University of Florida/Egm4313/s12.team6.reports/R1

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Report 1

Problem 1: Spring-dashpot system in parallel with a mass and applied force[edit]

Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force


Given[edit]

Spring-dashpot system in parallel

Solution[edit]

The kinematics of the system can be described as,

The kinetics of the system can be described as,

and,

Given that,


From (1), it can be found that,

and,

From (3), it can be found that,

Finally, it can be found that

Author[edit]

Solved by: EGM4313.s12.team6.hill 19:05, 1 February 2012 (UTC)

Problem 2: Spring-dashpot system in parallel with an applied force[edit]

Question[edit]

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.

Figure 53, page 85 K 2011

Solution[edit]

There are 2 possible cases that can occur in this problem, depending on the direction of the applied force.
In both cases,









Case 1[edit]
Case 1: Positive Force, Positive Displacement

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



Rearranging the equation, we get



Replacing Force variables, we get



Case 2[edit]
Case 2: Negative Force, Negative Displacement

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



Rearranging the equation, we get



Replacing Force variables, we get



Conclusion[edit]

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

Author[edit]

Solved by: Egm4313.s12.team6.hickey 20:04, 1 February 2012 (UTC)

Problem 3: Spring-dashpot-mass system[edit]

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

Given[edit]

spring-dashpot-mass system

Solution[edit]

Problem 3



The equation of motion


where:


is the inertial force


is the internal force


is the applied force
this analysis assumes:

Assumptions:[edit]

Motion in the horizontal direction
Massless spring
Massless dashpot
massless connections

Solution:[edit]

To analyze the system we look at the kinematics and kinetics of the system
Kinematics: This involves the displacement which affects the mass. The displacement, represented by:

: This is the total displacement of the spring plus the displacement of the dashpot.
: represents the displacement of the spring
: represents the displacement of the dashpot

Kinetics: this involves the forces associated with the displacements. The spring and dashpot are in series, therefore at any section in the series the internal for will be the same. This is denoted as:. The force in the dashpot is proportional to the first time derivative of displacement (velocity)

Where : is the spring constant.

is the damping coefficient.

and: is the velocity of the dashpot. From this we get : which presents : in terms of :


section cut
Free body diagram

The constitutive relations:
The spring force is equal to the spring constant times the displacement of the spring. The damping force from the dashpot is equal to the damping coefficient times the velocity.

this presents two unknown dependent variables by using the relation


Becomes:

now we can rewrite the equation of motion:


as :
However >:

so we get:


:

Author[edit]

Solved by: Hopeton87 19:42, 1 February 2012 (UTC)

Problem 4: RLC circuit[edit]

Derive (3) and (4) from (2) on pg. 2-2

Given:[edit]

Find:[edit]

Derive the below two equations from the given:

A)
and
B)

Solution:[edit]

Part A:[edit]

From Eq. 2-2(1):







substituting into we get:



substituting into we get:



So we now have:



Deriving the whole equation we get:







by multiplying by we get:







we can now sub for giving us:




Part B:[edit]

From Eq. 2-2(1):



multiplying by we get



Subbing into we get:







taking the first and second derivative of we get:



and



substitute into we get:



substitute into we get:




Author[edit]

Solved by:Egm4313.s12.team6.mcpherson 19:52, 1 February 2012 (UTC)

         Egm4313.s12.team6.jagolinzer 19:58, 1 February 2012 (UTC)

Problem 5:[edit]

This problem consists of finding a general solution to two linear, second order differential equations with constant coefficients. The problems are taken from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics and can be found as problems 4 and 5 of problem set 2.2 on page 59.

Problem 4[edit]

Given:[edit]

Find:[edit]

The homogeneous solution to the differential equation.

Solution:[edit]

Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let

Therefore:

and

So,


can be written as

where and

Since the exponential can never equal zero,

This equation can be solved for lambda using the quadratic equation :


Substituting values:

Which evaluates to:



or


Since the discriminant is less than zero we let and the homogeneous solution will be of the form:



Substituting values we have the homogeneous solution:


Problem 5[edit]

Given:[edit]

Find:[edit]

The homogeneous solution to the differential equation.

Solution:[edit]

Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let

Therefore:

and

So,


can be written as

where and

Since the exponential can never equal zero,

This equation can be solved for lambda using the quadratic equation :


Substituting values we can see that the discriminant is zero:


Therefore, only one solution can be found from this equation:

For the second solution reduction of order must be used:

Let:

Then,

and

Substituting these into the original equation gives:

Which simplifies to:

Since,

and

What remains is:

or

Therefore,

Now we let and so that:

Now,

Using and we have the general solution:

Author[edit]

Solved by:EGM4313.s12.team6.davis 02:50, 1 February 2012 (UTC)

Problem 6[edit]

For each ODE in Fig.2 in K 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.

Given[edit]

,

Solution[edit]

The order of an ODE is found by looking for the highest derivative. If an ODE is linear, than the principle of superposition can be used by finding the solution to the homogeneous equation and finding the particular solution and then adding up the two. If an equation is not linear, then the principle of superposition cannot be applied.

Order:2nd order. The highest derivative is a 2nd derivative on the y.
Linearity:Linear.
Superposition:Yes.


Order:1st order. The highest derivative is a 1st on the v.
Linearity:Non-linear.
Superposition:No.


Order:1st order
Linearity:Non-linear.
Superposition:No.


Order:2nd order.
Linearity:Linear.
Superposition:Yes.

,
Order:2nd order
Linearity:Linear.
Superposition:Yes.


Order:2nd order
Linearity:Linear.
Superposition:Yes.


Order:4th oder
Linearity:Linear.
Superposition:Yes.


Order:2nd order
Linearity:Non-linear.
Superposition:Yes.

Author[edit]

Solved by:Egm4313.s12.team6.berthoumieux 02:52, 1 February 2012 (UTC)

Contributing Members[edit]

Team Contribution Table
Problem Number Lecture Assigned To Solved By Typed By Proofread By
1.1 R1.1 in Sec 1 p. 1-5 Hill Hill Hill Davis
1.2 R1.2 in Sec 1 p. 1-4 Hickey Hickey Hickey Hill
1.3 R1.3 in Sec 1 p. 1-5 Nembhard Nembhard Nembhard Hill
1.4 R1.4 in Sec 2 p. 2-2 Jagolinzer/McPherson Jagolinzer/McPherson Jagolinzer/McPherson Hill
1.5 R1.5 in Sec 2 p. 2-5 Davis Davis Davis Hill
1.6 R1.6 in Sec 2 p. 2-7 Berthoumieux Berthoumieux Berthoumieux Hill