User:Egm4313.s12.team2/Report1

= R1.1 =

Problem Statement
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$f(t)$$. Figure 1.1: The spring is represented by K, the dashpot is represented by C, the mass is represented by M, and the applied force is represented by $$\bar{r}(t)$$. --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Kinematics

 * $$\bar{r}=y\bar{E}_y$$
 * $$\bar{v}={y}'\bar{E}_y$$
 * $$\bar{a}=y''\bar{E}_y$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 1. are derived, first, from the position function of the mass with respect to time ($$r(t)$$). The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$r(t)$$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

Kinetics- Free Body Diagram
Figure 1.2 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Using Newton's Second Law

 * $$F=m\bar{a}$$

Summing Forces
 * $$\bar{F}_k+\bar{F}_c+m\bar{g}=r(t)$$

$$ky\bar{E}_y+cy'\bar{E}_y+my''\bar{E}_y=r(t)$$ --Corey Williams 20:41, 27 January 2012 (UTC)

=R1.2=

Problem Statement
Derive the equation of motion of the spring-mass-dashpot in figure 53. on page 85 of Kreyszig's Advanced Engineering Mathematics (2011). Figure 2.1 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Kinematics

 * $$ \bar{r} = y \bar{E}_y $$
 * $$ \bar{v}=y' \bar{E_y}$$
 * $$ \bar{a} = \bar{v}' \bar{E_y}=y''\bar{E_y}$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 53 (also pictured above) are derived, first, from the position function of the mass with respect to time $$\bar{r}(t) = y\bar{E}_y $$. The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$\bar{r}(t) = y\bar{E}_y $$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

Kinetics-Free Body Diagram
Figure 2.2 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Using Newton's Second Law
Equations:
 * $$\bar{F} = m\bar{a} $$
 * $$\bar{F}_{spring} = ky\bar{E}_y$$
 * $$\bar{F}_{dashpot} = cy'\bar{E}_y $$
 * $$ mg = m\bar{a} = my''\bar{E}_y$$

Sum the Forces
 * $$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = r(t) $$

From Kreyszig pg. 86
 * $$ \bar{r}(t) = \bar{F}_0{cos}(wt) $$

$$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = \bar{F}_0{cos}(wt) $$

--Stephen Leshko 20:10, 30 January 2012 (UTC)

=R1.3=

Problem Statement
For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2)p.1-4. Figure 3.1 --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Free Body Diagrams
Figure 3.2 FBD of spring, dashpot, and mass --Rachel Dzadek 22:20, 31 January 2012 (UTC) Figure 3.3 --Rachel Dzadek 22:20, 31 January 2012 (UTC)

Position Function

 * $$ y(t)=y_k + y_c                    \ $$


 * $$ y'(t)=y'_k + y'_c                 \ $$


 * $$ y(t)=y_k + y''_c              \ $$

From Figure 3.2

$$ f(w)=f(k)=f(c)=:\ f(I)              $$

From Figure 3.3

$$ \overline{f}(t)-\overline{f}(I)=my'' $$

$$ \overline{f}(t)=\overline{f}(I)+my'' $$

--Corey Williams 18:29, 1 February 2012 (UTC)

=R1.4=

Problem Statement
Derive (3) and (4) on p.2-2 from (2) on the same page.


 * $$ LC\frac{{d^2}v_c }{dt^2} + RC\frac{dv_c}{dt} + v_c = V \ $$ .............(2)


 * $$ LI'' + RI' + \frac{1}{C}I = V' \ $$ .............(3)


 * $$ LQ'' + RQ' + \frac{1}{C}Q = V \ $$ ...........(4)

Useful Equations
The two equations can be derived using:
 * $$ Q = Cv_c ; Q' = Cv'_c; Q= Cv_c \ $$ ......(5)
 * $$ I = \frac{dQ}{dt}=Q' ; I' = Q ; I = Q''' \ $$ ..... (6)

Derivations
Also knowing that equation (2) can be written as:
 * $$ V = LCv_c'' +RCv_c' + v_c \ $$ .....(2')

Then by the simple substitutions in (5) equation (2) becomes: $$ V = LQ'' + RQ' + \frac{1}{C}Q \ $$ ....... (4) Then by taking the derivative with respect to time, and again treating R, L, and C as constants equation (2) becomes:
 * $$ V' = LQ' + RQ +\frac{1}{C}Q' \ $$ ......(7)

Then by using the substitutions in (6) equation (7) becomes: $$ V' = LI'' +RI' + \frac{1}{C}I \ $$ .....(3)

--Stephen Leshko 22:32, 31 January 2012 (UTC)

=R1.5=

Problem Statement
Do problems 4-5 on page 59 of Kreyszig's Advanced Engineering Mathematics (2011).

Problem (4)

 * $$ y''+ 4y'+(\pi^2+4)y=0 \ $$


 * $$ r^2+4r=(\pi^2+4)=0 \ $$


 * $$ r=(\frac{-4\pm \sqrt{16-4(\pi^2 +4)}}{2}) \ $$


 * $$r=-2\pm sqrt{4-(\pi^2+4)} \ $$


 * $$r=-2\pm\pi i \ $$

Therefore, the general solution takes the form of:
 * $$y_g=Ae^{-2t}cos(\pi t)+Be^{-2t}sin(\pi t) \ $$


 * $$y'=A[-2e^{-2t}cos(\pi t)-\pi e{^-2t}sin(\pi t)]+B[-2e^{-2t}sin(\pi t)+\pi e^{-2t}cos(\pi t)] \ $$


 * $$y''=A[4e^{-2t}cos(\pi t)+2\pi e{^-2t}sin(\pi t)+2\pi e^{-2t}sin(\pi t)-\pi^2 e^{-2t}cos(\pi t)] \ $$


 * $$ +B[4e^{-2t}sin(\pi t)-2\pi e^{-2t}cos(\pi t)-2\pi e^{-2t}cos(\pi t)-\pi^2 e{^-2t}sin(\pi t)] \ $$

Substitution back into the original equation should result in 0=0
 * $$ 0 = A[4e^{-2t}cos(\pi t)+2\pi e{^-2t}sin(\pi t)+2\pi e^{-2t}sin(\pi t)-\pi^2 e^{-2t}cos(\pi t)] \ $$
 * $$+B[4e^{-2t}sin(\pi t)-2\pi e^{-2t}cos(\pi t)-2\pi e^{-2t}cos(\pi t)-\pi^2 e{^-2t}sin(\pi t)]  \ $$
 * $$+(4A[-2e^{-2t}cos(\pi t)-\pi e{^-2t}sin(\pi t)]+B[-2e^{-2t}sin(\pi t)+\pi e^{-2t}cos(\pi t)] +(\pi^2+4)(Ae^{-2t}cos(\pi t)+Be^{-2t}sin(\pi t))                                           \ $$
 * $$ 0=0                                                                                            \ $$

Substitution of the solution and its derivatives into the original differential equation returns a value of zero, thereby proving that the solution is correct. --Nicolas Ellis 20:10, 30 January 2012 (UTC)

Problem (5)

 * $$ y'' +2\pi y' +\pi y = 0 \ $$


 * $$ r^2 +2\pi r +\pi ^2 = 0 \ $$


 * $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$


 * $$ r= \frac{-2\pi}{2} = -\pi\ $$

Therefore, the general solution takes the form of:
 * $$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$


 * $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$


 * $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$


 * $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0


 * $$ 0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$
 * $$ 0 = 0 \ $$

Thus the substitution checks out, and the form of the general solution is proven. --Stephen Leshko 20:10, 30 January 2012 (UTC)

=R1.6=

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 of), and show whether the principle of superposition can be applied.

Falling Stone

 * $$ y'' = g = const. \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h '' = 0                       \ $$
 * $$ y_p '' = g                       \ $$
 * $$ y(t) := (y_h + y_p)          \ $$
 * $$ y_h  + y_p  = 0 + g  \ $$
 * $$ y(t)'' = g                       \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Parachute

 * $$ mv' = mg - bv^2 \ $$

Order: 1st order Linearity (or lack of): not linear Superposition:
 * $$ mv_h ' + kv_h ^2 = 0                 \ $$
 * $$ mv_p ' + kv_p ^2 = mg                \ $$
 * $$ y(x) := y_h + y_p                          \ $$


 * $$ y(x) = mv_h ' + kv_h ^2 + mv_p ' + kv_p ^2 \ $$
 * $$ y_h + y_p = mg                             \ $$


 * $$ mv_h ' + kv_h ^2 + mv_p ' + kv_p ^2 = mg   \ $$
 * $$ m(v_h ' + v_p ') + k(v_h ^2 + v_p ^2) = mg \ $$
 * $$ m(v_h + v_p)' + k(v_h + v_p)^2 \neq mg     \ $$

Superposition false $$ \therefore $$ not linear. --Rachel Dzadek 19:53, 31 January 2012 (UTC)

Outflowing Water

 * $$ h' = -k \sqrt(h) \ $$

Order: 1st order Linearity (or lack of): not linear Superposition :
 * $$ y_h = h_h ' + k \sqrt h_h = 0         \ $$
 * $$ y_p = h_p ' = -k \sqrt h_p            \ $$
 * $$ y(x) := y_h + y_p                     \ $$


 * $$ y(x) = h_h ' + k \sqrt h_h + h_p '    \ $$
 * $$ y_h + y_p = -k \sqrt h_p              \ $$


 * $$ \frac {dh} {dt} = -kh^ \frac 1 2      \ $$
 * $$ \int h^ {- \frac 1 2} dh = - \int k dt \ $$
 * $$ 2h^ \frac 1 2 = -kt + c               \ $$
 * $$ h^ \frac 1 2 = - \frac {kt} 2 + c     \ $$
 * $$ h = ( - \frac {kt} 2 + c )^2          \ $$


 * $$ h(1) = ( - \frac k 2 + c )^2          \ $$
 * $$ h(2) = ( -k + c )^2                   \ $$
 * $$ h(3) = ( - \frac {3k} 2 + c )^2       \ $$
 * $$ h(1) \neq h(2) \neq h(3)              \ $$

Superposition false $$ \therefore $$ not linear. --Rachel Dzadek 23:00, 31 January 2012 (UTC)

Vibrating Mass on a Spring

 * $$ my'' + ky = 0 \ $$

Order: 2nd order Linearity (or lack of): linear Superposition :
 * $$ my'' + ky = 0                                                      \ $$
 * $$ y'' = - \frac k m y                                                \ $$

Let $$ y_1 \ $$ and $$ y_2 \ $$ be solutions.
 * $$ y_1 '' = - \frac k m y_1                                           \ $$
 * $$ y_2 '' = - \frac k m y_2                                           \ $$


 * $$ (ay_1 + by_2)'' = - \frac k m (ay_1 + by_2)                        \ $$
 * $$ ay_1  + by_2  = - \frac k m (ay_1 + by_2)                      \ $$
 * $$ a(- \frac k m y_1) + b(- \frac k m y_2) = - \frac k m (ay_1 + by_2) \ $$
 * $$ - \frac k m (ay_1 + by_2) = - \frac k m (ay_1 + by_2)              \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Beats of a Vibrating System

 * $$ y'' + \omega _0 ^2 y = \cos \omega t \ $$
 * $$ \omega_0 = \omega \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h = y_h '' + \omega _0 ^2 y_h = 0                                  \ $$
 * $$ y_p = y_p '' + \omega _0 ^2 y_p = \cos \omega t                      \ $$
 * $$ y(x) := y_h + y_p                                                    \ $$


 * $$ y(x) = y_h  + \omega _0 ^2 y_h + y_p  + \omega _0 ^2 y_p         \ $$
 * $$ y_h + y_p = 0 + \cos \omega t                                        \ $$


 * $$ y_h  + \omega _0 ^2 y_h + y_p  + \omega _0 ^2 y_p = \cos \omega t \ $$
 * $$ (y_h  + y_p ) + \omega _0 ^2 (y_h + y_p) = \cos \omega t         \ $$
 * $$ (y_h + y_p)'' + \omega _0 ^2 (y_h + y_p) = \cos \omega t             \ $$
 * $$ y'' + \omega _0 ^2 y = \cos \omega t                                 \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Current I in an RLC Circuit

 * $$ LI'' + RI' + \frac 1 C I = E' \ $$

Order: 2nd order Linearity (or lack of): linear Superposition:
 * $$ y_h = LI_h '' + RI_h ' + \frac 1 C I_h = 0                                        \ $$
 * $$ y_p = LI_p '' + RI_p ' + \frac 1 C I_p = E'                                       \ $$
 * $$ y(x) := y_h + y_p                                                                 \ $$


 * $$ y(x) = LI_h  + RI_h ' + \frac 1 C I_h + LI_p  + RI_p ' + \frac 1 C I_p        \ $$
 * $$ y_h + y_p = 0 + E'                                                                \ $$


 * $$ LI_h  + RI_h ' + \frac 1 C I_h + LI_p  + RI_p ' + \frac 1 C I_p = E'          \ $$
 * $$ L(I_h  + I_p ) + R(I_h ' + I_p ') + \frac 1 C (I_h + I_p) = E'                \ $$
 * $$ L(I_h + I_p)'' + R(I_h + I_p)' + \frac 1 C (I_h + I_p) = E'                       \ $$
 * $$ LI'' + RI' + \frac 1 C I = E'                                                     \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Deformation of a Beam

 * $$ EIy^{IX} = f(x) \ $$

Order: 4th order Linearity (or lack of): linear Superposition:
 * $$ y_h = EIy_h ^{IX} = 0           \ $$
 * $$ y_p = EIy_p ^{IX} = f(x)        \ $$
 * $$ y(x) := y_h + y_p               \ $$


 * $$ y(x) = EIy_h ^{IX} + EIy_p ^{IX} \ $$
 * $$ y_h + y_p = 0 + f(x)            \ $$


 * $$ EIy_h ^{IX} + EIy_p ^{IX} = f(x) \ $$
 * $$ EI(y_h ^{IX} + y_p ^{IX}) = f(x) \ $$
 * $$ EI(y_h + y_p)^{IX} = f(x)       \ $$
 * $$ EIy^{IX} = f(x)                 \ $$

Superposition true $$ \therefore $$ linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

Pendulum

 * $$ L \theta '' + g \sin \theta = 0 \ $$

Order: 2nd Linearity (or lack of): not linear Superposition:
 * $$ y_h = L \theta _h '' + g \sin \theta_h = 0                                      \ $$
 * $$ y_p = L \theta _p '' + g \sin \theta_p = 0                                      \ $$
 * $$ y(x) := y_h + y_p                                                               \ $$


 * $$ y(x) = L \theta _h  + g \sin \theta_h + L \theta _p  + g \sin \theta_p      \ $$
 * $$ y_h + y_p = 0 + 0                                                               \ $$


 * $$ L \theta _h  + g \sin \theta_h + L \theta _p  + g \sin \theta_p = 0         \ $$
 * $$ L( \theta _h  + \theta _p ) + g( \sin \theta_h + \sin \theta_p) = 0         \ $$
 * $$ L( \theta _h + \theta _p)'' + g( \sin \theta_h + \sin \theta_p) = 0             \ $$
 * $$ L \theta  + g( \sin \theta_h + \sin \theta_p) \neq L \theta  + g \sin \theta \ $$

Superposition false $$ \therefore $$ not linear. --Rachel Dzadek 00:15, 1 February 2012 (UTC)

=Effort Distribution=

--Corey Williams 15:58, 31 January 2012 (UTC)