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

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

Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $f(t)$

Given

Spring-dashpot system in parallel

Solution

The kinematics of the system can be described as,

$x=x_k=x_c$

$⟶(1)$

The kinetics of the system can be described as,

$m\ddot{x}+f_I=f(t)$

$⟶(2)$

and,

$f_I=f_k+f_c$

$⟶(3)$

Given that,

$f_k=k{x}_k$

$⟶(3a)$

$f_c=c\dot{x_c}$

$⟶(3b)$

From (1), it can be found that,

$\dot{x}=\dot{x_k}=\dot{x_c}$

and,

$\ddot{x}=\ddot{x_k}=\ddot{x_c}$

From (3), it can be found that,

$f_I=c\dot{x_c}+k{x}_k$

Finally, it can be found that

$m\ddot{x}+c\dot{x}+kx=f(t)$

Author

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

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

Question

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

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

$F_{applied}=r(t)$

$F_{inertia}=m{y}^{″}$

$F_{damping}=c{y}^{\prime }$

$F_{spring}=ky$

Case 1

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

$F_{applied}+F_{inertia}-F_{damping}-F_{spring}=0$

Rearranging the equation, we get

$F_{applied}=-F_{inertia}+F_{spring}+F_{damping}$

Replacing Force variables, we get

$r(t)=m{y}^{″}+c{y}^{\prime }+ky$

Case 2

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

$-F_{applied}-F_{inertia}+F_{damping}+F_{spring}=0$

Rearranging the equation, we get

$F_{applied}=-F_{inertia}+F_{spring}+F_{damping}$

Replacing Force variables, we get

$r(t)=m{y}^{″}+c{y}^{\prime }+ky$

Conclusion

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

$r(t)=m{y}^{″}+c{y}^{\prime }+ky$

Author

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

Problem 3: Spring-dashpot-mass system

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

Given

spring-dashpot-mass system

Solution

Problem 3

The equation of motion

$m{y}^{″}+f_i=f(t)$

where:

$m{y}^{″}$

is the inertial force

$f_i$

is the internal force

$f(t)$

is the applied force
this analysis assumes:

Assumptions:

Motion in the horizontal direction

Massless spring

Massless dashpot

massless connections

Solution:

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:

$y$: This is the total displacement of the spring plus the displacement of the dashpot.

$y_k$: represents the displacement of the spring

$y_c$: represents the displacement of the dashpot

$y=y_k+y_c$

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:$y_i$. The force in the dashpot is proportional to the first time derivative of displacement (velocity)

$f_k=f_c=f_i=k{y}_k=cy{\text{'}}_c$

Where :$k$ is the spring constant.

$c$ is the damping coefficient.

and:$y{\text{'}}_c$ is the velocity of the dashpot. From this we get :$y{\text{'}}_c=(k/c)y_k$ which presents :$y{\text{'}}_c$ in terms of :$y_k$

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.

$y=y_k+y_c$

this presents two unknown dependent variables by using the relation

$y^{″}=y_{k^{″}}+(y_{c^{\prime }}{)}^{\prime }$

Becomes:

$y^{″}=y_{k^{″}}+\frac{k}{k}{y}_{k^{\prime }}$

now we can rewrite the equation of motion:
$m{y}^{″}+f_i=f(t)$

as :$m[y{k}^{″}+\frac{k}{c}y{k}^{\prime }]+fi=f(t)$
However >:$f_i=f_k$

so we get: $m[y{k}^{″}+\frac{k}{c}y{k}^{\prime }]+fk=f(t)$

$m[y_{k^{″}}+\frac{k}{c}{y}_{k^{\prime }}]+f_k=f(t)$:

Author

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

Problem 4: RLC circuit

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

Given:

$V=LC\frac{d^2{V}_c}{d{t}^2}+RC\frac{d{V}_c}{dt}+V_c$

Find:

Derive the below two equations from the given:

A) $L{I}^{″}+R{I}^{\prime }+\frac{1}{C}I=V^{\prime }$

and

B) $L{Q}^{″}+R{Q}^{\prime }+\frac{1}{C}Q=V$

Solution:

Part A:

From Eq. 2-2(1): $Q=C{V}_c=\int Idt$

$\therefore \frac{dI}{dt}=C\frac{d^2{V}_c}{dt}$

substituting $Q=C{V}_c=\int Idt$ into $RC\frac{d{V}_c}{dt}$ we get: $RI$

substituting $\frac{dI}{dt}=C\frac{d^2{V}_c}{dt}$ into $LC\frac{d^2{V}_c}{d{t}^2}$ we get: $L{I}^{\prime }$

So we now have: $V=L{I}^{\prime }+RI+V_c$

Deriving the whole equation we get:

$V^{\prime }=L{I}^{″}+R{I}^{\prime }+\frac{d{V}_c}{dt}$

by multiplying $\frac{d{V}_c}{dt}$ by $/fracCC$ we get:

$V^{\prime }=L{I}^{″}+R{I}^{\prime }+\frac{Cd{V}_c}{Cdt}$

we can now sub $Q=C{V}_c=\int Idt$ for $\frac{Cd{V}_c}{dt}$ giving us:

$L{I}^{″}+R{I}^{\prime }+\frac{1}{C}I=V^{\prime }$

Part B:

From Eq. 2-2(1): $Q=C{V}_c=\int Idt$

multiplying $V_c$ by $\frac{C}{C}$ we get $\frac{C{V}_c}{C}$

Subbing $Q=C{V}_c=\int Idt$ into $\frac{C{V}_c}{C}$ we get:

$V=LC\frac{d^2{V}_c}{d{t}^2}+RC\frac{d{V}_c}{dt}+\frac{Q}{C}$

taking the first and second derivative of $Q=C{V}_c$ we get:

$Q^{\prime }=C\frac{d{V}_c}{dt}$ and $Q^{″}=C\frac{d^2{V}_c}{d{t}^2}$

substitute $Q^{\prime }=C\frac{d{V}_c}{dt}$ into $RC\frac{d{V}_c}{dt}$ we get: $R{Q}^{\prime }$

substitute $Q^{″}=C\frac{d^2{V}_c}{d{t}^2}$ into $LC\frac{d^2{V}_c}{d{t}^2}$ we get: $L{Q}^{″}$

$\therefore V=L{Q}^{″}+R{Q}^{\prime }+\frac{1}{C}Q$

Author

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

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

Problem 5:

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

Given:

$y^{″}+6y^{\prime }+({\pi }^2+4)y=0$

Find:

The homogeneous solution to the differential equation.

Solution:

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

Let $y=e^{\lambda x}$

Therefore:

$y^{\prime }=\lambda e^{\lambda x}$

and

$y^{″}={\lambda }^2{e}^{\lambda x}$

So,

$y^{″}+6y^{\prime }+({\pi }^2+4)y=0$

can be written as

$e^{\lambda x}({\lambda }^2+a\lambda +b)=0$ where $a=6$ and $b={\pi }^2+4$

Since the exponential $e^{\lambda x}$ can never equal zero,

${\lambda }^2+a\lambda +b=0$

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

$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$

Substituting values:

$\lambda =\frac{-(6)\pm \sqrt[]{(6{)}^2-4({\pi }^2+4)}}{2}$

Which evaluates to:

$\lambda =-3\pm \sqrt[]{-19.478}$

or

$\lambda =-3\pm 2.207i$

Since the discriminant is less than zero we let $\omega =2.207$ and the homogeneous solution will be of the form:

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

Substituting values we have the homogeneous solution:

$y=e^{-3x}(Acos(2.21x)+Bsin(2.21x))$

Problem 5

Given:

$y^{″}+2\pi y^{\prime }+{\pi }^{2}y=0$

Find:

The homogeneous solution to the differential equation.

Solution:

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

Let $y=e^{\lambda x}$

Therefore:

$y^{\prime }=\lambda e^{\lambda x}$

and

$y^{″}={\lambda }^2{e}^{\lambda x}$

So,

$y^{″}+2\pi y^{\prime }+{\pi }^{2}y=0$

can be written as

$e^{\lambda x}({\lambda }^2+a\lambda +b)=0$ where $a=2\pi$ and $b={\pi }^2$

Since the exponential $e^{\lambda x}$ can never equal zero,

${\lambda }^2+a\lambda +b=0$

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

$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$

Substituting values we can see that the discriminant is zero:

$\lambda =\frac{-2\pi \pm \sqrt[]{4{\pi }^2-4{\pi }^2}}{2}$

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

$y_1=e^{-\pi x}$

For the second solution reduction of order must be used:

Let:

$y_2=u{y}_1$

Then,

$y_{2^{\prime }}=u^{\prime }{y}_1+y_{1^{\prime }}u$

and

$y_{2^{″}}=u^{″}{y}_1+2u^{\prime }{y}_{1^{\prime }}+y_{1^{″}}u$

Substituting these into the original equation gives:

$(u^{″}{y}_1+2u^{\prime }{y}_{1^{\prime }}+y_{1^{″}}u)+a(u^{\prime }{y}_1+y_{1^{\prime }}u)+bu{y}_1=0$

Which simplifies to:

$u^{″}{y}_{1^{\prime }}+u^{\prime }(2y_{1^{\prime }}+a{y}_1)+u(y_{1^{″}}+a{y}_{1^{\prime }}+b{y}_1)=0$

Since,

$2y_{1^{\prime }}=-a{e}^{\frac{-a}{2}x}=-a{y}_1$

and

$y_{1^{″}}+a{y}_{1^{\prime }}+b{y}_1=0$

What remains is:

$u^{″}{y}_1=0$

or

$u^{″}=0$

Therefore,

$u=C_{1}x+C_2$

Now we let $C_1=1$ and $C_2=0$ so that:

$u=x$

Now,

$y_2=x{y}_1=x{e}^{-\pi x}$

Using $y_1$ and $y_2$ we have the general solution:

$y=(C_1+C_{2}x)e^{-\pi x}$

Author

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

Problem 6

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

$(1)y^{″}=g=constant$

$(2)m{v}^{\prime }=mg-b{v}^2$

$(3)h^{\prime }=-k\sqrt{h}$

$(4)m{y}^{″}+ky=0$

$(5)y^{″}+{\omega }_0^2=cos(\omega t)$, ${\omega }_0=\omega$

$(6)L{I}^{″}+R{I}^{\prime }+\frac{1}{c}I=E^{\prime }$

$(7)EI{y}^{iv}=𝑓\mathit{(}𝑥\mathit{)}$

$(8)𝐿{\mathit{\theta }}^{″}\mathit{+}𝑔𝑠𝑖𝑛\mathit{\theta }\mathit{=}0$

Solution

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.
$(1)y^{″}=g=constant$
Order:2^nd order. The highest derivative is a 2^nd derivative on the y.
Linearity:Linear.
Superposition:Yes.

$(2)m{v}^{\prime }=mg-b{v}^2$
Order:1^st order. The highest derivative is a 1^st on the v.
Linearity:Non-linear.
Superposition:No.

$(3)h^{\prime }=-k\sqrt{h}$
Order:1^st order
Linearity:Non-linear.
Superposition:No.

$(4)m{y}^{″}+ky=0$
Order:2^nd order.
Linearity:Linear.
Superposition:Yes.

$(5)y^{″}+{\omega }_0^2=cos(\omega t)$, ${\omega }_0=\omega$
Order:2^nd order
Linearity:Linear.
Superposition:Yes.

$(6)L{I}^{″}+R{I}^{\prime }+\frac{1}{c}I=E^{\prime }$
Order:2^nd order
Linearity:Linear.
Superposition:Yes.

$(7)EI{y}^{iv}=𝑓\mathit{(}𝑥\mathit{)}$
Order:4^th oder
Linearity:Linear.
Superposition:Yes.

$(8)𝐿{\mathit{\theta }}^{″}\mathit{+}𝑔𝑠𝑖𝑛\mathit{\theta }\mathit{=}0$
Order:2^nd order
Linearity:Non-linear.
Superposition:Yes.

Author

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

Contributing Members

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