University of Florida/Egm4313/s12.team6.reports/R1
Report 1
Problem 1: Springdashpot system in parallel with a mass and applied force
[edit  edit source]Derive the equation of motion of a springdashpot system in parallel, with a mass and applied force
Given
[edit  edit source]Springdashpot system in parallel
Solution
[edit  edit source]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  edit source]Solved by: EGM4313.s12.team6.hill 19:05, 1 February 2012 (UTC)
Problem 2: Springdashpot system in parallel with an applied force
[edit  edit source]Question
[edit  edit source]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
[edit  edit source]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  edit source]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  edit source]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  edit source]Since both cases return the same solution, the equation of motion is derived as:
Author
[edit  edit source]Solved by: Egm4313.s12.team6.hickey 20:04, 1 February 2012 (UTC)
Problem 3: Springdashpotmass system
[edit  edit source]For the springdashpotmass system on p.14, draw the FBDs and derive the equation of motion (2) p.14
Given
[edit  edit source]springdashpotmass system
Solution
[edit  edit source]
The equation of motion
where:
is the inertial force
is the internal force
is the applied force
this analysis assumes:
Assumptions:
[edit  edit source] Motion in the horizontal direction
 Massless spring
 Massless dashpot
 massless connections
Solution:
[edit  edit source]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 :
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  edit source]Solved by: Hopeton87 19:42, 1 February 2012 (UTC)
Problem 4: RLC circuit
[edit  edit source]Derive (3) and (4) from (2) on pg. 22
Given:
[edit  edit source]
Find:
[edit  edit source]Derive the below two equations from the given:
 A)
 and
 B)
Solution:
[edit  edit source]Part A:
[edit  edit source]From Eq. 22(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  edit source]From Eq. 22(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  edit source]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  edit source]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  edit source]Given:
[edit  edit source]
Find:
[edit  edit source]The homogeneous solution to the differential equation.
Solution:
[edit  edit source]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
 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  edit source]Given:
[edit  edit source]
Find:
[edit  edit source]The homogeneous solution to the differential equation.
Solution:
[edit  edit source]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  edit source]Solved by:EGM4313.s12.team6.davis 02:50, 1 February 2012 (UTC)
Problem 6
[edit  edit source]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  edit source]
,
Solution
[edit  edit source]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:2^{nd} order. The highest derivative is a 2^{nd} derivative on the y.
Linearity:Linear.
Superposition:Yes.
Order:1^{st} order. The highest derivative is a 1^{st} on the v.
Linearity:Nonlinear.
Superposition:No.
Order:1^{st} order
Linearity:Nonlinear.
Superposition:No.
Order:2^{nd} order.
Linearity:Linear.
Superposition:Yes.
,
Order:2^{nd} order
Linearity:Linear.
Superposition:Yes.
Order:2^{nd} order
Linearity:Linear.
Superposition:Yes.
Order:4^{th} oder
Linearity:Linear.
Superposition:Yes.
Order:2^{nd} order
Linearity:Nonlinear.
Superposition:Yes.
Author
[edit  edit source]Solved by:Egm4313.s12.team6.berthoumieux 02:52, 1 February 2012 (UTC)
Contributing Members
[edit  edit source]Team Contribution Table  
Problem Number  Lecture  Assigned To  Solved By  Typed By  Proofread By 
1.1  R1.1 in Sec 1 p. 15  Hill  Hill  Hill  Davis 
1.2  R1.2 in Sec 1 p. 14  Hickey  Hickey  Hickey  Hill 
1.3  R1.3 in Sec 1 p. 15  Nembhard  Nembhard  Nembhard  Hill 
1.4  R1.4 in Sec 2 p. 22  Jagolinzer/McPherson  Jagolinzer/McPherson  Jagolinzer/McPherson  Hill 
1.5  R1.5 in Sec 2 p. 25  Davis  Davis  Davis  Hill 
1.6  R1.6 in Sec 2 p. 27  Berthoumieux  Berthoumieux  Berthoumieux  Hill 