User:Egm4313.s12.team19.R1

R1.1 A Spring-Dashpot-Mass Parallel System

Contributed by Gabriel Arab.

Problem Statement:

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

File:R1a1a1.jpg
From our friends at team 18, R1.1

Solution:

Choosing a coordinate system with (+) y-direction being to the right, we develop the following equations from the analysis of forces occurring on the mass.

For the spring force where k is the spring constant,

$f_{k}=k\ y$	(Eq. 1)

For the dashpot damping force where c is the viscious friction,

$f_{c}=c\ y'$	(Eq. 2)

For the resultant force on the mass,

$F=m\ y''$	(Eq. 3)

Since from Calculus, acceleration is the second derivative of position.

Newton’s Second Law can then be applied to the mass, summing all forces at assumable equal distances,

$\Sigma F=m\ y''=-f_{k}-f_{c}+f(t)$	(Eq. 4)

Substituting (1) and (2) into (4) and rearranging into standard ODE form yields the equation of motion for a spring-dashpot-mass system:

$y''+{c \over m}y'+{k \over m}y={1 \over m}f(t)$ (Eq. 6)

R1.2 A Spring-Dashpot-Mass Series System

Contributed by Adam Burton.

Given:

We are given a mass-spring system, with a mass m on an elastic spring. This is shown in Figure 53 in Kreyszig 2011 page 85 (shown below). For this problem, $y(t)$ is a function of time t of the displacement of the mass m from rest.

File:R1a2a1.jpg
Figure 53 K2011 p. 85

Problem Statement:

We are asked to derive the equation of motion of the mass spring dashpot system shown in the figure, with an external force $r(t)$ applied on the ball.

Solution:

From Newton’s second law of motion:

$\Sigma F=ma=m{d^{2}y \over dt^{2}}$	(Eq. 1)

In this system, there are three internal forces and one external force. The internal forces include the inertial force $m{\ddot {y}}$, the damping force $c{\ddot {y}}$, and the spring force $ky$.

The external force is $r(t)$.

We also know that the force of a spring is given as:

$F_{s}=-k\ y$	(Eq. 2)

And that the force caused by damping is given as:

$F_{D}=c{dy \over dt}$	(Eq. 3)

For conventional purposes, ${\dot {y}}={dy \over dt}$, and ${\ddot {y}}={d^{2}y \over dt^{2}}$.

Using a Free Body Diagram:

$\Sigma F=m{d^{2}y \over dt^{2}}$	(Eq. 4)

$m{\ddot {y}}={-c{\dot {y}}-ky+r(t)}$	(Eq. 5)

$\ m{\ddot {y}}+c{\dot {y}}+ky=r(t)$ (Eq. 6)

R1.3 A spring–dashpot–mass system

Contributed by Joshua Campbell.

Given: A spring–dashpot(damper)–mass system

Sec 1, 1-4

Problem Statement:

Use the FBD to derive the equation of motion (2) from lecture slide 1–4.

Solution:

Assuming no gravitational acceleration in the downward direction, the FBD of the mass is:

File:R1a3a2.png

where f(t) is a force exerted on the mass as a function of time, and fx is the equal and opposite force asserted by Newton’s 3rd law.

An expression of Newton’s 2nd law states:

$\Sigma F={dv \over dt}=0$	(Eq. 1)

Since

$\ a=v'=y''$	(Eq. 2)

And

$\ \Sigma F=ma$	(Eq. 3)

Thus,

$\ my''=\Sigma F$	(Eq. 4)

or

$\ my''+f_{y}=f(t)$ (Eq. 5)

R1.4 RLC Voltage

Contributed by Charles Chiamchittrong.

Given:

Equation (2) p.2-2 which describes the voltage in an RLC circuit in series:

$V=LC\ {d^{2}v_{C} \over dt^{2}}+RC\ {dv_{C} \over dt}+v_{C}$	(Eq. 1)

Problem Statement:

Derive the following alternate forms of the given equation:

$LI''+RI'+{1 \over C}I=V'$	(Eq. 2) p. 2-2

$LQ''+RQ'+{1 \over C}Q=V$	(Eq. 3) p. 2-2

Solution:

Part 1

In order to derive equation (2) we start by taking the derivative of equation (1):

$V'=LC{d^{3}v_{C} \over dt^{3}}+RC{d^{2}v_{C} \over dt^{2}}+{dv_{C} \over dt}$	(Eq. 4)

The variables L (inductance), C (capacitance), and R (resistance) represent constant values in the circuit that do not change over time and therefore remain unaffected.

Next we use part of equation (1) p.2-2 which relates current and capacitance:

$I=C{dv_{C} \over dt}$	(Eq. 5)

By taking the derivative of this equation twice, we obtain the following:

$I'=C{d^{2}v_{C} \over dt^{2}}$	(Eq. 6)

and

$I'''=C{d^{3}v_{C} \over dt^{3}}$	(Eq. 7)

Also, we can manipulate equation (5) by dividing by C:

${I \over C}={dv_{C} \over dt}$	(Eq. 8)

Now, we can substitute equations (6), (7), and (8) back into equation (4), yielding the following:

$V'=LI''+RI'+{I \over C}$ (Eq. 9)

Rewriting the third term on the right-hand side and rearranging yields the given equation (2) that we were trying to derive.

Part 2

In order to derive equation (3) we use another part of equation (1) p.2-2 which relates charge to capacitance:

$\ Q=Cv_{C}$	(Eq. 10)

We manipulate this equation in a manner similar to that of equation (5). First we take the derivative twice, resulting in the following:

$Q'=C{dv_{C} \over dt}$	(Eq. 11)

and

$Q''=C{d^{2}v_{C} \over dt^{2}}$	(Eq. 12)

Next we divide equation (10) by C:

${Q \over C}=v_{C}$	(Eq. 13)

Finally, we can substitute equations (11), (12), and (13) back into the original given equation (1):

$V=LQ''+RQ'+{Q \over C}$ (Eq. 14)

Once again, rewriting the third term on the right-hand side and rearranging yields the equation (3) that we are looking for.

R1.5 Selected Problems from Kreyszig 2011

Contributed by Santiago Marin.

Problem Statement:

Solve the following questions from the book.

K 2011 p.59 pb. 4

Given: $\ y''+4y'+(\pi ^{2}+4)y=0$

Find: A general solution

Solution:

The equation is a second-order homogeneous ODE taking the standard form:

$\ y''+ay'+by=0$	(Eq. 1)

Using the discriminant: $\ a^{2}-4b$

$\ 4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}$	(Eq. 2)

Since $\ -4\pi ^{2}<0$ it gives a general solution:

$\ y=e^{\left({-ax \over 2}\right)}(A\cos(\omega x)+B\sin(\omega x))$ where $\ \omega ^{2}=b-{1 \over 4}a^{2}$

$\ \omega ^{2}=\pi ^{2}+4-{1 \over 4}(4^{2})$	(Eq. 3)

$\ \omega =\pi$	(Eq. 4)

Substituting back into the general solution gives:

$\ y=e^{(-2x)}(A\cos(\pi x)+B\sin(\pi x))$ (Eq. 5)

K 2011 p.59 Problem 5

Given: $\ y''+2\pi y'+\pi ^{2}y=0$

Find: A general solution

Solution:

The equation is a second-order homogeneous ODE taking the standard form:

$\ y''+ay'+by=0$	(Eq. 6)

Using the discriminant: $\ a^{2}-4b$

$\ 4\pi ^{2}-4\pi ^{2}=0$	(Eq. 7)

Since $\ 4\pi ^{2}-4\pi ^{2}=0$ it gives a general solution:

$\ y=(c_{1}+c_{2}x)e^{\left({-ax \over 2}\right)}$	(Eq. 8)

Substituting back into the general solution gives:

$\ y=(c_{1}+c_{2}x)e^{(-\pi x)}$ (Eq. 9)

R1.6 Selected Problems from Kreyszig 2011

Contributed by Grant (Kyle) Uppercue.

File:R1a6a1.jpg
Fig.2 K 2011 p. 3

Problem statement:

"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 there of), and show whether the principle of superposition can be applied." Sec 2 (d)

Definitions:

Order = Highest order of derivative

"A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable." [1]

Principle of Superposition

If y₁(t) and y₂(t) are two solutions to a linear, homogeneous differential equation then so is

$\ y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$	(Eq. 1)

Solutions:

1. Falling Stone: $\ y''=g=constant$
   1. Order = 2
   2. Linearity (or lack of): Linear.
   3. Show whether the principle of superposition can be applied:

$\ y''=g=constant$	(Eq. 2)
$\ y_{h}''=0$	(Eq. 3)
$\ y_{p}''=g$	(Eq. 4)
$\ y_{h}''+y_{p}''=g$	(Eq. 5)

${\bar {y}}(x)=(y_{h}+y_{p})''$ (Eq. 6)

Superposition can be applied

1. Parachutist: $\ mv'=mg-bv^{2}$
   1. Order = 1
   2. Linearity (or lack of): Non-linear
   3. Show whether the principle of superposition can be applied:

$\ mv'=mg-bv^{2}$	(Eq. 7)
$\ mv'+bv^{2}=mg$	(Eq. 8)
$\ mv_{h}'+bv_{h}^{2}=0$	(Eq. 9)
$\ mv_{p}'+bv_{p}^{2}=mg$	(Eq. 10)
$\ mv^{'}+bv^{2}=mg$	(Eq. 11)
$\ mv_{p}'+bv_{p}^{2}+mv_{h}'+bv_{h}^{2}=mg$	(Eq. 12)
$\ {m(v_{p}+v_{h})}'+b(v_{p}^{2}+v_{h}^{2})=mg$	(Eq. 13)

${m(v_{p}+v_{h})}'+b(v_{p}+v_{h})^{2}\neq mg$ (Eq. 14)

Superposition can NOT be applied

1. Outflowing Water: $\ h'=-k{\sqrt {h}}$
   1. Order = 1
   2. Linearity (or lack of): Non-linear.
   3. Show whether the principle of superposition can be applied:

$\ h'=-k{\sqrt {h}}$	(Eq. 15)

$\ h'+k{\sqrt {h}}=0$ (Eq. 16)

Since h involves a square root, the solution would be a complex solution (involving imaginary numbers), and Superposition can Not be applied.

1. Vibrating Mass on a Spring: $\ my''+ky=0$
   1. Order = 2
   2. Linearity (or lack of): Linear
   3. Show whether the principle of superposition can be applied:

$\ my''+ky=0$	(Eq. 17)
$\ my_{h}''+ky_{h}=0$	(Eq. 18)
$\ my_{p}''+ky_{p}=0$	(Eq. 19)
$\ m(y_{h}+y_{p})''+k(y_{h}+y_{p})$	(Eq. 20)

$m{\bar {y}}(x)''+k{\bar {y}}(x)=0$ (Eq. 21)

Super Position can be applied

1. Beats of a Vibrating System: $y''+\omega _{0}^{2}y=cos(\omega t),\ \omega _{0}\asymp \omega$
   1. Order = 2
   2. Linearity (or lack of): Linear
   3. Show whether the principle of superposition can be applied:

$\ y''+\omega _{0}^{2}y=cos(\omega t),\ \omega _{0}\asymp \omega$	(Eq. 22)
$\ (y_{h})''+\omega _{0}^{2}y_{h}=0$	(Eq. 23)
$\ (y_{p})''+\omega _{0}^{2}y_{p}=\cos(\omega t)$	(Eq. 24)
$\ (y_{h}+y_{p})''+\omega _{0}^{2}(y_{h}+y_{p})=\cos(\omega t)$	(Eq. 25)

${\bar {y}}(x)''+\omega _{0}^{2}{\bar {y}}(x)=\cos(\omega t)$ (Eq. 26)

Super Position can be applied

1. Current I in an RLC Circuit: $\ LI''+RI'+{1 \over C}I=E'$
   1. Order = 2
   2. Linearity (or lack of): Linear
   3. Show whether the principle of superposition can be applied:

$\ LI''+RI'+{1 \over C}I=E'$	(Eq. 27)
$\ L(I_{p})''+R(I_{p})'+{1 \over C}I_{p}=E'$	(Eq. 28)
$\ L(I_{h})''+R(I_{h})'+{1 \over C}I_{h}=0$	(Eq. 29)
$\ L(I_{p}+I_{h})''+R(I_{p}+I_{h})'+{1 \over C}(I_{p}+I_{h})=E'$	(Eq. 30)

$L{\bar {I}}''+R{\bar {I}}'+{1 \over C}{\bar {I}}=E'$ (Eq. 31)

Super Position can be applied

1. Deformation of a Beam: $\ EIy^{iv}=f(x)$
   1. Order = 4
   2. Linearity (or lack of): Linear
   3. Show whether the principle of superposition can be applied:

$\ EIy^{iv}=f(x)$	(Eq. 32)
$\ EI(y_{h})^{iv}=0$	(Eq. 33)
$\ EI(y_{p})^{iv}=f(x)$	(Eq. 34)

$EI(y_{p}+y_{h})^{iv}=EI{\bar {y}}^{iv}$ (Eq. 35)

Super Position can be applied

1. Pendulum: $\ L\theta ''+gsin(\theta )=0$
   1. Order = 2
   2. Linearity (or lack of): Non-linear
   3. Show whether the principle of superposition can be applied:

$\ L\theta ''+g\sin(\theta )=0$	(Eq. 36)
$\ L(\theta _{h})''+g\sin(\theta _{h})=0$	(Eq. 37)
$\ L(\theta _{p})''+g\sin(\theta _{p})=0$	(Eq. 38)
$\ L(\theta _{h}+\theta _{p})''+g\sin(\theta _{h}+\theta _{p})=0$	(Eq. 39)

$L{\bar {\theta }}''+g\sin({\bar {\theta }})\neq 0$ (Eq. 40)

Super Position can Not be applied

Microsoft Word source

Original Microsoft Word text compiled into one file.

Converted by Vladimir Horwitz.

Team member jobs

Team member	Contributed	Lecture	Proofread
Gabriel Arab	R1.1	Mtg2.sec1	R1.2
Adam Burton	R1.2	Sec. 1	R1.3
Joshua Campbell	R1.3	Sec.1 1-4	R1.4
Charles Chiamchittrong	R1.4	Sec 2 (d)	R1.5
Santiago Marin	R1.5	Sec 2 (d)	R1.6
Grant Uppercue	R1.6	Sec 2 (d)	This page
Vladimir Horwitz	This page	Help:Formula	R1.1

Page information

Page designed by Vladimir Horwitz on 00:56, 28 January 2012 (UTC).

Last edited by Egm4313.s12.team19.Horwitz on 20:39, 1 February 2012 (UTC).