User:Egm4313.s12.team14.report1

= Problem 1 =

Given
A spring-dashpot system in parallel, with a mass and an applied force f(t)

Find
Equation of motion

Solution
Free Body Diagram

Kinematics:


 * {| style="width:100%" border="0" align="left"

y=y_k=y_c $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Spring Force:


 * {| style="width:100%" border="0" align="left"

\vec{F}_s=k(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Dashpot Force:


 * {| style="width:100%" border="0" align="left"

\vec{F}_D=c(y')) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Applied Force:


 * {| style="width:100%" border="0" align="left"

\vec{F}(t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Forces:


 * {| style="width:100%" border="0" align="left"

\Sigma\vec{F}=\vec{F}(t)-\vec{F}_s-\vec{F}_D $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

m\vec{a}=\vec{F}_i=my'' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

From Newton's 2nd Law:


 * {| style="width:100%" border="0" align="left"

\Sigma\vec{F}=m\vec{a} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Substituting $$ (Eq. 1)$$ and $$ (Eq. 2)$$ into $$ (Eq. 3)$$:


 * {| style="width:100%" border="0" align="left"

\vec{F}(t)-\vec{F}_s-\vec{F}_D=\vec{F}_i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Solving for $$\vec{F}(t)$$:


 * {| style="width:100%" border="0" align="left"

\vec{F}(t)=\vec{F}_s+\vec{F}_D+\vec{F}_i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Substituting $$\vec{F}_s=ky$$, $$\vec{F}_D=cy'$$, and $$\vec{F}_i=my''$$ into $$ (Eq. 4)$$:


 * {| style="width:100%" border="0" align="left"

\vec{F}(t)=ky+Cy'+my'' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Putting into standard form:

$$my''+Cy'+ky=\vec{F}(t)$$

= Problem 2 =

Given
Spring-mass-dashpot in Fig. 53 with an applied force r(t) on the ball

Find
Equation of motion

Solution
FBD:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{F_k = ky}$$
 * $$\displaystyle (Eq. 1) $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{F_c = cy'}$$
 * $$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

Newton's Second Law:

$$\displaystyle{\Sigma\vec{F}=m\vec{a}}$$

From the FBD:


 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{ma = -F_k - F_c + r(t)}$$
 * $$\displaystyle (Eq. 3) $$
 * }
 * }
 * }

Plug (Eq. 1) and (Eq. 2) into (Eq. 3).

$${my'' + ky + cy' = r(t)}$$

= Problem 3 =

Find
Free body diagrams and equation of motion

Solution

 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{T_{1}=T_{2}=ky_{k}}$$
 * $$\displaystyle (Eq. 1) $$
 * }
 * {| style="width:100%" border="0" align="left"
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{T_{2}=T_{3}=cy{}'_{c}}$$
 * $$\displaystyle (Eq. 2) $$
 * }
 * {| style="width:100%" border="0" align="left"
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle{T_{3}=f(t) \; where\; T_{3}=ky_{k}=cy{}'_{c}}$$
 * $$\displaystyle (Eq. 3) $$
 * }
 * }
 * }

Thus, through Newton's 2nd Law F = ma

$$\displaystyle f(t)=my{}''+T_{3}$$

= Problem 4 =

Given


Circuit consisting of a Resistor with resistance R, an Inductor with inductance L, and a capacitor with capacitance C in series

Find
Derive Equations 1 and 2 from Equation 3.
 * {| style="width:100%" border="0" align="left"

{V}'=L{I}'' + R{I}' + \frac{1}{C}I $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

V=L{Q}'' + R{Q}' + \frac{1}{C}Q $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }


 * {| style="width:100%"

V=LC{v_c}'' + RC{v_c}' + v_c $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Solution
Begin with Equation 3
 * {| style="width:100%" border="0" align="left"

V=LC{v_C}'' + RC {v_C}' + v_C, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

and differentiate w.r.t. time to yield
 * {| style="width:100%" border="0" align="left"

{V}'=LC {v_C}' + RC {v_C} + {v_C}'. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Take the Equation for capacitance,
 * {| style="width:100%" border="0" align="left"

{v_C}'= \frac{1}{C}I, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

and differentiate w.r.t. time twice to yield
 * {| style="width:100%" border="0" align="left"

{v_C}''=\frac{1}{C}{I}' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

and
 * {| style="width:100%" border="0" align="left"

{v_C}'=\frac{1}{C}{I}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Substituting Equations 5, 6, and 7 into Equation 4 yields
 * {| style="width:100%" border="0" align="left"

{V}'=L{I}'' + R{I}' + \frac{1}{C}I. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (ans. Eq. 1)
 * }
 * }

Integrate Equation 1 w.r.t. time to yield:
 * {| style="width:100%" border="0" align="left"

V=L{I}' + R{I} + \frac{1}{C}\int I. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Take the Equation for Current in terms of charge
 * {| style="width:100%" border="0" align="left"

\int I = Q, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

and differentiate twice to yield
 * {| style="width:100%" border="0" align="left"

I = {Q}', $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

and
 * {| style="width:100%" border="0" align="left"

{I}' = {Q}''. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

Substituting Equations 9, 10, and 11 into Equation 8 yields


 * {| style="width:100%" border="0" align="left"

V=L{Q}'' + R{Q}' + \frac{1}{C}Q. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (ans. Eq. 2)
 * }
 * }

= Problem 5 =

Given
(a)
 * {| style="width:100%" border="0" align="left"

{y}'' + 4{y}' + ({\pi}^2 + 4)y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Given
(b)
 * {| style="width:100%" border="0" align="left"

{y}'' + 2{\pi}{y}' + {{\pi}^2}y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution for this homogeneous differential equation.

Solution
(a) Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle {{\lambda}^2} e^{\lambda x} + 4\lambda e^{\lambda x} + (\pi^2 + 4)e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + 4\lambda +(\pi^2 + 4)) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle (\lambda^2 + 4\lambda +(\pi^2 + 4)) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(\pi^2 + 4)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-4 \pm \sqrt{16 - 4\pi^2 - 16)}}{2}$$ $$\displaystyle \lambda = \frac{-4 \pm \sqrt{-4\pi^2)}}{2}$$ $$\displaystyle \lambda = -2 \pm i{\pi}$$

A solution of the form $$\displaystyle y_h = e^{\lambda z}$$ where z is a complex number of the form $$\displaystyle a + ib$$ requires the use of Euler's Formula, where $$\displaystyle e^{\lambda z} = e^{a}(\cos{b} + i\sin{b})$$

$$\displaystyle \therefore y_{h1} = e^{-2x}(\cos{\pi x} + i\sin{\pi x})$$ $$\displaystyle y_{h2} = e^{-2x}(\cos{-\pi x} + i\sin{-\pi x})$$ Finally $$\displaystyle y_h = C_1[e^{-2x}(\cos{\pi x} + i\sin{\pi x})] + C_2[e^{-2x}(\cos{-\pi x} + i\sin{-\pi x})]$$

(b) Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle \lambda^2 e^{\lambda x} + {2\pi}\lambda e^{\lambda x} + {\pi^2}e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + {2\pi}\lambda +{\pi^2}) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle (\lambda^2 + {2\pi}\lambda + \pi^2) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-(2\pi) \pm \sqrt{(2\pi)^2 - 4(1)(\pi^2)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-2\pi \pm \sqrt{4{\pi}^2 - 4{\pi^2}}}{2}$$ $$\displaystyle \lambda = -\pi$$ $$\displaystyle \therefore y_{h1} = e^{-\pi x}$$

Since we only obtain one independent solution from the repeated root, we must use the method of reduction of order and assume that $$\displaystyle y_{h2} = {x}e^{-\pi x}$$ $$\displaystyle \therefore y_h = {C_1}e^{-\pi x} + {C_2}{x}e^{-\pi x}$$ $$\displaystyle y_h = (C_1 + {C_2}{x})e^{-\pi x}$$

= Problem 6 =

Given
The following differential equations are provided for analysis.

(A) Falling stone:

$$\displaystyle y'' = g = C$$ Where C is a constant.

(B) Parachutist



$$\displaystyle mv' = mg - bv^{2}$$

(C) Outflowing water (where h is the water level)



$$\displaystyle h' = -k\sqrt{h}$$

(D) Vibrating Mass on a Spring



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

(E) Beats of a Vibrating System



$$\displaystyle y'' + \omega_{0} ^{2}y = cos(\omega t), \omega _{0} = \omega$$

(F) Current I in an RLC Circuit



$$\displaystyle LI'' + RI' + (1/C)I = E'$$

(G) Deformation of a Beam



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

(H) Pendulum



$$\displaystyle L\theta'' +gsin(\theta) = 0$$

Find
Determine the order and linearity of the equation. Next, show whether the principle of superposition can be applied.

(A) Falling Stone
($$ \displaystyle y $$ represents height above the ground)
 * {| style="width:100%" border="0" align="left"

This governing equation is SECOND order and LINEAR
 * $$\displaystyle{y'' = g}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y} = y_{h} + y_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle {y_h}'' = 0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle {y_p}'' = g $$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle {y_h} + {y_p} = g$$
 * }

Plugging $$\displaystyle \bar{y} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle ({y_h} + {y_p})'' = g $$
 * }

This result is equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is applicable and that the equation is LINEAR
 * $$\displaystyle {y_h} + {y_p} = (y_h + y_p)'', $$
 * }

(B) Parachutist
($$\displaystyle v $$ represents velocity in the vertical direction)
 * {| style="width:100%" border="0" align="left"

This governing equation is FIRST order and NON-LINEAR
 * $$\displaystyle{mv' = mg - bv^2}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{v} = v_{h} + v_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle m{v_h}' = - b{v_h}^2 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle m{v_p}' = mg - b{v_p}^2 $$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m({v_h}' + {v_p}') = mg -b({v_h}^2 + {v_p}^2)$$
 * }

Plugging $$\displaystyle \bar{v} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle m(v_h + v_p)' = mg - b(v_h + v_p)^2 $$
 * }

This result is not equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is not applicable and that the equation is NON-LINEAR
 * $$\displaystyle {y_h}^2 + {y_p}^2 \neq(y_h + y_p)^2, $$
 * }

(C) Outflowing Water
($$\displaystyle h $$ represents the water level)
 * {| style="width:100%" border="0" align="left"

This governing equation is FIRST order and NON-LINEAR
 * $$\displaystyle{h' = -k\sqrt{h}}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{h} = h_{h} + h_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle h'_h + k\sqrt{h_h} = 0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle h'_p + k\sqrt{h_p} = 0$$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle h'_h + h'_p +k(\sqrt{h_h} +\sqrt{h_p}) = 0$$
 * }

Plugging $$\displaystyle \bar{h} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


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

This result is not equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is not applicable and that the equation is NON-LINEAR
 * $$\displaystyle \sqrt{h_h} +\sqrt{h_p} \neq \sqrt{h_h + h_p}, $$
 * }

(D) Vibrating Mass on a Spring
(where $$\displaystyle y $$ represents displacement)
 * {| style="width:100%" border="0" align="left"

This governing equation is SECOND order and LINEAR
 * $$\displaystyle{m{y}''+ky = 0}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y} = y_{h} + y_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle my_{h}{}''+ky_{h}=0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle my_{p}{}''+ky_{p}=0 $$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle my_{h}{}+my_{p}{}+ky_{h}+ky_{p}=0$$
 * }

Plugging $$\displaystyle \bar{y} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


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

This result is equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is applicable and that the equation is LINEAR
 * $$\displaystyle (y_{h}{}+y_{p}{})=(y_h + y_p)'', $$
 * }

(E) Beats of a Vibrating System
($$\displaystyle y $$ represents displacement from the midpoint)
 * {| style="width:100%" border="0" align="left"

This governing equation is SECOND order and LINEAR
 * $$\displaystyle{{y}''+\omega _{0}^{2}y = \cos\omega t, \omega_{0}=\omega}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y} = y_{h} + y_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle y_{h}{}''+\omega _{0}^{2}y_{h} =0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle y_{p}{}''+\omega _{0}^{2}y_{p} =\cos\omega t $$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y_{h}{}+y_{p}{}+\omega _{0}^{2}(y_{h})+\omega _{0}^{2}(y_{p})=\cos\omega t$$
 * }

Plugging $$\displaystyle \bar{y} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle (y_{h}+y_{p}){}''+\omega _{0}^{2}(y_{h}+y_{p}) = \cos\omega t$$
 * }

This result is equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is applicable and that the equation is LINEAR
 * $$\displaystyle y_{h}{}+y_{p}{}=(y_{h}+y_{p}){}'', $$
 * }

(F) RLC Current
($$\displaystyle I $$ represents current)
 * {| style="width:100%" border="0" align="left"

This governing equation is SECOND order and LINEAR
 * $$\displaystyle{LI'' + RI' + (1/C)I = E'}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{I} = I_{h} + I_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle LI_h'' + RI_h' + (1/C)I_h = 0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle LI_p'' + RI_p' + (1/C)I_p = E' $$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L({I_{h}} + {I_{p}}) + R({I_{h}}' + {I_{p}}') + (1/c)(I_{h} + I_{p}) = {E}'$$
 * }

Plugging $$\displaystyle \bar{I} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L({I_{h}} + {I_{p}})'' + R({I_{h}} + {I_{p}})' + (1/c)(I_{h} + I_{p}) = {E}'$$
 * }

This result is equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle I_{h}{}+I_{p}{}=(I_{h}+I_{p}){}'', $$
 * }
 * {| style="width:100%" border="0" align="left"

which proves that superposition is applicable and that the equation is LINEAR
 * $$\displaystyle I_{h}{}'+I_{p}{}'=(I_{h}+I_{p}){}', $$
 * }

(G) Deformation of a Beam
($$\displaystyle y $$ represents displacement)
 * {| style="width:100%" border="0" align="left"

This governing equation is FOURTH order and LINEAR
 * $$\displaystyle{EIy^{iv}=f(x)}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{y} = y_{h} + y_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle EIy_h^{iv}=0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle EIy_p^{iv}=f(x) $$
 * }
 * {| style="width:100%" border="0" align="left"


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

Plugging $$\displaystyle \bar{y} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


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

This result is equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is applicable and that the equation is LINEAR
 * $$\displaystyle y_{h}^{iv}+I_{p}^{iv}=(y_{h}+y_{p})^{iv}, $$
 * }

(H) Pendulum
($$\displaystyle \theta $$ represents the angle with respect to vertical axis)


 * {| style="width:100%" border="0" align="left"

This governing equation is SECOND order and NON-LINEAR
 * $$\displaystyle{L{\theta}'' + g{\sin(\theta)} = 0}$$
 * }

We begin with the solution
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle \bar{\theta} = \theta_{h} + \theta_p. $$
 * }

Plugging the homogeneous and particular components individually into the governing equation yields
 * {| style="width:100%" border="0" align="left"

and
 * $$\displaystyle L{{\theta}_h}'' + g{\sin({\theta}_h)} = 0 $$
 * }
 * {| style="width:100%" border="0" align="left"

Adding these equations together, we obtain
 * $$\displaystyle L{{\theta}_p}'' + g{\sin({\theta}_p)} = 0 $$
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L{{\theta}_p} + L{{\theta}_h} + g{\sin({\theta}_p)} + g{\sin({\theta}_h)} = 0$$
 * }

Plugging $$\displaystyle \bar{\theta} $$ into the governing equation yields
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle L({\theta}_p +{\theta}_h)'' + g{\sin({\theta}_p + {\theta}_h)} = 0$$
 * }

This result is not equal to the previous equation because
 * {| style="width:100%" border="0" align="left"

which proves that superposition is not applicable and that the equation is NON-LINEAR
 * $$\displaystyle \sin({\theta}_p) + \sin({\theta}_h) \neq \sin({\theta}_p + {\theta}_h), $$
 * }

=Contributions=

Team Member Tasks
All team members contributed to the coding of this page.