User:Egm4313.s12.team15.r1

<Team 15

REPORT 1

= R1.1 =

Given
Derive the equation of motion of a spring-dashpot system in parallel (see Figure 1 below), with a mass and applied force $$f(t).$$ Figure 1

Find
The equation of motion of the system.

Solution
Kinematics, Kinetics, and Relations equations as interpreted from picture

Note: Notation of variables based upon p.1-4 notes.

Kinematics: $$y=y_{k}= y_{c}\!$$

Kinetics: $$f(t)=my'' + f_{I}\!$$ $$f_{I}\!$$ is the total damping force $$f_{I}=f_{k}+f_{c}\!$$

Relations:

-Force due to spring: $$f_{k}=ky_{k}\!$$ -Force due to dash-pot: $$f_{c}=cy'_{c}\!$$

Putting kinematics, kinetics, and relations together:

$$y=y_{k}=y''_{c}\!$$

$$f_{I}=f_{k}+f_{c}\!$$ Using kinetic relationship, replace variables to include relations equations: $$f(t)=my''_{c}+f_{k}+f_{c}\!$$

$$f_{k}=ky_{k}=ky_{c}\!$$ $$f_{c}=cy'_{c}\!$$

$$f(t)=my''_{c}+ky_{c}+cy'_{c}\!$$

'''Final Equation $$f(t)=my_{c}+ky_{c}+cy'_{c}\!$$'

Author
Solved and typed by Tim Pham 21:35, 25 January 2012 (UTC)

Reviewed By -Jenny Schulze

= R1.2 =

Given
Fig.53 K 2011 p.85



Find
Derive the equation of motion of the spring-mass-dashpot in Fig.52, in K 2011 p.85, with an applied force r(t) on the ball.

Solution
for ball:

$$F=ma=m\frac{\partial^2 y}{\partial t^2}=my''$$

for spring: (from hooke's law)

$$F=ky\!$$

for dashpot: (good approx for small velocities)

$$F=cy'\!$$

In this case, since there is an external force given it must be in equilibrum with the internal forces, it is called the driving force:

$$\sum F=r(t)$$

Thus:

$$my''+cy'+ky=r(t)\!$$

Author
Solved and typed by James Moncrief 01:28, 1 February 2012 (UTC)

Reviewed By - Jenny Schulze

= R1.3 =

Given
Spring-Dashpot-Mass System from Sec 1 lecture notes.



Find
Draw the free body diagrams for each component and derive the equation of motion for the mass. (Assume no additional force due to gravity.)

Solution
Free Body Diagrams

Equation of Motion for Mass

From Newton's Second Law:

$$\sum F = m\,\frac{\mathrm{d}v}{\mathrm{d}t} = ma$$

By definition:

$$a={y}''\!$$

So:

$$\sum F = m{y}''$$

From Mass FBD:

$$\sum F = f(t)- f_{I}$$

Combining the two equations:

$$m{y}'' = f(t)- f_{I}\!$$

'''Final equation after rearranging $$m{y} + f_{I} = f(t)\!$$'

Author
Solved and typed by Neil Tidwell 00:14, 29 January 2012 (UTC)

Reviewed By - Jenny Schulze

= R1.4 =

Given
$$ V=LC\frac{d^{2}v_{c}}{dt^{2}}+RC\frac{dv_{c}}{dt}+v_{c} \! $$ (1)

Find
Refer to Lecture 2 notes.

Derive (3) and (4) from (2).

Solution
Using $$I=C\frac{dv_{c}}{dt} \! $$ (2) & $$I'=C\frac{d^{2}v_{c}}{dt^{2}} \! $$(3)

After substituting equations (2 and (3) equation (1) becomes: $$V=LI'+RI+v_{c} \! $$ (4)

Deriving equation (4) yields: $$V'=LI''+RI'+\frac{1}{C}I\! $$

Note:

$$ \frac{dv_{c}}{dt}=\frac{1}{C}I\! $$

Using $$I=\frac{dQ}{dt} \! $$(5)

Deriving equation (5) becomes: $$I'=\frac{dQ^{2}}{dt^{2}} \! $$(6)

Substituting equations (5) and (6) into equation (4) yields: $$V=L\frac{dQ^{2}}{dt^{2}}+R\frac{dQ}{dt}+\frac{1}{C}Q\! $$

Note:

$$v_{c}=\frac{1}{C}Q=\frac{1}{C}\int I dt\! $$

Author
Solved and typed by - Jerry Shugart Reviewed By - Jenny Schulze

= R1.5 =

Given
Refer to the textbook Advanced Engineering Mathematics, 10th edition by Kreyszig, page 59.

$$ y''+ay'+by=0 \! $$

$$ \lambda ^2+a\lambda +b=0 \! $$

Roots:

$$ \lambda _1=\frac{1}{2}\sqrt{-a+(a^2-4b)} \!$$

$$ \lambda _2=\frac{1}{2}\sqrt{-a-(a^2-4b)} \!$$

Imaginary:

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

Double:

$$ y=(c_1+c_2 x)e^{-ax/2} \! $$

Find
4) $$ y''+4y'+(\pi ^{2}+4)y=0 \!$$

5) $$ y''+2\pi y'+\pi ^{2}y=0 \!$$

Solution
4) $$ y{}''+4y{}'+(\pi ^{2}+4)y=0 \!$$

$$y{}''+ay{}'+by=0 \!$$

$$a=4 \!$$

$$b=(\pi ^{2}+4) \!$$

$$\lambda^2+a\lambda+b=0 \!$$

$$\lambda^2+4\lambda+(\pi ^{2}+4)=0 \!$$

$$\lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b}) \!$$

$$\lambda_1=\frac{1}{2}(-4+\sqrt{4^2-4(\pi ^{2}+4)})=\frac{1}{2}(-4+\sqrt{-39.4784})=-2+\frac{1}{2}i(-39.4784) \!$$

$$\omega=39.4784 \!$$

$$\lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b}) \!$$

$$\lambda_2=\frac{1}{2}(-4-\sqrt{4^2-4(\pi ^{2}+4)})=\frac{1}{2}(-4-(-39.4784)^.5)=-2-\frac{1}{2}i(-39.4784) \!$$

$$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))=e^{-2x}(Acos(39.4784x)+Bsin(39.4784x)) \!$$

5) $$y{}''+2\pi y{}'+\pi ^{2}y=0 \!$$

$$y''+ay'+by=0 \!$$

$$a=2\pi \!$$

$$ b=\pi ^{2} \!$$

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

$$\lambda^{2}+(2\pi)\lambda+(\pi ^{2})=0 \!$$

$$\lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b}) \!$$

$$\lambda_1=\frac{1}{2}(-2\pi +\sqrt{(2\pi) ^{2}-4(\pi ^{2})})= -\pi\!$$

$$\omega=39.4784 \!$$

$$\lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b}) \!$$

$$\lambda_2=\frac{1}{2}(-2\pi -\sqrt{(2\pi) ^2-4(\pi^{2})})= -\pi\!$$

$$ y=(c_1+c_2 x)e^{-ax/2} \! $$

$$ y=(c_1+c_2 x)e^{-(2\pi)x/2}= (c_1+c_2 x)e^{-(\pi)x}\! $$

Author
Solved and typed by - Kristin Howe

Reviewed By - Jenny Schulze

= R1.6 =

Given
R1.6.1 $$ y''= g = const \!$$ R1.6.2 $$ mv'= mg -bv^2 \!$$ R1.6.3 $$ h'=-k\sqrt{h}\!$$ R1.6.4 $$ my''+ky = 0 \!$$ R1.6.5 $$ y'' +w_o^2y=coswt    \  \ \  \ \  \ \  \ \  \ \  \ \  \  w_o = w \!$$ R1.6.6 $$ LI'' + RI' + \frac{1}{c}I =E' \!$$ R1.6.7 $$ EI_y^{iv} = f(x) \!$$ R1.6.8 $$ L{\theta }'' + gsin\theta = 0 \!$$

Find
For each O.D.E. in Fig. 2 ( 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.

Solution
Refer to the lecture notes for additional information on the Superposition Principle: Section 2 Lecture Notes


 * Note : Lecture notes were referred to obtain the solutions below.

R1.6.1

Order: 2

Linearity: Linear

Superposition : Yes

$$ y''= g = const  \!$$   (Eq.1) $$ y''_h=0 \!$$       Homogeneous solution (Eq.2) $$ y''_p=g \!$$      Particular solution (Eq.3) $$ \bar{y}=y_h + y_p = (y_h + y_p) = 0 +g = g \!$$  (Eq.4) $$ \bar{y''}=g \!$$

Since (Eq.4) = (Eq.1) Therefore, superposition can be applied.

R1.6.2

Order: 1

Linearity: Non-linear

Superposition: No $$ mv'= mg - bv^2  \!$$  (Eq.0) $$ mv' + bv^2 = mg  \!$$  (Eq.0 Rearranged ) $$ mv_h'+b(v_h)^2=0 \!$$ Homogeneous solution (Eq. 1 ) $$ mv_p'+b(v_p)^2=mg \!$$ Particular solution (Eq. 2 )

By solving we obtain the following:

$$\bar{v}=v_h+v_p\!$$

Summing Eq.1 & Eq. 2 :

$$ m(v_h'+v_p')+ b[(v_h)^2 + (v_p)^2]=mg \!$$

$$m(v_h +v_p)' +b[(v_h)^2 +(v_p)^2]=mg\!$$

$$\bar{v'} + b(v_h+v_p)^2=mg\!$$

$$(v_h+v_p)^2\neq\bar{v}^2 \!$$

Therefore, superposition cannot be applied.

R1.6.3

Order: 1

Linearity:Non- linear

Superposition: No

$$ h'=-k\sqrt{h}\!$$  (Eq.0) $$ h'+k\sqrt{h}=0\!$$  (Eq.0 Rearranged) $${h_h}'+k\sqrt{h_h}= 0 \!$$ Homogeneous solution (Eq. 1 ) $${h_p}'+k\sqrt{h_p}= 0\!$$ Particular solution (Eq. 2)

By solving the following is obtained:

$$\bar{h}=h_h+h_p\!$$

Eq.1 & Eq. 2 are added:

$${h_h}'+k\sqrt{h_h}+{h_p}'+k\sqrt{h_p}=0\!$$

$$({h_h}'+{h_p}')+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$({h_h}+{h_p}) '+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$(\sqrt{h_h}+\sqrt{h_p})\neq \sqrt\bar{h}\!$$

Thus, superposition cannot be applied.

R1.6.4

Order:2

Linearity: Linear

Superposition: Yes

$$ my''+ky = 0 \!$$  (Eq.0) $$m{y}_h''+ky_h= 0\!$$ Homogeneous solution (Eq. 1 ) $$m{y}_p''+ky_p= 0\!$$ Particular solution (Eq. 2)

By solving the following is obtained:

$$\bar{y}=y_h+y_p\!$$

$$\bar{y}=y_h+y_p''\!$$

Summing Eq.1 & Eq. 2 :

$$m({y}_h+{y}_p)''+k(y_h+y_p) = 0\!$$

$$m\bar{y}''+k\bar{y}= 0\!$$

$$m\bar{y}+k\bar{y}$$ = $$ my+ky \!$$

Therefore, superposition can be applied.

R1.6.5

Order: 2

Linearity: Linear

Superposition: Yes

$$ y'' +w_o^2y=cos wt   \  \ \  \ \  \ \  \ \  \ \  \ \  \  w_o = w \!$$

$$ y_h'' +w_o^2y_h=0\!$$<p style="text-align:right">Homogeneous solution (Eq. 1 )

$$ y_p'' +w_o^2y_p=cos wt \!$$<p style="text-align:right">Particular solution (Eq. 2)

By solving we obtain the following:

$$\bar{y}=y_h+y_p\!$$

$$\bar{y}=y_h+y_p''\!$$

Adding Eq.1 & Eq.2 :

$$y_h+w_o^2y_h+y_p+ w_o^2y_p = cos wt\!$$

$$( y_h+y_p)''+w_o^2(y_h+y_p)= cos wt\!$$

$$\bar{y}''+w_o^2\bar{y}= cos wt\!$$

Therefore, superposition can be applied.

R1.6.6

Order: 2

Linearity : Linear

Superposition: Yes

$$ LI'' + RI' + \frac{1}{c}I =E' \!$$

$$ LI_h'' + RI_h' + \frac{1}{c}I_h =0 \!$$<p style="text-align:right">Homogeneous solution (Eq. 1 )

$$ LI_p'' + RI_p' + \frac{1}{c}I_p =E' \!$$<p style="text-align:right">Particular solution (Eq. 2 )

By solving we obtain the following:

$$\bar{I}= (I_h+I_p)\!$$

$$\bar{I}'= (I_h+I_p) '\!$$

$$\bar{I}= (I_h+I_p) \!$$

Adding Eq.1 & Eq. 2:

$$L(I_h+I_p) ''+R(I_h+I_p) ' + \frac{1}{c}(I_h+I_p) = E'\!$$

$$L\bar{I}''+R\bar{I}'+\frac{1}{c}\bar{I}= E'\!$$

Therefore, superposition can be applied.

R1.6.7

Order: 4

Linearity :Linear

Superposition : Yes

$$ EIy^{iv} = f(x) \!$$

$$ y_h= EIy_h^{iv}=0\!$$<p style="text-align:right"> Homogeneous solution (Eq. 1 )

$$ y_p= EIy_p^{iv}=f(x)\!$$<p style="text-align:right"> Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{y}=y_h+y_p\!$$

Summing Eq.1 & Eq. 2 :

$$ EIy_h^{iv} + EIy_p^{iv} = f(x)\!$$

$$ EI(y_h + y_p)^{iv}= f(x)\!$$

$$ EI\bar{y}^{iv} = f(x)\!$$

Therefore, superposition can be applied.

R1.6.8

Order: 2

Linearity: Non-linear

Superposition: No

$$ L{\theta }'' + gsin\theta = 0 \!$$ $$ L{\theta_h }'' + gsin\theta_h = 0 \!$$<p style="text-align:right"> Homogeneous solution (Eq. 1 ) $$ L{\theta_p }'' + gsin\theta_p = f(x) \!$$<p style="text-align:right"> Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{\theta}= {\theta}_h + {\theta}_p \!$$

Summing Eq. 1 & Eq. 2 :

$$L{\theta_h + \theta_p}''+gsin{\theta_h+\theta_p}=0\!$$

$$\bar{\theta}\neq\ {\theta}_h + {\theta}_p \!$$

Therefore, superposition cannot be applied.

Author
Solved and typed by - Cynthia Hernandez and Jenny Schulze

Reviewed By - Jenny Schulze

= Contributing Members =

Table code referenced from Team 1