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[1]
[edit | edit source]Problem Statement
[edit | edit source]Given a spring-dashpot system in parallel with an applied force, find the equation of motion.
Background Theory
[edit | edit source]For this problem, Newton's second law is used,
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and applied to the mass at the end of a spring and damper in parallel.
Solution
[edit | edit source]Assuming no rotation of the mass,
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Therefore, the spring and damper forces can be written as, respectively,
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and
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And the resultant force on the mass can be written as
Now, from equations (1-3) and (1-4) all of the forces can be substituted into Newton's second law, and since each distance is equal,
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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:
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References
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R.1.2: Equation of Motion for a Spring-mass-dashpot System with Applied Force[1]
[edit | edit source]Problem Statement
[edit | edit source]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
[edit | edit source]To solve this problem, note that the ODE of a damped mass-spring system is
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When there is a external formal added to the model r(t) on the right. This then gives up the equation
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Solution
[edit | edit source]In this problem referencing back to equation (1-2)
The resultant force for the system can be described as stated in equation (1-5)
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
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For the characteristic equation m is divided throughout the whole equation making it
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References
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R.1.3: Free-body Diagram and Equation of Motion for Spring-dashpot-mass System[1]
[edit | edit source]Problem Statement
[edit | edit source]For this problem we were to draw the FBDs and derive the equation of motion for the following system.
Background Theory
[edit | edit source]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 | |
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and | ||
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Next we take a look at what we already know.
We know from kinematics that:
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We also know from kinetics that:
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From the constitutive relations we know:
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We also know:
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by solving for we get
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Solution
[edit | edit source]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 (3-2)
and substitute in
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for . From equations (3-3), (3-4), and (3-5) we can substitute for
After the substitutions into equation (3-2) we have
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From the equation (3-8) we know
We can then substitute this into equation (3-10) getting
To get the answer in its final form we divide both sides of the equation by m, and our final answer is
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References
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R.1.4: Derivation of Voltage-Charge and Voltage-Current Relationships[1]
[edit | edit source]Problem Overview
[edit | edit source]For this problem, the goal is to derive two different equations from the circuit equation
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Each separate derivation is presented in Voltage-Charge Derivation and Voltage-Current Derivation
Background Theory
[edit | edit source]To solve these derivation, it is wise to note that capacitance is
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It can also be noted that
This means that
Completing the integration results in
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It should also be noted that (4-2) can be written in the form
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It should also be noted that (4-3) can be written in the form
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Voltage-Charge Derivation
[edit | edit source]Problem Statement
[edit | edit source]For this problem, the Voltage-Charge equation
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are derived from (4-1)
Solution
[edit | edit source]Taking the derivative of (4-2) are taken with respect to time
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Substituting (4-4), (4-7), and (4-8) into (4-1) results in (4-6) such that
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Voltage-Current Derivation
[edit | edit source]Problem Statement
[edit | edit source]For this problem, the following equation
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are derived from (4-1)
Solution
[edit | edit source]The first step is to take the derivative of (4-1) resulting in
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Taking the derivative of (4-3) are taken with respect to time
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Substituting (4-3), (4-12), and (4-13) into (4-11) results in (4-10) such that
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References
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R.1.5: Solutions of General 2nd Order ODEs[1]
[edit | edit source]Background Theory[2]
[edit | edit source]Consider a second-order homogeneous linear ODE with constant coefficients a and b.
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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:
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Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing and , and so we will take the derivatives (with respect to x) of (5-2).
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We will now substitute (5-2), (5-3), and (5-4) into (5-1) to obtain the relationship:
- Simplifying, we have:
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Since (5-5) follows the same form as the quadratic equation, we can solve for and as follows:
Referring back to algebra, we know that the solution to these two equations can be one of three cases:
Case 1
[edit | edit source]Two real roots if...
These two roots give us two solutions:
and
The corresponding general solution then takes the form of the following:
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Case 2
[edit | edit source]A real double root if...
This yields only one solution:
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:
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Since is a solution of (5-1), the last set of parentheses is zero.
Similarly, the first parentheses is zero as well, because
From integration, we get the solution:
If we set and
Thus, the general solution is:
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Case 3
[edit | edit source]Complex conjugate roots if...
In this case, the roots of (5-5) are complex:
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Thus, the corresponding general solution is of the form:
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P.2.2.4
[edit | edit source]Problem Statement
[edit | edit source]For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:
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Solution
[edit | edit source]Following the process that yields (5-5), we find the equation:
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To find the form of the general solution, we must solve for the values of using the quadratic formula:
Since the solutions to are complex numbers, our solution takes the form of (5-8):
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A and B are constant coefficients of unknown value. Had we been given initial values of and , we would solve for those values as well.
Confirmation of Solution
[edit | edit source]To confirm that the solution is true, we will substitute , , and into (5a-1). If the final result is a tautology[3] then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.
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Substituting (5a-3), (5a-4), and (5a-5) into (5a-1), we are left with:
Combining similar terms allows us to clean up this solution and check our answer:
Final Solution
[edit | edit source]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:
P.2.2.5
[edit | edit source]Problem Statement
[edit | edit source]For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:
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Solution
[edit | edit source]Following the process that yields (5-5), we find the equation:
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To find the form of the general solution, we must solve for the values of . 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 (5-7), and we have the following:
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where and are unknown constant coefficients. If this were an initial value problem with values of and , we would solve for those values as well.
Confirmation of Solution
[edit | edit source]To confirm that the solution is true, we will substitute , , and into (5a-1) and check to see if the final result is a tautology or a contradiction.
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Substituting (5b-3), (5b-4), and (5b-5) into (5b-1), we are left with:
After combining similar terms, we have the form:
Final Solution
[edit | edit source]Since the above result is a tautology for all values of and , (5-3) is confirmed and the general solution to (5-1) is:
References
[edit | edit source]- ↑ EGM4313: Section 2 Notes (.djvu)
- ↑ Kreyszig, Erwin; Herbert Kreyszig; Edward J. Norminton (2011). Advanced Engineering Mathematics. John Wily & Sons, Inc. ISBN 978-0-470-45836-5. Pg 52-58
- ↑ Introduction to Propositional Logic
R.1.6: Determination of ODE Order, Linearity, and Application of the Superposition Principle[1]
[edit | edit source]Problem Statement
[edit | edit source]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[2][1]
[edit | edit source]Consider a function , defined as the sum of the homogeneous solution, , and the particular solution, :
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Example:
Consider the Differential Equation in standard form:
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The homogeneous solution is then:
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and the particular solution is:
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Now we sum up equations (6-3) and (6-4) to get:
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If the Equation is Linear then some reductions can take place:
and
Now we can make some substitutions into (6-5)using (6-3)and (6-4):
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Which is the same as equation (6-2):
Solution
[edit | edit source]Part A
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Order: Order ODE
Linearity: Linear
Superposition: Yes
This can be shown using the method from the background theory.
Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6a-2) and (6a-3):
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We can use (6-1) to simplify (6a-4):
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Since (6a-1) and (6a-6) are the same then we can apply Superposition.
Part B
[edit | edit source]-
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Order: 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:
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Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6b-3) and (6b-4):
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We can use (6-1) to simplify (6b-5):
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Note that:
So we cannot apply Super position.
Part C
[edit | edit source]-
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Order: 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:
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Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6c-3) and (6c-4):
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We can use (6-1) to simplify (6c-5):
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Note that:
So we cannot apply Super position.
Part D
[edit | edit source]-
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Order: Order ODE
Linearity: Linear
Superposition: Yes
This can be shown using the method from the background theory.
Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6d-2) and (6d-3):
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We can use (6-1) to simplify (6d-4):
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Since (6d-1) and (6d-6) are the same then we can apply Superposition.
Part E
[edit | edit source]-
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Order: Order ODE
Linearity: Linear
Superposition: Yes
This can be shown using the method from the background theory.
Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6e-2) and (6e-3):
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We can use (6-1) to simplify (6e-4):
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Since (6e-1) and (6e-6) are the same then we can apply Superposition.
Part F
[edit | edit source]-
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Order: Order ODE
Linearity: Linear
Superposition: Yes
This can be shown using the method from the background theory.
Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6f-2) and (6f-3):
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We can use (6-1) to simplify (6f-4):
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Since (6f-1) and (6f-6) are the same then we can apply Superposition.
Part G
[edit | edit source]-
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Order: Order ODE
Linearity: Linear
Superposition: Yes
This can be shown using the method from the background theory.
Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6g-2) and (6g-3):
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We can use (6-1) to simplify (6g-4):
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Since (6g-1) and (6g-6) are the same then we can apply Superposition.
Part H
[edit | edit source]-
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Order: Order ODE
Linearity: Non-Linear
Superposition: No
This can be shown using the method from the background theory.
Recall:
The homogeneous solution is:
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The particular solution is:
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Now we add Equations (6h-2) and (6h-3):
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We can use (6-1) to simplify (6h-4):
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Note that:
So we cannot apply Super position.
References
[edit | edit source]- ↑ 1.0 1.1 EGM4313: Section 2 Notes (.djvu)
- ↑ Kreyszig, Erwin; Herbert Kreyszig; Edward J. Norminton (2011). Advanced Engineering Mathematics. John Wily & Sons, Inc. ISBN 978-0-470-45836-5. Pg 46-48
Contributing Members
[edit | edit source]
Team Contribution Table
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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 |