User:Egm4313.s12.team13.r1

=Report 1=

Problem R1


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

First we examine the figure, and declare constants. $$ y \,\!$$ is the distance the coil expands, and k is the spring constant. Therefore, we are able to define a force $$ F_1\,\!$$ as

(1.0) $$ F_1= -ky\,\!$$.

The force is negative because in this model, the up position is negative and the down position is positive. This is a force opposing the motion of the mass, thus it is negative. Thereafter, we know Newton's second law as

$$ Force= mass* acceleration \,\!$$, or otherwise known as

$$ F=m\,\!$$$$ \frac{dV}{dt}\,\!$$

From calculus, it is also known that $$ \frac{dy}{dx}\,\!$$ can also be written as $$ y',\!$$. Furthermore, acceleration is known as

$$ \frac{dV}{dt}\,\!$$ $$ =\!$$ $$ \frac{d^2y}{dx^2}\,\!$$ $$ = \,\!$$ $$ y'',\!$$ (1.1)

If we sum the forces and equal them to zero we ca write Newtons second law for this system as shown below

$$ \sum{F}\!$$ $$ = \!$$$$ m\,\!$$$$ \frac{dV}{dt}\,\!$$ (1.2)

$$ my''+ky=0 \!$$ (Notice it is important to have all terms on the same side of the)

It is important to note, this force only includes the restoring force of the spring. It is now necessary to include the damper term from the dashpot. The damping constant will be defined as $$ C\!$$. It is also assumed the damping force is proportional to the velocity (Kreyszig, 2011). Therefore, we are able to write

$$ F_2= -c y',\!$$ (1.3)

and by summing forces one more time:

$$ \sum{F}\! = $$ $$ \frac{dV}{dt}\,\!$$ $$ \longrightarrow\!$$ $$F_1 + F_2=my \!$$ $$ \longrightarrow,\!$$$$ my+ cy' +ky =0 \!$$ (1.5)

To get the differential equation in standard form, the only thing required is to divide the equation by the mass.

Solution
$$ y''+ \frac{c}{m}y' +\frac{c}{m}y =0 \!$$

Since this is a parallel system, we are able to say the spring force relation and the damper force relation are equal to each other, which is the reason we don't differ from the $$ y's,\!$$ with subscripts.

Picture obtained from here Section 1 Lecture Notes

Problem R1.2
Derive the equation of motion of the spring-mass-dashpot in fig. 5.3 in K 2011 p 85 with an applied force r(t) on the ball.

In order to derive the equation of motion of the spring-mass-dashpot with an applied force of $$ \displaystyle r(t)$$, the formula of a non-homogenous equation must first be understood.

In the non homogenous equation:
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$$  \displaystyle y'' + p(x)y' + q(x)y = r(x) $$     (2.0) The $$\displaystyle y $$ represents position, $$ \displaystyle y' $$ represents velocity, and $$ \displaystyle y'' $$ represents acceleration.
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Figure 5.3 consist of a mass $$\displaystyle m$$, a spring constant $$\displaystyle c$$, and a damper coefficent $$\displaystyle k$$ These components create 3 internal forces- intertia $$\displaystyle my''$$, damping force $$\displaystyle cy'$$, and restoring force $$\displaystyle ky$$. Along with these 3 forces is also a driving force of $$ \displaystyle r(t)$$.

Solution
Using these 4 components, equation 1 becomes:
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$$\displaystyle my'' + cy' + ky = r(t)$$
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Section 1 Lecture Notes

Problem Statement
For the spring-dashpot-mass system on p.1-4 (Section 1-4 Notes) draw the FBDs and derive the equation of motion.

A free body diagram of the spring-mass-dashpot found in Sec 1 of the notes must be broken up into three different parts. The spring, the dashpot, and the mass.

The positive y direction equals $$ \overrightarrow{y+} \! $$

The spring's free body diagram is: $$ F_k \longleftarrow\! $$ $$ \longrightarrow\! F_k $$

The dashpot's free body diagram is: $$ F_k \longleftarrow\! $$ $$ \longrightarrow\! F_c $$

The free body diagram of the mass is: $$ F_c \longleftarrow\! $$ $$ \longrightarrow\! F(t) $$

Since there are only two opposing forces on the dashpot those two forces must be equal. :{| style="width:100%" border="0" $$  \displaystyle F_c = F_k = F_i $$     (3.0)
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By definition
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$$  \displaystyle ma=my'' $$     (3.1)
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Using Newtons second law :
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$$  \displaystyle \sum F = ma $$ (3.2)
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using substitution and equations (3.2), (3.1), and (3.0) we get
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$$  \displaystyle \sum F_y = F_(t) - F_i = my'' $$     (3.3)
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Rearranging equation (3.3) gives us the solution

Solution
$$ \displaystyle F(t) = F_i + my''$$

Section 1-4 Notes

Problem Statement
Using the circuit given, the governing circuit equation (4.0), and the capacitance equation (4.1), derive alternate equations (4.2) and (4.3).

Circuit Equation


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$$  \displaystyle V=LC\frac{{d}^{2}{v}_{c}}{d{t}^{2}}+RC\frac{d{v}_{c}}{dt}+{v}_{c} $$     (4.0)
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Capacitance Equation


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$$  \displaystyle Q = C{v}_{c} \rightarrow \int idt = C{v}_{c} \rightarrow i=C\frac{d{v}_{c}}{dt} $$     (4.1)
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Alternate Equation (4.2),


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$$  \displaystyle LCI''+RCI'+\frac{1}{C}I = V' $$ (4.2)
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Alternate Equation (4.3),


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$$  \displaystyle LCQ''+RCQ'+\frac{1}{C}Q = V $$ (4.3)
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Solution
First take the derivative of the governing circuit equation (4.0) to get


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$$  \displaystyle V' = LC\frac{{d}^{3}v_{c}}{d{t}^{3}}+RC\frac{{d}^{2}v_{c}}{d{t}^{2}}+\frac{{d}^{}v_{c}}{d{t}^{}} $$     (4.4)
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Since L, R, and C are constants with respect to time

Then, since


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$$  \displaystyle I = C\frac{dv_{c}}{dt} $$     (4.1)
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that means


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$$  \displaystyle I' = C\frac{{d}^{2}v_{c}}{d{t}^{2}} $$     (4.5)
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And


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$$  \displaystyle I'' = C\frac{{d}^{3}v_{c}}{d{t}^{3}} $$     (4.6)
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Substituting these into the derivative of the circuit equation (4.4) gives the first alternate equation (4.2)

$$  \displaystyle V' = LCI''+RCI'+\frac{1}{C}I $$

Then, since


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$$  \displaystyle Q=Cv_{c} $$     (4.1)
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$$  \displaystyle Q' = C\frac{d{v}_{c}}{dt} $$     (4.7)
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$$  \displaystyle Q'' = C\frac{{d}^{2}{v}_{c}}{d{t}^{2}} $$     (4.8)
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Now substituting these into the appropriate places in the circuit equation (4.0) gives the second alternate equation (4.3)

$$ LQ''+RQ'+\frac{1}{C}Q = V $$

IEA notes section 2

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

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

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is

$$ \lambda^2 + a\lambda + b = 0\,\!$$<p style="text-align:right">(5.1a)

For the ODE in this problem, the characteristic equation becomes

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

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$<p style="text-align:right">(5.2a)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$<p style="text-align:right">(5.3a)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$<p style="text-align:right">(5.4a)

$$ \lambda_{2} = -2 - \pi i \,\!$$<p style="text-align:right">(5.5a)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$<p style="text-align:right">(5.6a)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$<p style="text-align:right">(5.7a)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$<p style="text-align:right">(5.8a)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$<p style="text-align:right">(5.9a)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$<p style="text-align:right">(5.10a)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$<p style="text-align:right">(5.11a)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$<p style="text-align:right">(5.12a)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)}$$

$$ + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} +

\cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 2 Lecture Notes

Problem Statement
5. Find a general solution. Check by substitution.

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

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi ^{2}y=0   $$ <p style="text-align:right">(5.0b)

If $$ \; \frac{d}{dx}\ = \lambda $$,

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$ <p style="text-align:right">(5.1b)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$ <p style="text-align:right">(5.2b)

$$ \; \lambda =-\pi, -\pi $$

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2xe ^ { \lambda _2 x}  $$ <p style="text-align:right">(5.3b)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$ <p style="text-align:right">(5.4b)

Check by Substitution:

To check by substitution, you find $$  \; y, y', y'' $$ and plug into the original ODE.

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$<p style="text-align:right">(5.5b)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$ <p style="text-align:right">(5.6b)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 2 Notes

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.

Background Information
Superposition Principle

An ordinary differential equation in standard form:

$$ y''+p(x)y'+q(x)y=r(x)\!$$

The homogeneous solution of the function is in the form:

$$ y_h''+p(x)y_h'+q(x)y_h=0 \!$$

The particular solution of the function is in the form:

$$ y_p''+p(x)y_p'+q(x)y_p=r(x) \!$$

The function, $$ \bar y(x) \!$$, is the sum of the homogeneous solution, $$ y_h(x)\!$$ and the particular solution $$y_p(x)\!$$:

$$ \bar y(x) = y_h(x)+y_p(x) \!$$

To prove the superposition principle, we must prove that $$ \bar y \!$$ is a solution. We then plug in $$ \bar y \!$$ into the standard form of the ordinary differential equation:

$$ (y_h+y_p)+p(x)(y_h'+y_p')+q(x)(y_h+y_p)=r(x) \!$$

If the function is linear:

$$ y_h'+y_p'=\bar y' \!$$ and $$ y_h+y_p=\bar y'' \!$$

If the superposition principle holds, the function should yield:

$$ \bar y''+p(x)\bar y'+q(x)\bar y=r(x) \!$$

Falling Stone Problem
$$ y''=g=constant \!$$

Order: 2nd  Linearity: Linear

The homogeneous equation is:

$$ y_h''=0\!$$<p style="text-align:right"> (6.1)

The particular equation is:

$$ y_p''=g \!$$<p style="text-align:right"> (6.2)

Adding equations 6.1 and 6.2 yields:

$$ y_h+y_p=g \!$$<p style="text-align:right"> (6.3)

Simplifying equation 6.3:

$$ (y_h+y_p)''=g \!$$<p style="text-align:right"> (6.4) Which equals:

$$ \bar y''=g \!$$<p style="text-align:right"> (6.5)

Superposition principle can be applied

Parachutist Problem
$$ mv'=mg-bv^2 \!$$

Order:1st Linearity Non-linear

The homogeneous equation is:

$$ mv_h'+bv_h^2=0 \!$$<p style="text-align:right"> (6.6)

The particular equation is:

$$ mv_p'+bv_p^2=mg \!$$<p style="text-align:right"> (6.7)

Add equations 6.6 and 6.7:

$$ (mv_h'+mv_p')+(bv_h^2+bv_p^2)=mg \!$$<p style="text-align:right">(6.8)

Simplify equation 6.8:

$$ m(v_h+v_p)'+b(v_h^2+v_p^2)=mg \!$$<p style="text-align:right"> (6.9)

Which equals: $$ m(\bar v)'+b(v_h^2+v_p^2)=mg \!$$<p style="text-align:right"> (6.10)

Since $$ b(v_h^2+v_p^2)\ne b(v_h+v_p)^2\!$$ the superposition principle can not be applied.

Water Level Problem
$$ h'=-k\sqrt{h} \!$$

Order:1st  Linearity: Non-linear

The homogeneous equation is:

$$ h_h'+k\sqrt{h_h}=0 \!$$<p style="text-align:right"> (6.11)

The particular equation is:

$$ h_p'+k\sqrt{h_p}=0 \!$$<p style="text-align:right"> (6.12)

Adding equations 6.11 and 6.12 yields:

$$ (h_p'+h_h')+(k\sqrt{h_p}+k\sqrt{h_h})=0 \!$$<p style="text-align:right"> (6.13)

Since $$ k\sqrt{h_h}+k\sqrt{h_p}\ne k\sqrt{h_h+h_p} \!$$ the superposition principle can not be applied

Vibrating Mass on a Spring Problem
$$ my''+ky=0 \!$$

Order: 2nd  Linearity: Linear

The homogeneous equation is:

$$ my_h''+ky_h=0 \!$$<p style="text-align:right"> (6.14)

The particular equation is:

$$ my_p''+ky_p=0 \!$$<p style="text-align:right"> (6.15)

Adding equations 6.14 and 6.15 yields:

$$ m(y_h+y_p)+k(y_p+y_h)=0 \!$$<p style="text-align:right"> (6.16)

Simplifying the equation 6.16:

$$ m(y_h+y_p)''+k(y_p+y_h)=0 \!$$<p style="text-align:right"> (6.17)

Which equals:

$$ m(\bar y'')+k(\bar y)=0 \!$$<p style="text-align:right"> (6.18)

Superposition principle can be applied.

Beats of a Vibrating System Problem
$$ y''+w_0y=\!$$$$ \cos(wt) \!$$

Order: 2nd  Linearity: Linear

The homogeneous equation is:

$$ y_h''+w_0y_h=0\!$$<p style="text-align:right"> (6.19)

The particular equation is:

$$ y_p''+w_0y_p=\!$$$$ \cos(wt) \!$$<p style="text-align:right"> (6.20)

Adding equations 6.19 and 6.20 yields:

$$ (y_h+y_p)+w_0(y_h+y_p)=\!$$$$ \cos(wt) \!$$<p style="text-align:right"> (6.21)

Simplifying equation 6.21:

$$ (y_h+y_p)''+w_0(y_h+y_p)=\!$$$$ \cos(wt) \!$$<p style="text-align:right"> (6.22)

Which equals:

$$ (\bar y'')+w_0(\bar y)=\!$$$$ \cos(wt) \!$$<p style="text-align:right"> (6.23)

The superposition principle can be applied

Current (I) in an RLC Circuit Problem


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

Order: 2nd  Linearity: Linear

The homogeneous equation is: $$ LI_h''+RI_h'+\frac{1}{C}I_h=0 \!$$<p style="text-align:right"> (6.24)

The particular equation is: $$ LI_p''+RI_p'+\frac{1}{C}I_p=E' \!$$<p style="text-align:right"> (6.25)

Adding equations 6.24 and 6.25 yields: $$ L(I_h+I_p)+R(I_h'+I_p')+\frac{1}{C}(I_h+I_p)=E' \!$$<p style="text-align:right"> (6.26)

Simplifying equation 6.26:

$$ L(I_h+I_p)''+R(I_h+I_p)'+\frac{1}{C}(I_h+I_p)=E' \!$$<p style="text-align:right"> (6.27)

Which equals:

$$ L(\bar I)''+R(\bar I)'+\frac{1}{C}(\bar I)=E' \!$$<p style="text-align:right"> (6.28)

The superposition principle can be applied

Deformation of a Beam Problem


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

Order: 4th  Linearity: Linear

The homogeneous equation is:

$$ EIy_h^{iv}=0 \!$$<p style="text-align:right"> (6.29)

The particular equation is:

$$ EIy_p^{iv}=f(x) \!$$<p style="text-align:right"> (6.30)

Adding equations 6.29 and 6.30 yields:

$$ EI(y_h^{iv}+y_p^{iv})=f(x) \!$$<p style="text-align:right"> (6.31)

Simplifying equation 6.31:

$$ EI(\bar y^{iv})=f(x) \!$$<p style="text-align:right"> (6.32)

The superposition principle can be applied.

Pendulum Problem


$$ L\theta''+g\sin(\theta)=0\!$$

Order: 2nd  Linearity: Non-linear

The homogeneous equation is:

$$ L\theta_h''+g\sin(\theta_h)=0 \!$$<p style="text-align:right"> (6.33)

The particular equation is:

$$ L\theta_p''+g\sin(\theta_p)=0 \!$$<p style="text-align:right"> (6.34)

Adding equations 6.33 and 6.34 yields:

$$ L(\theta_h+\theta_p)+g(\sin(\theta_h)+\sin(\theta_p))=0 \!$$<p style="text-align:right"> (6.35)

Simplifying equation 6.35: $$ L(\theta_h+\theta_p)''+g(\sin(\theta_h)+\sin(\theta_p))=0 \!$$<p style="text-align:right"> (6.36)

Since $$\sin(\theta_h)+\sin(\theta_p) \ne \sin(\theta_h+\theta_p) \!$$ the superposition principle does not apply.

Reference:

Section 2 Lecture Notes Photos taken from Kreyszig, Erwin. Advanced Engineering Mathematics. 10th. Hoboken: John Wiley & Sons, Inc., 2011. 3. eBook.

= Contributing Members =

Table code referenced from Team 1