar{y}=y_h+y_p\!$$ $$\bar{y}=y_h+y_p''\!$$

Summing Eq.1 & Eq. 2 :

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

Since:

$$m\bar{y}+k\bar{y}$$ = $$ my+ky \!$$

Then superposition can be applied.

R1.6.5

Order: 2

Linearity: Linear

Superposition: Yes

$$ y'' +w_o^2y=cos wt   \  \ \  \ \  \ \  \ \  \ \  \ \  \  w_o = w \!$$

$$ y_h'' +w_o^2y_h=0\!$$
 Homogeneous solution (Eq. 1 ) $$ y_p'' +w_o^2y_p=cos wt \!$$
  Particular  solution (Eq. 2) By solving we obtain the following: $$\bar{y}=y_h+y_p\!$$

$$\bar{y}=y_h+y_p''\!$$ Adding Eq.1 & Eq.2 : $$y_h+w_o^2y_h+y_p+ w_o^2y_p = cos wt\!$$

$$( y_h+y_p)''+w_o^2(y_h+y_p)= cos wt\!$$

$$\bar{y}''+w_o^2\bar{y}= cos wt\!$$

Therefore, superposition can be applied.

R1.6.6

Order: 2

Linearity : Linear

Superposition: Yes

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

$$ LI_h'' + RI_h' + \frac{1}{c}I_h =0 \!$$
 Homogeneous solution (Eq. 1 )

$$ LI_p'' + RI_p' + \frac{1}{c}I_p =E' \!$$
 Particular  solution (Eq. 2 )

By solving we obtain the following:

$$\bar{I}= (I_h+I_p)\!$$

$$\bar{I}'= (I_h+I_p) '\!$$ $$\bar{I}= (I_h+I_p) \!$$ Adding Eq.1 & Eq. 2:

$$L(I_h+I_p) ''+R(I_h+I_p) ' + \frac{1}{c}(I_h+I_p) = E'\!$$

$$L\bar{I}''+R\bar{I}'+\frac{1}{c}\bar{I}= E'\!$$

Therefore, superposition can be applied.

R1.6.7 Order: 4

Linearity :Linear

Superposition : Yes

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

$$ y_h= EIy_h^{iv}=0\!$$
 Homogeneous solution (Eq. 1 ) $$ y_p= EIy_p^{iv}=f(x)\!$$
 Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{y}=y_h+y_p\!$$

Summing Eq.1 & Eq. 2 :

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

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

$$ EI\bar{y}^{iv} = f(x)\!$$ Therefore, superposition can be applied.

R1.6.8

Order: 2

Linearity: Non-linear

Superposition: No

$$ L{\theta }'' + gsin\theta = 0 \!$$ $$ L{\theta_h }'' + gsin\theta_h = 0 \!$$
 Homogeneous solution (Eq. 1 ) $$ L{\theta_p }'' + gsin\theta_p = f(x) \!$$
 Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{\theta}= {\theta}_h + {\theta}_p \!$$

Summing Eq. 1 & Eq. 2 :

$$L{\theta_h + \theta_p}''+gsin{\theta_h+\theta_p}=0\!$$ $$\bar{\theta}\neq\ {\theta}_h + {\theta}_p \!$$

Therefore, superposition cannot be applied.

Report 2: 4a)

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

a=2\pi\!

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

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

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

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

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

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 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.

title: User:Egm4313.s12.team15.cynthiah


text:
User:Egm4313.s12.team4.friedman

test

title: User:Egm4313.s12.team4.friedman


text:
Sport coaching pedagogy

Welcome
Welcome to the Home Page of the Sport Coaching Pedagogy unit.

This unit will take place in Semester 1, 2013, at the University of Canberra.

Schedule
The scheduled times for this class are:

Lectures: Friday 12.30 – 14.30. Room 02B07. Tutorials: Friday 9.30 – 11.30, 14.30 – 16.30.

Students will choose one tutorial session from the two available. The tutorials will be practical sessions and students are encouraged to wear appropriate athletic clothing and footwear for these sessions.

(Note that two Computer Laboratory Tutorials are included in the timetable (weeks 2 and 7). Arrangements for these tutorials will be confirmed at our week 1 meeting, Friday 8 February.)

Topics to be covered during the unit include:
 * Week 1 (8 February) Welcome to Sport Coaching Pedagogy.
 * Week 2 (15 February) Personal Development Pathways.
 * Week 3 (22 February) Planning Programs.
 * Week 4 (1 March) Observation and Augmented Information.
 * Week 5 (8 March) Sharing and Presenting Information 1.
 * Week 6 (15 March) Mentoring 1.
 * Week 7 (22 March) Using Digital Media: Produsing Resources
 * Week 8 (29 March) Class Free Week.
 * Week 9 (5 April) The Expert Pedagogue.
 * Week 10 (12 April) Sharing and Presenting Information 2.
 * Week 11 (19 April) Managing Data.
 * Week 12 (26 April) Mentoring 2.
 * Week 13 (3 May) Long Term Athlete Development.
 * Week 14 (10 May) Reflecting on Coaching as a Vocation.

Learning Outcomes
On completion of this unit a student will be able to:

 * 1) Plan and deliver sport and physical activity initiatives in the community.
 * 2) Review, evaluate and communicate evidence-based resources relevant to sport coaching pedagogy and teaching.
 * 3) Assess their sport coaching and teaching performance through a reflective framework.
 * 4) Use an e-portfolio to share their work with others.

Generic skills
This unit is planned to support the diversity of attributes and experiences that students bring to it. The nature of the subject matter lends itself unequivocally to the development of critical thinking and reflective practice. Students will be encouraged to use foresight, initiative and leadership, and be open to alternative perspectives. The unit aims to make a significant contribution to the University’s five generic skills and attributes for graduates:

 * 1) Communication. The ability to present knowledge, ideas and opinions effectively and communicate within and across professional and cultural boundaries.
 * 2) Analysis and inquiry. The ability to gather information, and to analyse and evaluate information and situations in a systematic, creative and insightful way.
 * 3) Problem solving. The ability to apply problem-solving processes in novel situations; to identify and analyse problems then formulate and implement solutions.
 * 4) Working independently and with others. The ability to plan their own work, be self-directed, and use interpersonal skills and attitudes to work collaboratively.
 * 5) Professionalism and social responsibility. The capacity and intention to use professional knowledge and skills ethically and responsibly, for the benefit of others and the environment.

Assessment
Assessment item 1: Presentation
 * Due date: 12 April (Week 10)
 * Weighting: 30%
 * Addresses learning outcome: 2
 * Related generic skills: 1, 2, 4, 5

The Presentation will be submitted in tutorial classes on Friday, 12 April (Week 10). The Presentation should be designed to review the literature surrounding an area of personal interest in coaching or teaching. The choice of topic for the presentation should be confirmed with the unit convenor by Week 7 (22 March). The format for the presentation should be compatible with uploading the presentation to SlideShare. The presentation should take a maximum of 15 minutes. The mark for this assessment item will represent 30% of the final grade.

Assessment item 2: Coaching/Teaching Practical
 * Due date: 27 April (Week 12)
 * Weighting: 40%
 * Addresses learning outcomes: 1, 3
 * Related generic skills: 1, 3, 4, 5

The Coaching/Teaching Practical is due on Friday, 26 April (Week 12) by 5pm. There are three parts to this assessment item. Students are required to submit a video (complete with audio) showing them coaching or teaching a technical or tactical aspect of a sport or physical activity. The video should be a maximum of 5 minutes in length and should be submitted as a file on a USB memory stick. The USB should also contain a file of the session or teaching plan and a self-assessment of the coaching or teaching effectiveness and strategies for improvement (no more than 500 words). No other files should be on the USB memory stick. Templates for the session plan and self-assessment will be provided to students in the tutorials. The mark for this assessment item will represent 40% of the final grade.

Assessment item 3: E-Portfolio The E-Portfolio is due on Friday, 10 May (Week 14) by 5pm. The aim of this assessment item is to share a record of your learning journey through the unit. The style and format are open but must reflect your participation and engagement in the unit's learning activities. Guidelines for the development of the e-portfolio will be shared in lectures and tutorials at the start of the unit. The e-portfolio should demonstrate evidence of regular and continuous reporting of and reflection on your involvement in the unit. 33.3% of the marks available in this assessment item will be allocated to evidence of consistent and continuous engagement in learning activities throughout the unit. 66.7% of the marks available in this assessment item will be allocated to evidence of in depth, thoughtful, questioning, insightful reflection on the unit's learning activities. The mark for this assessment item will represent 30% of the final grade.
 * Due date: 10 May (Week 14)
 * Weighting: 30%
 * Addresses learning outcomes: 3, 4
 * Related generic skills: 1, 2, 3, 4, 5

Unit Resources
This course will use Internet based material. Reading suggestions and guides will be provided for each session.

This is an Introduction to the Unit (using SlideShare)

You might like to read about Jeffrey Wright as another introduction to the unit.

Week 1 (8 February) Welcome to Sport Coaching Pedagogy.

Sport Coaching Pedagogy: Introductions (SlideCast)

Week 2 (15 February) Personal Development Pathways.

A Guide to Using Training Needs Analysis and Personal Development Plans (2011). Leeds: National Coaching Foundation

Week 3 (22 February) Planning Programs.

Week 4 (1 March) Observation and Augmented Information.

Week 5 (8 March) Sharing and Presenting Information 1.

Week 6 (15 March) Mentoring 1.

Week 7 (22 March) Using Digital Media: Produsing Resources

David Worlick (2013), It's the Information (Revisited)

Digital Research Tools (Bamboo Project wiki)

Week 8 (29 March) Class Free Week.

Week 9 (5 April) The Expert Pedagogue.

Week 10 (12 April) Sharing and Presenting Information 2.

Week 11 (19 April) Managing Data.

Week 12 (26 April) Mentoring 2.

Week 13 (3 May) Long Term Athlete Development.

Identifying Excellent Coaching Practice along the Sporting Pathway (2012). London: sportscoachUK.

Long-term athlete development: Foundations and challenges for coaches, scientists & policy-makers

Week 14 (10 May) Reflecting on Coaching as a Vocation.

Important references are:

For Coaches

 * Australian Sports Commission Presenter, Assessor and Mentor resources:
 * Presenter Training
 * Assessor Training
 * Mentor Training

For Teachers

 * Personal Development, Health and Physical Education K-6 Syllabus
 * Personal Development, Health and Physical Education Years 7–10 Syllabus

Other online resources include:

 * Moodle site for unit 8914
 * Wikiversity page for Sport Coaching Pedagogy (this page) and an e-portfolio page [here].
 * Postillion SlideShare
 * The Facebook page for Sport Coaching Pedagogy

Tools

A to Z of Social Media

You might think about using:

 * Diigo to help compile resources.

Background Reading

 * Cassidy, T., Jones, R., and Potrac, P. (2009). Understanding Sports Coaching: The Social, Cultural and Pedagogical Foundations of Coaching Practice. Abingdon: Routledge.
 * Jones, R.L. (ed) (2006). The Sports Coach as Educator. Abingdon: Routledge.
 * Jones, R., Armour, K., and Potrac, P. (2004). Sports Coaching Cultures. London: Routledge.
 * Jones, R., Potrac, P., Cushion, C., and Ronglan, L.T. (eds) (2011). The Sociology of Sports Coaching. Abingdon: Routledge.
 * Kramer, M. and Montgomery, R. (2012). The Best-Laid Plans of a High School CEO. Indianapolis: Dog Ear Publishers.
 * Lyle, J. (2002). Sports Coaching Concepts. London: Routledge.
 * Phillips, M. (2005). From Sidelines to Centre Field. A History of Sports Coaching in Australia. Sydney: University of New South Wales Press.

title: Sport coaching pedagogy


text:
User:Marshallsumter/Radiation astronomy1/First yellow source in Aquila

The first yellow source in Aquila is unknown.

This is a lesson in map reading, coordinate matching, and searching. It is also a project in the history of yellow astronomy looking for the first astronomical yellow source discovered in the constellation of Aquila.

Nearly all the background you need to participate and learn by doing you've probably already been introduced to at a secondary level.

Some of the material and information is at the college or university level, and as you progress in finding yellow sources, you'll run into concepts and experimental tests that are an actual search.

First step
To succeed in finding a yellow source in Aquila is the first step.

Over the history of yellow astronomy a number of sources have been found, many as point sources in the night sky. These points are located on the celestial sphere using coordinate systems. Familiarity with these coordinate systems is not a prerequisite. Here the challenge is geometrical, astrophysical, and historical. The coordinates are usually supplied by the yellow source observers.

Next, you'll need to determine the time stamp of its discovery and compare it with any that have already been discovered.

Control groups
Every experimental study has a control group. Here, a control group may be a yellow star already determined to be the first discovered in a specific part of the sky.

Astronomical sources
Def. a natural source usually of radiation in the sky especially at night is called an astronomical source.

A source of astronomical information on older detections of yellow sources may be included in the Science section of the lecture yellow astronomy.

Backgrounds
To introduce yourself to some aspects of the challenge may I suggest reading the highlighted links mentioned above, and if you're curious, those listed under the section "See also" below.

Yellow rays are a form of radiation that is currently part of electromagnetic radiation intersecting the Earth. More information about radiation is in radiation astronomy, and specifically in yellow astronomy.

Traveling yellow sources
Many yellow sources do not remain in a constellation for lengthy periods. Some of these are the Sun and sources apparently in orbit around the Sun. The Sun travels through the 13 constellations along the ecliptic (the plane of the Earth's orbit around the Sun): the 12 of the Zodiac and the constellation Ophiuchus. These are described in source astronomy. Is Aquila one of the constellations of the Zodiac?

Constellations
The Wikipedia article about the constellation Aquila contains a high school level description. The figure at right shows the sky map of Aquila. Around the edges of the map are coordinates related to longitude and latitude, but with the Earth rotating on its axis every 24 hours the celestial coordinates must remain fixed. How has this been accomplished?

Also, in the Wikipedia article is a list of stars in Aquila.

From the Wikipedia article on the Zodiac, is Aquila a zodiacal constellation?

Aquila is a constellation catalogued by Ptolemy before 1800 b2k.

"In Greek mythology, Aquila is identified as the eagle that carried Zeus’ thunderbolts and was once dispatched by the god to carry Ganymede, the young Trojan boy Zeus desired, to Olympus to be the cup bearer of the gods."

"In anothery story, the eagle is found guarding the arrow of Eros (represented by the constellation Sagitta), which hit Zeus and made him love-struck."

"In yet another myth, Aquila represents Aphrodite disguised as an eagle, pretending to pursue Zeus in the form of a swan, so that Zeus’ love interest, the goddess Nemesis, would give him shelter. In the story, Zeus later placed the images of the eagle and the swan among the stars to commemorate the event."

Testing a source
There are many web sites that may have a yellow source listed for the constellation Aquila.

Wikipedia sources
A. Constellation article

Under "Notable features" in the Wikipedia article on the constellation Aquila is the list of stars in Aquila. Click on this link. In the table of this Wikipedia article is α Aql. To the right are coordinates:
 * 1) [Right ascension] (RA): 19h 50m 46.99855s and
 * 2) [Declination] (Dec): +08° 52' 05.9563".

Find these coordinates on the Aquila map at the right. Is alpha Aquilae really inside the boundaries of the constellation?

To evaluate the star as a yellow source, skip ahead to section "Yellow source".

B. Wikipedia search

Another way to look for yellow sources in the constellation is to perform a search on Wikipedia. Try "Aquila yellow" without the outside quotes. This yields about 361 returns which include Xi Aquilae b, 15 Aquilae, Nu Aquilae, HD 192699, Chi Aquilae, COROT-3, and Pi Aquilae.

To evaluate each of these as a yellow source, skip ahead to section "Yellow source".

SIMBAD sources
Another way to find possible yellow sources in Aquila is to use search queries on SIMBAD.

Click on the SIMBAD link under "External links" below, then click on "Criteria query".

In the tan box, type in "region(19 50 46.99855 +08 52 05.9563, 10m)", without the quotes. This tells the SIMBAD computer you are interested in a circular region of the celestial sphere centered on the coordinates for alpha Aquilae, with a radius of 10 arcminutes (m).

Notice on the page over at the right from the tan colored box: "Return". The default is "object count". Click on "submit query". In a few moments a result something like "Number of objects:	   	 10" should appear. Click "Back" to see the tan box again.

To see if you have found at least one object, change "Return" to "display" by clicking on the circle to its left, then "submit query".

SIMBAD should display a list of objects. Read through the resource yellow astronomy for clues that may indicate whether a particular spectral type (Sp type) is a yellow source. If none of the objects listed seem to be described as a yellow source, try going "Back" and increasing the arcminutes from "10m" to "20m", and repeat until a yellow source is found.

Once you believe you have discovered a yellow source, proceed to the section "Yellow source".

Yellow sources
There are several ways to evaluate a yellow source for the constellation Aquila.

Wikipedia source
Click on the link to the Wikipedia article. After you've enjoyed reading about the source, use the 'find' command of your browser to see if this Wikipedia page mentions anything about "yellow", or "yellow rays". Does the article mention whether or not the source is a yellow source?

What is the current time stamp for the Wikipedia article on the source? [Hint]: look for something like "This page was last modified on 12 January 2012 at 06:47." very near the bottom of the page. For now this is an adequate time stamp.

From reading the Wikipedia article on the source, if you believe the text demonstrates that the source is not a yellow source in Aquila edit the "Non-yellow source in Aquila" section near the bottom of the page with an entry similar to "# Alpha Aquilae 12 January 2012 at 06:47 Wikipedia article "Alpha Aquilae".", without the outer quotes, and finish the entry with four "~"s without the quotes after the period. The date included with your designation or username is a time stamp for the entry. The last portion of the entry is the source of your information.

On the other hand, if there are one or more sentences in the article that you believe demonstrates that the source is a yellow source in Aquila edit the section below "Yellow source in Aquila" with a similar entry.

Go to the section entitle, "Challenging an entry".

SIMBAD source
To check any source (even one from Wikipedia) on SIMBAD, click of the "External link" to the "SIMBAD Astronomical Database".

At the lower right side of the SIMBAD Astronomical Database page is a "Basic search" box. There are several ways to try your target:
 * 1) source name: without the quotes or
 * 2) source coordinates: without the quotes, for example, "19 50 46.99855 +08 52 05.9563".

If you are looking at a SIMBAD generated table which lists possible targets, click on one.

On its SIMBAD page move over to the right side until you see an Aladin visual photograph of the object. Is it a yellow source?

Even if the source does not look yellow, look down the left hand side of the page for "Spectral type:". From your reading of yellow astronomy, do you believe the source is a yellow source, or not? Noting that SIMBAD does, or does not consider the source to be a yellow source is important, so skip down to the "SIMBAD time stamp" section.

If you have already found a yellow source (or a table of them) using SIMBAD, click on the blue link identifier for the first.

SIMBAD time stamp
Peruse the SIMBAD page for a time stamp or date of last revision. [Hint: it may look something like "2012.01.09CET20:10:02" and be in the upper right.]

If the entry at SIMBAD convinces you that the source is not a yellow source, edit the "Non-yellow source in Aquila" section near the bottom of this page and type in an entry similar to "# Source Name 2012.01.09CET20:10:02 SIMBAD article "SIMBAD source name".", without the first set of quotes, followed by four ~s.

If your SIMBAD analysis convinces you that you have found a yellow source in Aquila (did you check the coordinates vs. the map of Aquila?), make an entry something like the ones in the section "Yellow source in Aquila".

Challenging an entry
Any entry in either the section "Yellow source in Aquila" or "Non-yellow source in Aquila" can be challenged. The time stamp can be challenged to see if there is an earlier one. The source can be challenged by an earlier source.

Wikipedia challenge
Is Wikipedia a 'primary source', or does the Wikipedia article cite a source?

Even though Wikipedia has an article on the source, is it a good place to stop in testing whether the source has been detected as an astronomical yellow source?

If the Wikipedia article cites a primary source, skip down to the section on "Primary sources".

SIMBAD challenge
Is SIMBAD a 'primary source'?

SIMBAD is an astronomical database provided by the Centre de Données astronomiques de Strasbourg. It is an authoritative source, but they do occasionally make a mistake.

If you find a yellow source within the constellation on SIMBAD, the next step is to find the earliest time stamp of discovery.

Wikipedia test source
For a Wikipedia article that cites a primary source, scroll down to the reference and open the reference. Read through the article looking for where the source mentioned in the Wikipedia article occurs. Some primary source authors may use source designations that are not mentioned in the Wikipedia article. To look for other designations, click on the link to SIMBAD in the "External links" on this page, enter the source name from the Wikipedia article, and see if other names are mentioned in the article.

When none of the names are mentioned, click on the link for "Google Advanced Search" in the list of "External links", enter the source name or designation(s) such as "Gliese 866", with "yellow" to see if the source has a reference indicating it is a yellow source. And, look for the earliest one. Compose an entry using the primary source.

SIMBAD test source
Further down the SIMBAD page is a list of "Identifiers". Click on the blue bold portion.

On the page that appears, there should be a primary source listed after Ref:. Click on the blue link with the oldest year. This yields an earlier time stamp and entry citation like the current one in the section "Yellow source in Aquila". If you find another source or an earlier time stamp, compose a similar entry and edit the section. Additional information to add into the reference can be found by clicking on "ADS services" from the SIMBAD page.

Changing an entry
From your analysis of the source so far, is it a yellow source?

If you have found an earlier time stamp for the source than the one listed in the section below "Non-yellow source in Aquila" and the answer to the above question is "no", you can edit the section with your result. Or, you can leave the entries as is and try another star or object.

If you have found an earlier time stamp for the source than the one listed in the section below "Yellow source in Aquila", edit the section with your result. Or, if you found another yellow source with a comparable or earlier time stamp, edit the section with your result.

Yellow sources in Aquila

 * 1) Eta Aquilae 20 August 2011 at 06:54 Wikipedia article "Eta Aquilae". "It is a yellow-white supergiant, and is about 3000 times more luminous than the Sun, with a diameter about 60 times that of the Sun." "It is a Cepheid variable star, with its apparent visual magnitude varying between 3.5 and 4.4 with a period of 7.176641 days. Eta Aquilae is one of the easiest Cepheids to distinguish by the naked eye." Marshallsumter 20:12, 25 January 2012 (UTC)
 * 2) Beta Aquilae 21 March 2012 at 16:35 Wikipedia article "Beta Aquilae", does not include the word "yellow" in the text but by the star classification above and spectral type G8 IVvar in the Starbox it is a yellow star. Marshallsumter (talk) 00:16, 27 March 2012 (UTC)
 * 3) Alshain (Beta Aquilae) SIMBAD has spectral type as G9.5IV, followed by a reference . The earliest time stamp is March 28, 2006. Since 1943, the spectrum of this star has served as one of the stable anchor points by which other stars are classified. Marshallsumter (talk) 00:36, 27 March 2012 (UTC). But, "The Atlas of Stellar Spectra was published fifty years ago by W.W. Morgan, P.C. Keenan, and E. Kellman(1943). Since then, there have been supplementary lists of standard stars and atlases published by Morgan and/or Keenan in 1953, 1973, 1976, and 1978. In these later publications, some of the types for the standard stars were modified in the light of better data."  In 1953, β Aql is listed as the standard star for G8 IV. A G8 IV is a yellow (or orange-yellow) star. --Marshallsumter (talk) 23:41, 13 August 2012 (UTC)

Non-yellow sources in Aquila

 * 1) Alpha Aquilae 12 January 2012 at 06:47 Wikipedia article "Alpha Aquilae". Marshallsumter 20:12, 25 January 2012 (UTC).
 * 2) Beta Aquilae "The variable and comparison stars are, in the Harvard system, ... . H.R. 7602, β Aquilae, magnitude 3.90, spectrum K. The Mt. Wilson spectral types are ... β Aquilae, G7." "[T]oward the end of the series [of observations] β Aquilae was fainter than at the beginning." Class K are orangish stars that are slightly cooler than our Sun. Apparently, in 1922, β Aquilae was an orangish star. --Marshallsumter (talk) 23:24, 13 August 2012 (UTC). In 1908, β Aql is listed as a spectral class K II with a suspected period of variability of two months. --Marshallsumter (talk) 03:44, 14 August 2012 (UTC). "It is well known that [β Aquilae] has sensibly diminished in brightness within memory of persons living; but no change has taken place in this respect for many years past, nor is there any reason to suppose that it was ever brighter than γ [Aquilae]. ... Flamsteed rated β Aquilae as 3 1/2; whilst Hevelius, remarkably enough, estimated it as only (what it is now) the fourth, although later observers reckoned it, like Flamsteed, of the 3 1/2 magnitude." --Marshallsumter (talk) 04:11, 14 August 2012 (UTC).

Oldest records

 * 1) Beta Aquilae: Since 1943, the spectrum of this star has served as one of the stable anchor points by which other stars are classified.

SIMBAD annotations
For any particular source, the SIMBAD record may indicate that it is not a yellow source yet above you may have found at least two refereed journal articles to indicate that it is. Use the second SIMBAD External links to directly display the SIMBAD database in France.

Enter the name source you have found into the search box. Scroll down to the Annotations :. Look for the link "add an annotation to this object". With browser open to the literature citations available, click on this link. You may need to register as a user. It's free. Post your annotation containing the literature references.

Hypotheses

 * 1) The first yellow source in Aquila may have been discovered around 42,000 b2k.


text:
User:Egm4313.s12.team4/Report 1

= Problem 1: Equation of Motion of a Spring-Dashpot System =

Problem Statement
Derive the equation of motion of a spring-dashpot system in parallel with a mass and applied force, $$\displaystyle f(t)$$

Solution
Using a force balance, the following equation can be derived:

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

$$\displaystyle f(t) = my'' + f_I$$ Where $$\displaystyle f_I = f_c = f_k$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.1)
 * }

Adding the displacements due to the spring $$\displaystyle y_k$$ and the dashpot $$\displaystyle y_c$$ we can write the following equation:

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

$$\displaystyle y=y_k+y_c$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.2)
 * }

Taking two time derivatives of (1.1.2) gives:

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

$$\displaystyle y = y_k + y''_c$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.3)
 * }

Since $$\displaystyle f_k=f_c$$, we can write:

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

$$\displaystyle ky_k = cy'_c$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.4)
 * }

Now, solving for $$\displaystyle y'_c $$ in (1.1.4), we find:

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

$$\displaystyle y'_c = \frac{k}{c}y_k$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.5)
 * }

We can substitute our new value for $$\displaystyle y'_c$$ into (1.1.3) to obtain:

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

$$\displaystyle y = y_k + (y'_c)'$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.6)
 * }

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

$$\displaystyle y = y_k + (\frac{k}{c}y_k)' $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.7)
 * }

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

$$\displaystyle y = y_k + \frac{k}{c}y'_k$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.8)
 * }

We can now substitute for $$\displaystyle y''$$ and $$\displaystyle f(I)$$ in equation (1.1.1) to obtain an equation in terms of $$\displaystyle y_k$$:

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

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

= Problem 2: Equation of Motion of a Spring-Mass-Dashpot System =

Problem Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in K2011 pg. 85 with an applied force r(t) on the ball.

Solution
In order to derive an equation of motion, all of the forces must be analyzed. In the spring-mass-dashpot model there are forces from the spring, inertia, and a damping force caused by the dashpot.

The force of a spring oriented in the y-direction is given by Hooke's Law:
 * {| style="width:100%" border="0"

$$\displaystyle F=-ky$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.1)
 * }

The damping force of a dashpot is proportional to the mass' velocity. Relating velocity to position and time:
 * {| style="width:100%" border="0"

$$\displaystyle V=\frac{dy}{dt}$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.2)
 * }

Proportionality is related by a dashpot coefficient $$\displaystyle c$$
 * {| style="width:100%" border="0"

$$\displaystyle F_{d}=c\frac{dy}{dt}$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.3)
 * }

According to Newton's Second Law of Motion
 * {| style="width:100%" border="0"

$$\displaystyle \Sigma F_{y}=ma_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.4)
 * }

Relating acceleration to position and time
 * {| style="width:100%" border="0"

$$\displaystyle a_{y}=\frac{\mathrm{d^2}y}{\mathrm{d} t^2}$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.5)
 * }

Summing spring and damping forces and substituting new acceleration term
 * {| style="width:100%" border="0"

$$\displaystyle m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}=-c\frac{dy}{dt}-ky$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.6)
 * }

Both forces are negative because they resist the motion of the spring.

Moving all terms to one side
 * {| style="width:100%" border="0"

$$\displaystyle m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}+c\frac{dy}{dt}+ky=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.7)
 * }

Setting the equation equal to $$\displaystyle r(t)$$
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |
 *
(1.2.8)
 * }

=Problem 3: FBDs and Equation of Motion of a Spring-Dashpot System =

Problem Statement
For the spring-dashpot-mass system on p.1-4, draw the free body diagrams and derive the equation of motion (2) p.1-4.

Solution
Free body diagram:

Force balance equation of spring-dashpot-mass system:
 * {| style="width:100%" border="0"

$$\displaystyle \sum F_x=f(t)-f_I=ma$$ where $$\displaystyle a={y}''\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.1)
 * }

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

$$\displaystyle m{y}''+f_I=f(t)\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.2)
 * }

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

$$\displaystyle f_I=f_k=f_c\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.3)
 * }

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

$$\displaystyle f_k=ky_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.4)
 * }

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

$$\displaystyle f_c={y_c}''\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.5)
 * }

Where

$$\displaystyle c = $$ dashpot constant

$$\displaystyle k = $$ spring constant

Total displacement is displacement from the spring plus displacement from the dashpot:
 * {| style="width:100%" border="0"

$$\displaystyle y=y_k+y_c\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.6)
 * }

Taking the double integral of $$\displaystyle y$$ you get
 * {| style="width:100%" border="0"

$$\displaystyle {y}={y_k}+{y_c}''\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.7)
 * }

Get $$\displaystyle y_c$$ in terms of $$\displaystyle y_k$$:
 * {| style="width:100%" border="0"

$$\displaystyle f_k=f_c\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.8)
 * }

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

$$\displaystyle ky_k=c{y_c}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.9)
 * }

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

$$\displaystyle {y_c}''=\frac{k}{c}y_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.10)
 * }

Plug back into $$\displaystyle {y}''\,\!$$:
 * {| style="width:100%" border="0"

$$\displaystyle {y}={y_k}+{({y_c}')}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.11)
 * }

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

$$\displaystyle {({y_c}')}'=\frac{k}{c}{y_k}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.12)
 * }

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

$$\displaystyle \rightarrow {y}={y_k}+\frac{k}{c}{y_k}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.13)
 * }

Now get $$\displaystyle f_I$$ in terms of $$\displaystyle y_k$$:
 * {| style="width:100%" border="0"

$$\displaystyle f_I=f_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.14)
 * }

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

$$\displaystyle f_k=ky_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.15)
 * }

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

$$\displaystyle \rightarrow f_I=ky_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.16)
 * }

When plugging in all variables into original equation you get:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |
 *
(1.3.17)
 * }

= Problem 4: Deriving Capacitance Equations =

Problem Statement
Derive equations (1.4.2) and (1.4.3) from (1.4.1).

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

$$\displaystyle V=LC \frac{d^2 v_C}{dt^2} + RC \frac{d v_C}{dt} + v_C$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.1)
 * }

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

$$\displaystyle L I'' + R I' + \frac{1}{C} I = V'$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.2)
 * }

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

$$\displaystyle L Q'' + R Q' + \frac{1}{C} Q = V$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.3)
 * }

Background Information

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

$$\displaystyle C = \frac{Q}{V}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.4)
 * }

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

$$\displaystyle I = C \frac{d v_C}{d t}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.5)
 * }

Part A
Taking the derivative of (1.4.1)
 * {| style="width:100%" border="0"

$$\displaystyle V' = LC \frac{d^3 v_C}{d t^3} + RC \frac{d^2 v_C}{d t} + \frac{d v_C}{d t}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.6)
 * }

Rewrite equation (1.4.5)
 * {| style="width:100%" border="0"

$$\displaystyle I^{(n-1)} = C \frac{d^n v_C}{d t^n}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.7)
 * }

Substituting (1.4.7) into (1.4.6)
 * {| style="width:100%" border="0"

$$\displaystyle V' = L I^{(2)} + R I^{(1)} + \frac{1}{C} I^{(0)}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.8)
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.2)
 * }

Part B
Rewrite equation (1.4.4)
 * {| style="width:100%" border="0"

$$\displaystyle Q^{(n)} = C V^{(n)}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.9)
 * }

Substituting (1.4.9) into (1.4.1)
 * {| style="width:100%" border="0"

$$\displaystyle V = L Q^{(2)} + R Q^{(1)} + \frac{1}{C} Q^{(0)}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.10)
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 *
(1.4.3)
 * }

= Problem 5a: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problem 4. Find a general solution. Check your answer by substitution.

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

$$\displaystyle y'' + 4y' + (\pi^2 +4)y = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.1)
 * }

Solution
The equation above has a characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2 + 4\lambda + (\pi^2 +4) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.2)
 * }

which is of the form:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2 + a\lambda + b = 0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.3)
 * }

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

$$\displaystyle a^2 - 4b < 0$$ making it negative.
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.4)
 * }

The roots of the characteristic equation are:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -\frac{1}{2}a\pm wi $$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.5)
 * }

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

$$\displaystyle w^2 = b - \frac{1}{4}a^2$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.6)
 * }

Therefore real solutions can be written as:
 * {| style="width:100%" border="0"

$$\displaystyle y_1 = e^{-\frac{ax}{2}}\cos wx$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.7)
 * }

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

$$\displaystyle y_2 = e^{-\frac{ax}{2}}\sin wx$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.8)
 * }

This will yield a general solution of:
 * {| style="width:100%" border="0"

$$\displaystyle y = e^{-\frac{ax}{2}}(A \cos wx + B \sin wx)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.9)
 * }

where A and B are arbitrary.

Solving for $$\displaystyle w$$, the roots become:
 * {| style="width:100%" border="0"

$$\displaystyle \lambda = -2 \pm \pi i $$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.10)
 * }

The final general solution for this problem is:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.11)
 * }

Check by substitution:

In order to check the solution we must find $$\displaystyle y' $$ and $$\displaystyle y'' $$ :

Now that we have
 * {| style="width:100%" border="0"

$$\displaystyle y= e^{-2x}(A \cos \pi x + B \sin \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.12)
 * }

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

$$\displaystyle y'= -2e^{-2x}(A \cos \pi x + B \sin \pi x) + e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.13)
 * }

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

$$\displaystyle y''= 4e^{-2x}(A \cos \pi x + B \sin \pi x) - 2e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x)$$ $$\color{White}y''= \color{Black} - 2e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x) + e^{-2x}(-A \pi ^2 \cos \pi x - B \pi ^2 \sin \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.14)
 * }

we can substitute these equations into the original equation,
 * {| style="width:100%" border="0"

$$\displaystyle y'' + 4y' + (\pi^2 +4)y = 0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.15)
 * }

After substitution we see that
 * {| style="width:100%" border="0"

$$\displaystyle 0 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5a.16)
 * }

Therefore the solution is correct.

= Problem 5b: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problem 5. Find a general solution. Check your answer by substitution.

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

$$\displaystyle {y}''+2\pi {y}'+\pi^{2}y=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.1)
 * }

Solution
Using the following:

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

$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.2)
 * }

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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.3)
 * }

The given equation can be rewritten as follows, using (1.5b.2) and (1.5b.3):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}+2\pi\frac{dy}{dx}+\pi^{2}y=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.4)
 * }

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

$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.5)
 * }

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

$$\displaystyle (D^{2}+2\pi D+\pi^{2})y=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.6)
 * }

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

$$\displaystyle (D^{2}+2\pi D+\pi^{2})=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.7)
 * }

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

$$\displaystyle (D+\pi)^{2}=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.8)
 * }

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

$$\displaystyle D=-\pi,-\pi$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.9)
 * }

The final general solution for this problem is:

Reach the general solution :

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

 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.10)
 * }

Check by substitution:

From (1.5b.10), we find,
 * {| style="width:100%" border="0"

$$\displaystyle {y}'=-\pi(C_{1}+C_{2}x)e^{-\pi x}+C_{2}e^{-\pi x}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.11)
 * }

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

$$\displaystyle {y}''=-\pi(-\pi C_{1}+-\pi C_{2}x)e^{-\pi x}-\pi C_{2}e^{-\pi x}-\pi C_{2}e^{-\pi x}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.12)
 * }

Substituting (1.5b.11) and (1.5b.12) in (1.5b.1), we reach:
 * {| style="width:100%" border="0"

$$\displaystyle (\pi^{2}(C_{1}+C_{2}x)e^{-\pi x}-2\pi C_{2}e^{-\pi x})+2\pi(-\pi(C_{1}+C_{2}x)e^{-\pi x}+C_{2}e^{-\pi x})+\pi^{2}((C_{1}+C_{2}x)e^{-\pi x})=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.13)
 * }

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

$$\displaystyle \pi^{2}C_{1}e^{-\pi x}+\pi^{2}C_{2}xe^{-\pi x} -2\pi C_{2}e^{-\pi x}-2\pi^{2}C_{1}e^{-\pi x} -2\pi^{2} C_{2}xe^{-\pi x}+2\pi C_{2}e^{-\pi x} +\pi^{2}C_{1}e^{-\pi x} +\pi^{2}C_{2}xe^{-\pi x}=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.14)
 * }

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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.5b.15)
 * }

Therefore the solution is correct.

= Problem 6: Classification of ODEs and the Superposition Prcinciple =

Problem Statement
For each ODE in Fig.2 in Kreyszig 2011 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 can be applied.

Order
The order of an equation is determined by the highest derivative. In this report, the first derivative of the $$\displaystyle y$$ variable is denoted as $$\displaystyle y'$$, and the second derivative is denoted as $$\displaystyle y''$$, and so on. This can be determined upon observation of the equation.

Linearity
An ordinary differential equation (ODE) is considered linear if it can be brought to the form :

 * $$y'+p(x)y=q(x)\!$$

Superposition
Superposition can be applied if when a homogeneous and particular solution of an original equation are added, that they are equivalent to the original equation. In this report, variables with a bar over them represent the addition of the homogeneous and particular solution's same variable.

 * For example:

 * $$ y_p+y_h=\overline y \!$$

Given
The following equations were given in the textbook on p. 3 :
 * {| style="width:100%" border="0"

$$\displaystyle y''=g=constant$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6a.1)
 * }

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

$$\displaystyle mv'=mg-bv^2$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6b.1)
 * }

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

$$\displaystyle h'=-k\sqrt{h}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6c.1)
 * }

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

$$\displaystyle my''+ky=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6d.1)
 * }

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

$$\displaystyle y''+\omega_0^2 y=\cos\omega t, \omega_0 = \omega$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6e.1)
 * }

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

$$\displaystyle LI''+RI'+\frac{1}{C} I=E'$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6f.1)
 * }

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

$$\displaystyle EIy^{\omega}=f(x)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6g.1)
 * }

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

$$\displaystyle L\theta''+g\sin\theta=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6h.1)
 * }

1.6a

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

$$\displaystyle y''=g=constant$$ Order: 2nd Linear: yes Superposition: yes The given equation can be algebraically modified as the following:
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6a.1)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y''=g\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6a.1)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}''=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6a.2)
 * }

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

$$\displaystyle y_{p}''=g\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6a.3)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}+y_{p})=\overline{y}''\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6a.4)
 * }

The solution resembles the original equation, therefore superposition is possible

1.6b

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

$$\displaystyle mv'=mg-bv^2$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6b.1)
 * }

Order: 1st Linear: no Superposition: no The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle mv'+bv^2=mg\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6b.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle mv_{h}'+bv_{h}^2=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6b.3)
 * }

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

$$\displaystyle mv_{p}'+bv_{p}^2=mg\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6b.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle m(v_{h}'+v_{p}')+b(v_{h}^2+v_{p}^2)\not= m\overline{v}'+b(\overline{v}^2)\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6b.5)
 * }

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible as also proven in the class notes

1.6c

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

$$\displaystyle h'=-k\sqrt{h}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6c.1)
 * }

Order: 1st Linear: no Superposition: no The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle h'+k\sqrt{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6c.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle h_{h}'+k\sqrt{h_{h}}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6c.3)
 * }

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

$$\displaystyle h_{p}'+k\sqrt{h_{p}}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6c.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (h_{h}'+h_{p}')+k(\sqrt{h_{h}}+\sqrt{h_{p}})\not=\overline{h}'+k\sqrt{\overline {h}}\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6c.5)
 * }

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible.

1.6d

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

$$\displaystyle my''+ky=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6d.1)
 * }

Order: 2nd Linear: yes Superposition: yes The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle y''+\frac{k}{m}y=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6d.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}''+\frac{k}{m}y_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6d.3)
 * }

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

$$\displaystyle y_{p}''+\frac{k}{m}y_{p}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6d.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}+y_{p})+\frac{k}{m}(y_{h}+y_{p})=\overline{y}''+\frac{k}{m}\overline{y}\!$$ The solution resembles the original equation, therefore superposition is possible
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6d.5)
 * }

1.6e

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

$$\displaystyle y''+\omega_0^2 y=\cos\omega t, \omega_0 = \omega$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6e.1)
 * }

Order: 2nd Linear: yes Superposition: yes

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle y''+\omega _{o}^{2}y=\cos (\omega t)\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6e.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}''+\omega _{o}^{2}y_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6e.3)
 * }

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

$$\displaystyle y_{p}''+\omega _{o}^{2}y_{p}=\cos (\omega t)\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6e.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}+y_{p})+\omega _{o}^{2}(y_{h}+y_{p})=\overline{y}''+\omega _{o}^{2}\overline{y}\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6e.5)
 * }

The solution resembles the original equation, therefore superposition is possible

1.6f

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

$$\displaystyle LI''+RI'+\frac{1}{C} I=E'$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6f.1)
 * }

Order: 2nd linear: yes superposition: yes

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle I''+\frac{R}{L}I'+\frac{1}{LC}I=\frac{E'}{L}\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6f.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle I_{h}''+\frac{R}{L}I_{h}'+\frac{1}{LC}I_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6f.3)
 * }

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

$$\displaystyle I_{p}''+\frac{R}{L}I_{p}'+\frac{1}{LC}I_{p}=\frac{E'}{L}\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6f.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (I_{h}+I_{p})+\frac{R}{L}(I_{h}'+I_{p}')+\frac{1}{LC}(I_{h}+I_{p})=\overline{I}''+\frac{R}{L}\overline{I}'+\frac{1}{LC}\overline{I}\!$$ The solution resembles the original equation, therefore superposition is possible
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6f.5)
 * }

1.6g

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

$$\displaystyle EIy=f(x)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6g.1)
 * }

Order: 4th Linear: yes Superposition: yes

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle y-\frac{1}{EI}y=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6g.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}-\frac{1}{EI}y_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6g.3)
 * }

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

$$\displaystyle y_{p}-\frac{1}{EI}y_{p}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6g.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}'+y_{p}')-\frac{1}{EI}(y_{h}+y_{p})=\overline{y}-\frac{1}{EI}\overline{y}\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6g.5)
 * }

The solution resembles the original equation, therefore superposition is possible

1.6h

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

$$\displaystyle L\theta''+g\sin\theta=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6h.1)
 * }

Order: 2nd Linear: no Superposition:no

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle \theta''+\frac{g}{L}\sin\theta=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6h.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \theta_{h}''+\frac{g}{L}\sin\theta_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6h.3)
 * }

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

$$\displaystyle \theta_{p}''+\frac{g}{L}\sin\theta_{p}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6h.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (\theta_{h}+\theta_{p})+\frac{g}{L}(\sin\theta_{h}+\sin\theta_{p})\not=\overline{\theta}''+\frac{g}{L}\sin\overline{\theta}\!$$
 * style="width:95%" |
 * style="width:95%" |
 *
(1.6h.5)
 * }

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible.

=Notes and References=

title: User:Egm4313.s12.team4/Report 1


text: File:Augsburg Canoe Slalom 01.jpg

title: File:Augsburg Canoe Slalom 01.jpg


text:
User:Egm4313.s12.team8.lucas/R1.4

=R1.4=

Problem Statement
Derive (i) Equation 3 ($$\displaystyle LI+RI'+\frac{1}{C}I=V'$$) and (ii) Equation 4 ($$\displaystyle LQ+RQ'+\frac{1}{C}Q=V$$) using Equation 2 ($$\displaystyle V=LC\frac{\mathrm{d^2v_{c}} }{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_c$$) on page 2-2 in the lecture notes.

Part i
Equation 2:

Equation 1:

Equation 1 can be rewritten as Eqn 1-b:

Differentiating Eqn 1-b yields:

Differentiating Eqn 1-c gives:

Then we differentiate Equation 2 to get:

Then we subsitute Eqn. 1-b, Eqn. 1-c, and Eqn. 1-d into Equation 2-b to get:

By cancelling like terms we derive Equation 3:

Which can be rewritten as:

Part ii
To derive Equation 4, we again start with Equation 2:

Then we will use another version of Equation 1 on p. 2-2 in the lecture notes.

Then rearranging the variables, we get:

Differentiating this yields:

And differentiating again yields:

Substituting these derivatives into Equation 2 gives:

Cancelling like terms gives Equation 4:

Authorship:

Solved and typed by: Brian Lucas Egm4313.s12.team8.lucas 20:14, 30 January 2012 (UTC)

Checked by:

title: User:Egm4313.s12.team8.lucas/R1.4


text:
User:Egm4313.s12.team4.blj/editpage

= Problem 3: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problems 3 and 4. Find a general solution. Check your answer by substitution.

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

a) $$\displaystyle {y}''+6 {y}'+8.96y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.1)
 * }

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

b) $$\displaystyle {y}''+4{y}'+(pi^{2}+4)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3b.1)
 * }

Solution (2.3a)
Using the following:

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

$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.2)
 * }

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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.3)
 * }

The given equation can be rewritten as follows, using (2.3a.2) and (2.3a.3):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}+6\frac{dy}{dx}+8.96y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.4)
 * }

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

$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.5)
 * }

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

$$\displaystyle (D^{2}+6D+8.96)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.6)
 * }

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

$$\displaystyle (D^{2}+6D+8.96)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.7)
 * }

Use the quadratic formula

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

$$\displaystyle \frac{6\pm\sqrt{6^{2}-4(1)(8.96)}}{2(1)} $$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.8)
 * }

Arriving at

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

$$\displaystyle (D+3.2)(D+2.8)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.9)
 * }

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

$$\displaystyle D=-3.2,-2.8$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.10)
 * }

The final general solution for this problem is:

Reach the general solution :

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

 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.11)
 * }

Check by substitution:

From (2.3a.11), we find,
 * {| style="width:100%" border="0"

$$\displaystyle {y}'=-3.2C_{1}e^{-3.2x}-2.8C_{2}e^{-2.8x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.12)
 * }

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

$$\displaystyle {y}''=10.4C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.13)
 * }

Substituting (2.3a.11), (2.3a.12), and (2.3a.13) in (2.3a.1), we reach:
 * {| style="width:100%" border="0"

$$\displaystyle (10.4C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x})+6(10.4C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x})+8.96(C_{1}e^{-3.2x}+C_{2}e^{-2.8x})=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.14)
 * }

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

$$\displaystyle 19.2C_{1}e^{-3.2x}+16.8C_{2}e^{-2.8x} -19.2C_{1}e^{-3.2x}-16.8C_{2}e^{-2.8x}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.3a.15)
 * }

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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3a.16)
 * }

Therefore the solution is correct.

Solution (2.3b)
Using the following:

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

$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.2)
 * }

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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.3)
 * }

The given equation can be rewritten as follows, using (2.3b.2) and (2.3b.3):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+(\pi^2 +4)y=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.4)
 * }

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

$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.5)
 * }

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

$$\displaystyle (D^{2}+4D+(\pi^2 +4))y=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.6)
 * }

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

$$\displaystyle (D^{2}+4D+(\pi^2 +4))=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.7)
 * }

Use the quadratic formula

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

$$\displaystyle \frac{4\pm\sqrt{4^{2}-4(1)(\pi^2 +4)}}{2(1)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.8)
 * }

Arriving at

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

$$\displaystyle (D+(2+\pi i))(D+(2-\pi i)=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.9)
 * }

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

$$\displaystyle D=-(2+\pi i),-(2-\pi i)$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.10)
 * }

The final general solution for this problem is:

Reach the general solution :

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

 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.11)
 * }

Check by substitution:

From (2.3b.11), we find,
 * {| style="width:100%" border="0"

$$\displaystyle {y}'= -2(A \cos (\pi x) + B \sin (\pi x))e^{-2x} + (-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.12)
 * }

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

$$\displaystyle {y}''=4(A \cos (\pi x) + B \sin (\pi x))e^{-2x} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x}$$ $$\color{White}y''= \color{Black} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x} + (-A \pi ^2 \cos (\pi x) - B \pi ^2 \sin (\pi x))e^{-2x}$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.13)
 * }

Substituting (2.3b.11), (2.3b.12), and (2.3b.13) in (2.3b.1), we reach:
 * {| style="width:100%" border="0"

$$\displaystyle 4(A \cos (\pi x) + B \sin (\pi x))e^{-2x} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x} $$ $$\displaystyle+ (-A \pi ^2 \cos (\pi x) - B \pi ^2 \sin (\pi x))e^{-2x} +4 -2(A \cos (\pi x) + B \sin (\pi x))e^{-2x}) + (-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x}$$ $$\displaystyle+(\pi^{2}+4)((A\cos(\pi x) + B\sin(\pi x))e^{-2x})=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.14)
 * }

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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.3b.15)
 * }

Therefore the solution is correct.

title: User:Egm4313.s12.team4.blj/editpage


text:
User:Egm4313.sp12.team4.diroma/report1

= Problem 1 =

=Problem Statement=

Derive the equation of motion of a spring-dashpot system in parallel with a mass and applied force, $$   f(t)$$

Force Balance
Using a force balance, the following equation can be derived:

$$ f(t) = my'' + f_I $$ (1) $$\Rightarrow $$ where$$(f_I = f_c = f_k) $$

Adding the displacements due to the spring $$ (y_k)$$ and the dashpot $$ (y_c)$$ we can write the following equation:

$$ y=y_k+y_c $$ (2)

Taking two time derivatives of (2) gives:

$$ y = y_k + y''_c $$                                              (3)

Since $$f_k=f_c $$ ; We can write:

$$ ky_k = cy'_c $$ (4)

Now, solving for $$ y'_c $$ in (4), we find:

$$ y'_c = \frac{k}{c}y_k $$

We can substitute our new value for $$ y'_c $$ into (3) to obtain:

$$ y = y_k + (y'_c)' $$ ==> $$ y = y_k + (\frac{k}{c}y_k)' $$ ==> $$ y = y_k + \frac{k}{c}y'_k $$

We can now substitute for $$ y'' $$ and $$ f(I) $$ in equation (1) to obtain an equation in terms of $$ y_k $$ :

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

title: User:Egm4313.sp12.team4.diroma/report1


text:
User:Egm4313.s12.team9.lebo

Report 1

Report 2

title: User:Egm4313.s12.team9.lebo


text:
User:Egm4313.s12.team6.hickey

Question
Derive the equation of motion of the spring - mass - dashpot in Fig. 53, in K 2011 p. 85, with an applied force r(t) on the ball.

Solution
There are 2 possible cases that can occur in this problem, depending on the direction of the applied force. In both cases, $$ F_{applied}=r(t) $$ $$ F_{inertia}=my'' $$ $$ F_{damping}=cy' $$ $$ F_{spring}=ky $$

Case 1

The applied force is in the positive direction, and therefore the displacement is in the positive direction.

From the Free Body Diagram, we get the equation

$$ F_{applied}+F_{inertia}-F_{damping}-F_{spring}=0 $$

Rearranging the equation, we get

$$ F_{applied}=-F_{inertia}+F_{spring}+F_{damping} $$

Replacing Force variables, we get

$$ r(t)=my''+cy'+ky $$

Case 2

The applied force is in the negative direction, and therefore the displacement is in the negative direction.

From the Free Body Diagram, we get the equation

$$ -F_{applied}-F_{inertia}+F_{damping}+F_{spring}=0 $$

Rearranging the equation, we get

$$ F_{applied}=-F_{inertia}+F_{spring}+F_{damping} $$

Replacing Force variables, we get

$$ r(t)=my''+cy'+ky $$

Conclusion
Since both cases return the same solution, the equation of motion is derived as:

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

$$ r(t)=my''+cy'+ky $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

Question
Find and plot the solution for the Linear Second order ODE with constant coefficients corresponding to equation (1). Equation (1):

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

Initial Conditions: $$ y(0)=1, y'(0)=0 $$

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

In another Figure, superpose 3 Figures: (a) This figure, (b)The figure in R2.6 p. 5-6, (c) The Figure in R2.1 p. 3-7.

Solution
The question implied by the problem statement is that we are solving the following ODE: $$ y''+4y'+13=r(x) $$   Where we know that $$ r(x)=0 $$   and the corresponding characteristic equation is   $$ \lambda ^{2}+4\lambda +13=0 $$   To find the roots of this equation, we must use the Quadratic Equation, giving us   $$ \lambda _{1,2}= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} $$   Inserting our corresponding values for these variables, we have $$ \lambda _{1,2}= \frac{-4 \pm \sqrt{4^{2}-4(1)(13)}}{2(1)} = \frac{-4\pm \sqrt{16-52}}{2} = -2\pm \frac{\sqrt{-36}}{2} = -2\pm 3i $$   Knowing that we have complex roots, we can now insert the roots of our characteristic equation into the proper form for the homogeneous solution: $$ y(x)=e^{-2x}[C_{1}cos{3x}+C_{2}sin{3x}] $$   And taking the first derivative of this equation, we get $$ y'(x)=-2e^{-2x}[C_{1}cos{3x}+C_{2}sin{3x}]+3e^{-2x}[-C_{1}sin{3x}+C_{2}cos{3x}] $$

Now, we can use our initial conditions to find the values of our C coefficients: $$ y(0)=1=e^{0}[C_{1}cos{0}+C_{2}sin{0}]=1[C_{1}+0]=C_{1} $$ $$C_{1}=1$$ $$y'(0)=0= -2e^{0}[C_{1}cos{0}+C_{2}sin{0}]+3e^{0}[-C_{1}sin{0}+C_{2}cos{0}] = -2[C_{1}+0]+3[0+C_{2}] $$ $$-2C_{1}+3C_{2}=0 $$ $$3C_{2}=2 $$ $$C_{2}= \frac{2}{3} $$

Therefore, we now know that our final solution is:
 * {| style="width:100%" border="0"

$$ y(x)=e^{-2x}[cos{3x}+\frac{2}{3}sin{3x}] $$  The plot for this equation is given as shown in plot 1:
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

Comparing the graphs between problems R2.1, R2.6, and R2.9, we get the following plot: Where the solution for R2.1 is given in red, the solution for R2.6 is given in green, and the solution for R2.9 is given in blue.

title: User:Egm4313.s12.team6.hickey


text:
University of Florida/Egm4313/s12.team6

Members:
Alyssia Berthoumieux

A. Scott Davis

Patrick Hickey

Paul Hill

Scott Jagolinzer

Chad McPherson

Hopeton Nembhard

title: University of Florida/Egm4313/s12.team6


text:
User:Egm4313.s12.team14.turano

=R2.1= Given the two roots and the initial conditions:

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

\lambda_1=-2, \lambda_2=+5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1a)
 * }
 * }

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

y(0)=1,y'(0)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1b)
 * }
 * }

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$\displaystyle r(x) $$

Consider no excitation:

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

r(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Plot the solution.

Characteristic equation:

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

(\lambda-\lambda_1)(\lambda-\lambda_2)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Substituting $$\displaystyle(Eq. 1a)$$ into $$\displaystyle(Eq. 3)$$:

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

(\lambda+2)(\lambda-5)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

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

\lambda^2-3\lambda-10=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Non-homogeneous solution:

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

y''-3y'-10y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Homogeneous solution:

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

y_h(x)=C_1e^{-2x}+C_2e^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Overall solution:

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

y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

No excitation:

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

r(x)=0 => y_p(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

From intitial conditions:

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

y(0)=1=C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

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

y'(0)=0=-2C_1+5C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

Solving for $$\displaystyle C_1 $$ and $$\displaystyle C_2 $$:

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

C_1=\frac{5}{7} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and

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

C_2=\frac{2}{7} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

So, the final solution is:

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

=R2.2=

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

y'' - 10y' +25y = r(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Initial conditions:

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

y(0) = 1, y'(0) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

No excitation:

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

r(x) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Find and plot the solution for $$\displaystyle (Eq. 1) $$

Due to no excitation, $$\displaystyle (Eq. 1) $$ becomes:

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

y'' - 10y' +25y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Substituting $$\displaystyle \lambda $$ into $$\displaystyle (Eq. 4) $$:

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

\lambda^2-10\lambda+25=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Factoring, and solving for $$\displaystyle \lambda $$:

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

(\lambda-5)^2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

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

\lambda_2=\lambda_1=\lambda=5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Since $$\displaystyle \lambda $$ is a double root, the general solution:

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

y=C_1e^{5x}+C_2xe^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

From intitial conditions:

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

y(0)=1=C_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }

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

y'(0)=0=5C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * }

Solving for $$\displaystyle C_1 $$ and $$\displaystyle C_2 $$:

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

C_1=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

and

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

C_2=-5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

So, the final solution is:

$$\displaystyle y(x)=e^{5x}-5xe^{5x} $$

=R1.6G= G) Deformation of a Beam

$$\displaystyle{EIy''=f(x)}$$

$$\cdot$$ This equation is of SECOND order and LINEAR.

$$\cdot$$ To prove the applicability of Superposition, the following is done:

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

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

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

Particular:

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

 * $$\displaystyle{EIy_p''=f(x)}$$
 *
$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

Add (Eq. 1) and (Eq. 2):

$$\displaystyle{EIy_h+EIy_p=f(x}$$

It could be said that:

$$\displaystyle{y_{h}+y_{p}=Y}$$

and due to the fact that,

$$\displaystyle{EI(y_h''+y_p)=EIY}$$

=R1.1=

Team14P1R16Diagram1.png

$$y=y_k=y_c$$

Spring Force, $$\vec{F}_s=k(y)$$

Dashpot Force, $$\vec{F}_D=c(y')$$

Applied Force, $$\vec{F}(t)$$

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

\Sigma\vec{F}=\vec{F}(t)-\vec{F}_s-\vec{F}_D $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

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

m\vec{a}=\vec{F}_i=my'' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

From Newton's 2nd Law:

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

\Sigma\vec{F}=m\vec{a} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Substituting $$ (Eq. 1)$$ and $$ (Eq. 2)$$ into $$ (Eq. 3)$$:

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\vec{F}(t)-\vec{F}_s-\vec{F}_D=\vec{F}_i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Solving for $$\vec{F}(t)$$:

$$\vec{F}(t)=\vec{F}_s+\vec{F}_D+\vec{F}_i$$

Substituting $$\vec{F}_s=ky$$, $$\vec{F}_D=cy'$$, and $$\vec{F}_i=my''$$ into $$ (Eq. 4)$$:

$$\vec{F}(t)=ky+Cy'+my''$$

Putting into standard form:

$$my''+Cy'+ky=\vec{F}(t)$$

title: User:Egm4313.s12.team14.turano


text:
User:Egm4313.s12.team8.kyle/R1.3

=R1.3=

Problem Statement:
For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2) on p.1-4.

Solution:
Break the system into its three components: spring, dashpot, and mass.

Derivation of Equation of Motion
Looking at the FBD of the mass, $$f_I$$ represents the internal forces in the system from both the spring and dashpot. $$my''$$ represents the inertial force of the mass. $$f(t)$$ represents the excitation on the spring-dashpot-mass system. Summing the forces on this FBD yields the desired equation of motion for the mass, $$my''+f_{I}=f(t)$$

title: User:Egm4313.s12.team8.kyle/R1.3


text:
User:EGM4313.s12.team4.jones

 Andrew Jones

Egm4313.s12.team4

title: User:EGM4313.s12.team4.jones