Intermediate engineering analysis/Problem set 1

R.1.1: Equation of Motion for a Spring-dashpot Parallel System in Series with Mass and Applied Force[1]

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Problem Statement

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

 

Background Theory

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

 

 

 

 

 

(1-1)

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

Solution

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Assuming no rotation of the mass,

 

 

 

 

 

(1-2)

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

 

 

 

 

 

(1-3)

and

 

 

 

 

 

(1-4)

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,

 

 

 

 

 

(1-5)

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:

 

 

 

 

 

(1-6)

References

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  1. EGM4313: Section 1 Notes (.djvu)


R.1.2: Equation of Motion for a Spring-mass-dashpot System with Applied Force[1]

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Problem Statement

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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

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

 

 

 

 

 

(2-1)

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

 

 

 

 

 

(2-2)

Solution

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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

 

 

 

 

 

(2-3)

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

 

 

 

 

(2-4)

References

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  1. EGM4313: Section 1 Notes (.djvu)


R.1.3: Free-body Diagram and Equation of Motion for Spring-dashpot-mass System[1]

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Problem Statement

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

Background Theory

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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
   

and
 

 

 

 

 


Next we take a look at what we already know.

We know from kinematics that:

 

 

 

 

 

(3-1)

We also know from kinetics that:

 

 

 

 

 

(3-2)

 

 

 

 

 

(3-3)

From the constitutive relations we know:

 

 

 

 

 

(3-4)

 

 

 

 

 

(3-5)

We also know:

 

 

 

 

 

(3-6)

 

 

 

 

 

(3-7)

by solving for   we get

 

 

 

 

 

(3-8)

Solution

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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

 

 

 

 

 

(3-9)

for  . From equations (3-3), (3-4), and (3-5) we can substitute   for

 

 

 

 

 

After the substitutions into equation (3-2) we have

 

 

 

 

 

(3-10)

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

 

 

 

 

 

(3-11)

References

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  1. EGM4313: Section 1 Notes (.djvu)


R.1.4: Derivation of Voltage-Charge and Voltage-Current Relationships[1]

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Problem Overview

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

 

 

 

 

 

(4-1)

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

Background Theory

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

 

 

 

 

 

(4-2)

It can also be noted that

 

This means that

 

Completing the integration results in

 

 

 

 

 

(4-3)

It should also be noted that (4-2) can be written in the form

 

 

 

 

 

(4-4)

It should also be noted that (4-3) can be written in the form

 

 

 

 

 

(4-5)

Voltage-Charge Derivation

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Problem Statement

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For this problem, the Voltage-Charge equation

 

 

 

 

 

(4-6)

are derived from (4-1)

Solution

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Taking the derivative of (4-2) are taken with respect to time

 

 

 

 

 

(4-7)

 

 

 

 

 

(4-8)

Substituting (4-4), (4-7), and (4-8) into (4-1) results in (4-6) such that

 

 

 

 

 

(4-9)

Voltage-Current Derivation

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Problem Statement

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For this problem, the following equation

 

 

 

 

 

(4-10)

are derived from (4-1)

Solution

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The first step is to take the derivative of (4-1) resulting in

 

 

 

 

 

(4-11)

Taking the derivative of (4-3) are taken with respect to time

 

 

 

 

 

(4-12)

 

 

 

 

 

(4-13)

Substituting (4-3), (4-12), and (4-13) into (4-11) results in (4-10) such that

 

 

 

 

 

(4-16)

References

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  1. EGM4313: Section 2 Notes (.djvu)


R.1.5: Solutions of General 2nd Order ODEs[1]

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Background Theory[2]

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

 

 

 

 

 

(5-1)

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:

 

 

 

 

 

(5-2)

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).

 

 

 

 

 

(5-3)

 

 

 

 

 

(5-4)

We will now substitute (5-2), (5-3), and (5-4) into (5-1) to obtain the relationship:

 

 

 

 

 

Simplifying, we have:
 

 

 

 

 

 

 

 

 

 

(5-5)


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

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Two real roots if...

 

 

 

 

 

These two roots give us two solutions:

  and  

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

 

 

 

 

 

(5-6)


Case 2

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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:

 
 
and
 
yeilding
 
 


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:

 

 

 

 

 

(5-7)

Case 3

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Complex conjugate roots if...

 

 

 

 

 

In this case, the roots of (5-5) are complex:

 
and
 


Thus, the corresponding general solution is of the form:

 

 

 

 

 

(5-8)


P.2.2.4

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Problem Statement

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

 

 

 

 

 

(5a-1)

Solution

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Following the process that yields (5-5), we find the equation:

 

 

 

 

 

(5a-2)

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):

 

 

 

 

 

(5a-3)

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

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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.

 

 

 

 

 

 

 

 

 

 

(5a-4)

 

 

 

 

 

 

 

 

 

 

(5a-5)


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

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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

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Problem Statement

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

 

 

 

 

 

(5b-1)


Solution

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Following the process that yields (5-5), we find the equation:

 

 

 

 

 

(5b-2)


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:

 

 

 

 

 

(5b-3)

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

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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.

 

 

 

 

 

(5b-4)

 

 

 

 

 

(5b-5)


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

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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

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  1. EGM4313: Section 2 Notes (.djvu)
  2. Kreyszig, Erwin; Herbert Kreyszig; Edward J. Norminton (2011). Advanced Engineering Mathematics. John Wily & Sons, Inc. ISBN 978-0-470-45836-5.  Pg 52-58
  3. Introduction to Propositional Logic


R.1.6: Determination of ODE Order, Linearity, and Application of the Superposition Principle[1]

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Problem Statement

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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]

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Consider a function  , defined as the sum of the homogeneous solution,  , and the particular solution,  :

 

 

 

 

 

(6-1)

Example:

Consider the Differential Equation in standard form:

 

 

 

 

 

(6-2)

The homogeneous solution is then:

 

 

 

 

 

(6-3)

and the particular solution is:

 

 

 

 

 

(6-4)

Now we sum up equations (6-3) and (6-4) to get:

 

 

 

 

 

(6-5)

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):

 

 

 

 

 

(6-6)

Which is the same as equation (6-2):

 

Solution

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Part A

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(6a-1)

Order:   Order ODE
Linearity: Linear
Superposition: Yes

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

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6a-2)

The particular solution is:

 

 

 

 

 

(6a-3)

Now we add Equations (6a-2) and (6a-3):

 

 

 

 

 

(6a-4)

We can use (6-1) to simplify (6a-4):

 

 

 

 

 

(6a-5)

 

 

 

 

 

(6a-6)

Since (6a-1) and (6a-6) are the same then we can apply Superposition.

Part B

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(6b-1)

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:

 

 

 

 

 

(6b-2)

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6b-3)

The particular solution is:

 

 

 

 

 

(6b-4)

Now we add Equations (6b-3) and (6b-4):

 

 

 

 

 

(6b-5)

We can use (6-1) to simplify (6b-5):

 

 

 

 

 

(6b-6)

Note that:

 

So we cannot apply Super position.

Part C

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(6c-1)

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:

 

 

 

 

 

(6c-2)

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6c-3)

The particular solution is:

 

 

 

 

 

(6c-4)

Now we add Equations (6c-3) and (6c-4):

 

 

 

 

 

(6c-5)

We can use (6-1) to simplify (6c-5):

 

 

 

 

 

(6c-6)

Note that:

 

So we cannot apply Super position.

Part D

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(6d-1)

Order:   Order ODE
Linearity: Linear
Superposition: Yes

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

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6d-2)

The particular solution is:

 

 

 

 

 

(6d-3)

Now we add Equations (6d-2) and (6d-3):

 

 

 

 

 

(6d-4)

We can use (6-1) to simplify (6d-4):

 

 

 

 

 

(6d-5)

 

 

 

 

 

(6d-6)

Since (6d-1) and (6d-6) are the same then we can apply Superposition.

Part E

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(6e-1)

Order:   Order ODE
Linearity: Linear
Superposition: Yes

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

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6e-2)

The particular solution is:

 

 

 

 

 

(6e-3)

Now we add Equations (6e-2) and (6e-3):

 

 

 

 

 

(6e-4)

We can use (6-1) to simplify (6e-4):

 

 

 

 

 

(6e-5)

 

 

 

 

 

(6e-6)

Since (6e-1) and (6e-6) are the same then we can apply Superposition.

Part F

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(6f-1)

Order:   Order ODE
Linearity: Linear
Superposition: Yes

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

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6f-2)

The particular solution is:

 

 

 

 

 

(6f-3)

Now we add Equations (6f-2) and (6f-3):

 

 

 

 

 

(6f-4)

We can use (6-1) to simplify (6f-4):

 

 

 

 

 

(6f-5)

 

 

 

 

 

(6f-6)

Since (6f-1) and (6f-6) are the same then we can apply Superposition.

Part G

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(6g-1)

Order:   Order ODE
Linearity: Linear
Superposition: Yes

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

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6g-2)

The particular solution is:

 

 

 

 

 

(6g-3)

Now we add Equations (6g-2) and (6g-3):

 

 

 

 

 

(6g-4)

We can use (6-1) to simplify (6g-4):

 

 

 

 

 

(6g-5)

 

 

 

 

 

(6g-6)

Since (6g-1) and (6g-6) are the same then we can apply Superposition.

Part H

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(6h-1)

Order:   Order ODE
Linearity: Non-Linear
Superposition: No

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

Recall:

 

The homogeneous solution is:

 

 

 

 

 

(6h-2)

The particular solution is:

 

 

 

 

 

(6h-2)

Now we add Equations (6h-2) and (6h-3):

 

 

 

 

 

(6h-4)

We can use (6-1) to simplify (6h-4):

 

 

 

 

 

(6h-5)

Note that:

 

So we cannot apply Super position.

References

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  1. 1.0 1.1 EGM4313: Section 2 Notes (.djvu)
  2. Kreyszig, Erwin; Herbert Kreyszig; Edward J. Norminton (2011). Advanced Engineering Mathematics. John Wily & Sons, Inc. ISBN 978-0-470-45836-5.  Pg 46-48