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.

File:Spring-dashpot-parallel.png

Background Theory

For this problem, Newton's second law is used,
Template:NumBlk and applied to the mass at the end of a spring and damper in parallel.

Solution

Assuming no rotation of the mass,
Template:NumBlk

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

And the resultant force on the mass can be written as Template:NumBlk

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

Substituting these exact force equations,
Template:NumBlk

A little algebraic manipulation yields
Template:NumBlk

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

R.1.2: Equation of Motion for a Spring-mass-dashpot System with Applied Force

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).

File:Spring-ball-dashpot.PNG

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 (Template:EquationNote)

The resultant force for the system can be described as stated in equation (Template:EquationNote)

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

R.1.3: Free-body Diagram and Equation of Motion for Spring-dashpot-mass System

Problem Statement

For this problem we were to draw the FBDs and derive the equation of motion for the following system.
File:Spring-dashpot-ball2.PNG

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.

Free-Body Diagram		Derived Equation
File:BallFBD.png		$\sum F{}_y=m\frac{dv}{dt}\to ma=m{y}^{″}$ and $m{y}^{″}=f(t)-f_I$
File:SpringFBD.png		$k{y}^{\prime }=f_c(t)-f_k(t)$
File:DashpotFBD.png		$cy=f_I-f_c(t)$

Next we take a look at what we already know.

We know from kinematics that: Template:NumBlk We also know from kinetics that: Template:NumBlk Template:NumBlk From the constitutive relations we know: Template:NumBlk Template:NumBlk We also know: Template:NumBlk Template:NumBlk by solving for $y_{c^{\prime }}$ we get Template:NumBlk

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 (Template:EquationNote) and substitute in Template:NumBlk

for $y^{″}$. From equations (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) we can substitute $f{}_i$ for Template:NumBlk After the substitutions into equation (Template:EquationNote) we have Template:NumBlk From the equation (Template:EquationNote) we know Template:NumBlk We can then substitute this into equation (Template:EquationNote) getting Template:NumBlk To get the answer in its final form we divide both sides of the equation by m, and our final answer is Template:NumBlk

R.1.4: Derivation of Voltage-Charge and Voltage-Current Relationships

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

$\int idt=C{v}_C$

Completing the integration results in
Template:NumBlk

It should also be noted that (Template:EquationNote) can be written in the form
Template:NumBlk

It should also be noted that (Template:EquationNote) can be written in the form
Template:NumBlk

Voltage-Charge Derivation

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Problem Statement

For this problem, the Voltage-Charge equation

are derived from (Template:EquationNote)

Solution

Taking the derivative of (Template:EquationNote) are taken with respect to time

Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote) results in (Template:EquationNote) such that

Voltage-Current Derivation

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Problem Statement

For this problem, the following equation

are derived from (Template:EquationNote)

Solution

The first step is to take the derivative of (Template:EquationNote) resulting in

Taking the derivative of (Template:EquationNote) are taken with respect to time

Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote) results in (Template:EquationNote) 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. Template:NumBlk

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

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

Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing $y^{\prime }$ and $y^{″}$, and so we will take the derivatives (with respect to x) of (Template:EquationNote). Template:NumBlk Template:NumBlk

We will now substitute (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote) to obtain the relationship: Template:NumBlk

Simplifying, we have:

Template:NumBlk Template:NumBlk

Since (Template:EquationNote) follows the same form as the quadratic equation, we can solve for ${\lambda }_1$ and ${\lambda }_2$ as follows: Template:NumBlk Template:NumBlk

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...
Template:NumBlk These two roots give us two solutions:

$y_1=e^{{\lambda }_{1}x}$

and

$y_2=e^{{\lambda }_{2}x}$

The corresponding general solution then takes the form of the following: Template:NumBlk

Case 2

A real double root if...
Template:NumBlk 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:

$y_2=u{y}_1$

$y{\text{'}}_2=u^{\prime }{y}_1+uy{\text{'}}_1$

and

$y^{\prime }{\text{'}}_2=u^{″}{y}_1+2u^{\prime }y{\text{'}}_1+u{y}^{\prime }{\text{'}}_1$

yeilding

$(u^{″}{y}_1+2u^{\prime }y{\text{'}}_1+u{y}^{\prime }{\text{'}}_1)+a(u^{\prime }{y}_1+uy{\text{'}}_1)+bu{y}_1=0$

$u^{″}{y}_1+u^{\prime }(2y{\text{'}}_1+a{y}_1)+u(y^{\prime }{\text{'}}_1+a^{\prime }{y}_1+b{y}_1)=0$

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

$2y{\text{'}}_1=-a{e}^{ax/2}=-a{y}_1$

From integration, we get the solution:

$u=c_{1}x+c_2$

If we set $c_1=1$ and $c_2=0$

$y_2=x{y}_1$

Thus, the general solution is: Template:NumBlk

Case 3

Complex conjugate roots if...
Template:NumBlk In this case, the roots of (Template:EquationNote) 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: Template:NumBlk

P.2.2.4

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Problem Statement

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

Solution

Following the process that yields (Template:EquationNote), we find the equation: Template:NumBlk

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 (Template:EquationNote):
Template:NumBlk

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

Confirmation of Solution

To confirm that the solution is true, we will substitute $y$, $y^{\prime }$, and $y^{″}$ into (Template:EquationNote). 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 (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote), 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:

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

P.2.2.5

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Problem Statement

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

Solution

Following the process that yields (Template:EquationNote), we find the equation: Template:NumBlk
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. Template:NumBlk

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of (Template:EquationNote), and we have the following: Template:NumBlk where $c_1$ and $c_2$ are unknown constant coefficients. If this were an initial value problem with values of $y(x)$ and $y^{\prime }(x)$, we would solve for those values as well.

Confirmation of Solution

To confirm that the solution is true, we will substitute $y$, $y^{\prime }$, and $y^{″}$ into (Template:EquationNote) and check to see if the final result is a tautology or a contradiction.
Template:NumBlk Template:NumBlk
Substituting (Template:EquationNote), (Template:EquationNote), and (Template:EquationNote) into (Template:EquationNote), we are left with: Template:NumBlk After combining similar terms, we have the form: Template:NumBlk Template:NumBlk

Final Solution

Since the above result is a tautology for all values of $c_1$ and $c_2$, (Template:EquationNote) is confirmed and the general solution to (Template:EquationNote) is: Template:NumBlk

R.1.6: Determination of ODE Order, Linearity, and Application of the Superposition Principle

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 $\overline{y}(x)$, 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 (Template:EquationNote) and (Template:EquationNote) to get:

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

${y_h}^{″}+{y_p}^{″}=(y_h+y_p{)}^{″}={\overline{y}}^{″}$

and

${y_h}^{\prime }+{y_p}^{\prime }=(y_h+y_p{)}^{\prime }={\overline{y}}^{\prime }$

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

Which is the same as equation (Template:EquationNote):

$y^{″}+p(x)y^{\prime }+q(x)y=r(x)$

Solution

Part A

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

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

Recall:

$\overline{y}(x)=y_p(x)+y_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Since (Template:EquationNote) and (Template:EquationNote) 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:

$\overline{v}(x)=v_p(x)+v_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Note that:

$(v_h^2+v_p^2)\ne {\overline{v}}^2$

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:

$\overline{h}(x)=h_h(x)+h_p(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Note that:

$(\sqrt{h_h}+\sqrt{h_p})\ne \sqrt{\overline{h}}$

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:

$\overline{y}(x)=y_p(x)+y_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Since (Template:EquationNote) and (Template:EquationNote) 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:

$\overline{y}(x)=y_p(x)+y_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Since (Template:EquationNote) and (Template:EquationNote) 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:

$\overline{I}(x)=I_p(x)+I_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Since (Template:EquationNote) and (Template:EquationNote) 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:

$\overline{y}(x)=y_p(x)+y_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Since (Template:EquationNote) and (Template:EquationNote) 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:

$\overline{\theta }(x)={\theta }_p(x)+{\theta }_h(x)$

The homogeneous solution is:

The particular solution is:
Template:NumBlk

Now we add Equations (Template:EquationNote) and (Template:EquationNote):

We can use (Template:EquationNote) to simplify (Template:EquationNote):

Note that:

$(\sin {\theta }_h+\sin {\theta }_p)\ne \sin \overline{\theta }$

So we cannot apply Super position.

Contributing Members

Template:Center topTeam Contribution TableTemplate:Center bottom
Problem Number	Solved and Typed By	Proofread By
1.1	Maxwell Shuman	Yamil Herrera
1.2	Dalwin Marrero	Maxwell Shuman
1.3	Ron D'Amico	Dalwin Marrero
1.4	Jennifer Melroy	Ron D'Amico
1.5	Mark James and Avery Cornell	Jennifer Melroy
1.6	Yamil Herrera	Avery Cornell
Diagrams and Wiki Set-up	Mark James	Team