User:Egm4313.s12.team20/R1

Statement
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$\displaystyle f(t). $$ The spring-dashpot system is shown in Figure 1.

Figure 1

Solution
Kinematics:
 * {| style="width:100%" border="0"|-

y = y_{k} = y_c$$
 * style="width:95%" | $$ \displaystyle
 * (1.0)
 * }

Kinetics:
 * {| style="width:100%" border="0"|-

f(t) = my'' + f_1 $$
 * style="width:95%" | $$ \displaystyle
 * (1.1)
 * }


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


 * style="width:95%" | || (1.2)
 * }

By Substitution of $$ \displaystyle f_1 $$ into equation (1.1) we obtain
 * {| style="width:100%" border="0"|-

f(t) = my'' + f_k + f_c $$ Constitutive Relations: -Force of Spring
 * style="width:95%" | $$ \displaystyle
 * (1.3)
 * }
 * {| style="width:100%" border="0"|-

f_k = ky_k $$
 * style="width:95%" | $$ \displaystyle
 * (1.4)
 * }

-Force of Dash-Pot
 * {| style="width:100%" border="0"|-

f_c = cy'_c$$ Because of the kinematic relations where $$ \displaystyle y_c = y_k $$ we manipulate equation (1.4) and obtain
 * style="width:95%" | $$\displaystyle
 * (1.5)
 * }
 * {| style="width:100%" border="0"|-

f_k = ky_k = ky_c$$
 * style="width:95%" | $$\displaystyle
 * (1.6)
 * }

Therefore, by subsitution into equation (1.2)
 * {| style="width:100%" border="0"|-


 * style="width:95%" | || (1.7)
 * }

After finally substituting equation (1.7) into the original f(t) equation (1.1), the final equation equals

Given
Figure 2

Solution
Kinematics:
 * {| style="width:100%" border="0"|-

y=y_k=y_c$$
 * style="width:95%" | $$ \displaystyle
 * (2.0)
 * }

Kinetics:


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

r(t)=-my''-f_k-f_c$$
 * style="width:95%" | $$ \displaystyle
 * (2.1)
 * }


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

f_k=ky_k$$
 * style="width:95%" | $$ \displaystyle
 * (2.2)
 * }


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

f_c=cy_c'$$
 * style="width:95%" | $$ \displaystyle
 * (2.3)
 * }

Given
Free body diagram:



Solution
By definition:

$$\displaystyle a=y'' $$

(3.1)

$$\displaystyle f_k=ky_k$$ (3.2)

$$\displaystyle f_c=cy_c' $$ (3.3)

Since the spring, dashpot and mass are aligned, the forces on the dashpot are equal and opposite and can be described as the internal forces.

$$ \displaystyle f_k=f_c=f_I $$ (3.4)

The net force of the system is the difference between the applied force and the internal force.

$$ \displaystyle F_s=f(t)-f_I $$ (3.5)

Applying Newtons 2nd law:

$$\displaystyle F_s=m*a $$  (3.6)

Plug in equation (3.1) and (3.5) into equation (3.6):

$$\displaystyle

my''=f(t)-f_I $$  (3.7)

Rearrange:

Given
Circuit Equation:
 * {| style="width:100%" border="0"


 * style="width:95%" | $$ V = LC\frac{d^2v_{c}}{dt^2} + RC\frac{dv_{c}}{dt} + v_{c}$$
 * (4.0)
 * }

Capacitance:
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ Q=Cv_{c} \rightarrow \int idt=Cv_{c} \rightarrow i = C\frac{dv_{c}}{dt}$$
 * (4.1)
 * }

Solution
Alternate Circuit Equation 1

Current Equation:
 * {| style="width:100%" border="0"


 * style="width:95%" | $$ I = C\frac{dv_{c}}{dt}$$
 * <p style="text-align:right">(4.2)
 * }

Derive Equation 4.2:
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ I' = C\frac{d^2v_{c}}{dt^2}$$
 * <p style="text-align:right">(4.3)
 * }

Rearrange the Current Intergral from Equation 4.1:
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ \int idt=Cv_{c} \rightarrow v_{c}=\frac{1}{C}\int idt $$
 * <p style="text-align:right">(4.4)
 * }

Substitute Equations 4.2, 4.3 and 4.4 into Equation 4.0:
 * {| style="width:100%" border="0"|-

$$ \Downarrow $$
 * style="width:95%" | $$ V = LC\frac{d^2v_{c}}{dt^2} + RC\frac{dv_{c}}{dt} + v_{c}$$
 * }
 * {| style="width:100%" border="0"|-
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ V = LI'+ RI + \frac{1}{C}\int idt $$
 * <p style="text-align:right">(4.5)
 * }

Derive Equation 4.5 to find Alternate Circuit Equation 1:
 * {| style="width:100%" border="0"|-

$$ \Downarrow $$
 * style="width:95%" | $$ V = LI'+ RI + \frac{1}{C}\int idt $$
 * }
 * }

Alternate Circuit Equation 2

Charge Equation:
 * {| style="width:100%" border="0"


 * style="width:95%" | $$ Q = Cv_{c} \ $$
 * <p style="text-align:right">(4.7)
 * }

Derive Equation 4.7:
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ Q'=C\frac{dv_{c}}{dt}$$
 * <p style="text-align:right">(4.8)
 * }

Derive Equation 4.8:
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ Q''=C\frac{d^2v_{c}}{dt^2}$$
 * <p style="text-align:right">(4.9)
 * }

Rearrange the Charge equation from Equation 4.1:
 * {| style="width:100%" border="0"|-


 * style="width:95%" | $$ Q = Cv_{c} \rightarrow v_{c}=\frac{Q}{C}$$
 * <p style="text-align:right">(4.10)
 * }

Substitute Equations 4.8, 4.9 and 4.10 into Equation 4.0:
 * {| style="width:100%" border="0"|-

$$ \Downarrow $$
 * style="width:95%" | $$ V = LC\frac{d^2v_{c}}{dt^2} + RC\frac{dv_{c}}{dt} + v_{c}$$
 * }
 * }

R 1.5
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.4-5

Problem 4
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+4y'+(\pi^2 + 4)y = 0$$ (5.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Rewrite as


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

$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y=0 $$ (5.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (5.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+4\lambda+(\pi^2+4)=0$$ (5.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The formula for the Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha \pm \beta i$$ (5.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the quadratic formula, the roots are determined to be
 * {| style="width:100%" border="0"

$$\lambda =-2\pm (\pi)i$$ (5.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The General Solution formula for Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1\cos \beta x+c_2\sin \beta x)$$ (5.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 5
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+2\pi y'+\pi^2y = 0$$ (5.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Rewrite as


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

$$\frac{d^2 y}{dx^2} + 2\pi \frac{dy}{dx} + (\pi^2 y)=0 $$ (5.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Set value for lambda (refer to equation 5.2)
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+2\pi \lambda+\pi^2 y=0$$ (5.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The formula for the Real Double root is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha$$ (5.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the quadratic formula, the root is determined to be
 * {| style="width:100%" border="0"

$$\displaystyle\lambda=-\pi$$ (5.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The General Solution formula for Real Double root is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (5.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Given
For each ODE in Fig.2 in Kreyszig 2011 p.3 (except the last one involving a system of 2 ODE's), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied:

Falling stone
 * {| style="width:100%" border="0"

$$  \displaystyle y''=g=const. $$     (6.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle mv'=mg-bv^2 $$     (6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Outflowing water
 * {| style="width:100%" border="0"

$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Vibrating mass on a spring
 * {| style="width:100%" border="0"

$$  \displaystyle my''+ky=0 $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Beats of a vibrating system
 * {| style="width:100%" border="0"

$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0}\approx w $$ (6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Current I in an RLC circuit
 * {| style="width:100%" border="0"

$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Deformation of a beam
 * {| style="width:100%" border="0"

$$  \displaystyle EIy^{iv}=f(x) $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
The order of an ODE is determined by the order of the highest-order derivatives present in the equation.

A linear ODE is one in which the dependent variable, y in this case, and its derivatives in additive combinations of their first powers with coefficients that are constants or that depend on an independent variable. If the equation does not fit the above specifications, the equation is said to be nonlinear.

The superposition principle states that if $$y_p$$ (particular) and $$y_h$$ (homogeneous) are both solutions to a differential equation:

$$ \displaystyle ay_p'' + by_p' + cy_p=f_p(x)$$

$$ \displaystyle ay_h'' + by_h' + cy_h=f_h(x)$$

Then when the two equations are added together, substituting in:

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

You will obtain the original ODE.

6.0


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

$$  \displaystyle y''=g=const. $$     (6.0)     Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Linear

Superposition: Yes

Examining both particular and homogeneous solutions: Particular solution: $$ \displaystyle y_p'' = g$$

Homogeneous solutions: $$ \displaystyle y_{h1}'' = 0$$

$$ \displaystyle y_{h2}'' = 0$$

Adding the two homogeneous solutions together, results in the following: $$ \displaystyle y_h = (y_{h1} + y_{h2}) = 0$$

Let: $$ \displaystyle \bar{y} = y_h + y_p''$$

Adding $$y_p$$ and $$y_h$$ equations together results in: $$ \displaystyle (y_p + y_h) = g $$

Which simplifies to: $$ \displaystyle \bar{y}'' = g $$

Thus, the superposition principle can be applied.

6.1


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

$$  \displaystyle mv'=mg-bv^2 $$     (6.1)     Order: $$1^{st}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Nonlinear

Superposition: No

Homogeneous solution: $$ \displaystyle mv_h' + bv_h^2 = 0 $$ <p style="text-align:right"> (6.1.1)

Particular solution: $$ \displaystyle mv_p' + bv_p^2 = mg $$ <p style="text-align:right"> (6.1.2)

Let $$ \displaystyle \bar{v} = v_h + v_p $$

Adding (6.1.1) and (6.1.2): $$ \displaystyle m(v_h' + v_p') + b(v_h^2 + v_p^2) = mg $$

Likewise $$ \displaystyle m\bar{v} + b\bar{v}^2 = mg $$

However $$ \displaystyle \bar{v}^2 = (v_h + v_p)^2 \not= (v_h^2 + v_p^2) $$

Thus $$ \displaystyle \bar{v} $$ is not a solution and superposition does not apply.

6.2


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

$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)     Order: $$1^{st}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Nonlinear

Superposition: No

Homogeneous solution: $$ \displaystyle h_{h}' + k\sqrt{h_{h}} = 0 $$

Particular solution: $$ \displaystyle h_{p}' + k\sqrt{h_{p}} = 0 $$

Adding $$h_p'$$ and $$h_h'$$ equations together results in: $$ \displaystyle h_{h}' + k\sqrt{h_{h}} + h_{p}' + k\sqrt{h_{p}} = 0 $$

Which can be rearranged to: $$ \displaystyle (h_{h}' + h_{p}') + k(\sqrt{h_{h}} + \sqrt{h_{p}}) = 0 $$

Simplify using: $$ \displaystyle \bar{h} = h_{p} + h_{h} $$

To: $$ \displaystyle \bar{h'} + k(\sqrt{h_{h}} + \sqrt{h_{p}}) = 0 $$

Since: $$ \displaystyle k(\sqrt{h_{h}} + \sqrt{h_{p}}) \not= k\sqrt{h_{h} + h_{p}} $$

Then: $$ \displaystyle \bar{h'} + k(\sqrt{h_{h}} + \sqrt{h_{p}}) \not= \bar{h'} + k\sqrt{\bar{h}} $$

So superposition can not be applied.

6.3


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

$$  \displaystyle my''+ky=0 $$     (6.3)     Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Linear

Superposition: Yes

$$ \displaystyle y_1'' + \frac{k}{m}y_1 = 0 $$ <p style="text-align:right"> (6.3.1)

$$ \displaystyle y_2'' + \frac{k}{m}y_2 = 0 $$ <p style="text-align:right"> (6.3.2)

$$ \displaystyle c_1 * (6.3.1) + c_2 * (6.3.2) = (c_1y_1 + c_2y_2)'' + \frac{k}{m}(c_1y_1 + c_2y_2) = 0 $$

Therefore, $$ \displaystyle y = c_1y_1 + c_2y_2 $$

Thus the principle of superposition applies.

6.4


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

$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0} \approx w $$ (6.4)    Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Linear

Superposition: Yes

Homogeneous solution: $$ \displaystyle y_h'' + w_{0}^2 y_h = 0 $$

Particular solution: $$ \displaystyle y_p'' + w_{0}^2 y_p = \cos{wt} $$

Examining homogeneous solution: $$ \displaystyle y_{h1}'' + w_{0}^2y_{h1} = 0 $$ <p style="text-align:right"> (6.4.1)

$$ \displaystyle y_{h2}'' + w_{0}^2y_{h2} = 0 $$ <p style="text-align:right"> (6.4.2)

$$ \displaystyle c_1 * (6.4.1) + c_2 * (6.4.2) = (c_1y_{h1} + c_2y_{h2})'' + w_{0}^2(c_1y_{h1} + c_2y_{h2}) = 0 $$

Therefore $$ \displaystyle y_h = c_1y_{h1} + c_2y_{h2} $$

And also $$ \displaystyle y(t) = y_h + y_p = c_1y_{h1} + c_2y_{h2} + y_p $$

Superposition is valid.

6.5


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

$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)     Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Linear

Superposition: Yes

Homogeneous solution: $$ \displaystyle LI_h'' + RI_h' + (1/C)I_h = 0 $$

Particular solution: $$ \displaystyle LI_p'' + RI_p' + (1/C)I_p = E' $$

Examining the homogeneous solution: $$ \displaystyle LI_{h1}'' + RI_{h1}' + (1/C)I_{h1} = 0 $$ <p style="text-align:right"> (6.5.1)

$$ \displaystyle LI_{h2}'' + RI_{h2}' + (1/C)I_{h2} = 0 $$ <p style="text-align:right"> (6.5.2)

$$ \displaystyle c_1 * (6.5.1) + c_2 * (6.5.2) = L(c_1I_{h1} + c_2I_{h2})'' + R(c_1I_{h1} + c_2I_{h2})' + (1/C)(c_1I_{h1} + c_2I_{h2})$$

Therefore $$ \displaystyle I_h = c_1I_{h1} + c_2I_{h2} $$

And $$ \displaystyle I = I_h + I_p = c_1I_{h1} + c_2I_{h2} + I_p $$

Superposition is valid.

6.6


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

$$  \displaystyle EIy^{iv}=f(x) $$     (6.6)     Order: $$4^{th}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Linearity: Linear

Superposition: Yes

Homogenous solution: $$ \displaystyle EIy_h^{iv} = 0 $$

Particular solution: $$ \displaystyle EIy_p^{iv} = f(x) $$

Examining the homogeneous solution:

$$ \displaystyle EIy_{h1}^{iv} = 0 $$ <p style="text-align:right"> (6.6.1)

$$ \displaystyle EIy_{h2}^{iv} = 0 $$ <p style="text-align:right"> (6.6.2)

$$ \displaystyle EIy_{h3}^{iv} = 0 $$ <p style="text-align:right"> (6.6.3)

$$ \displaystyle EIy_{h4}^{iv} = 0 $$ <p style="text-align:right"> (6.6.4)

$$ \displaystyle c_1 * (6.6.1) + c_2 * (6.6.2) + c_3 * (6.6.3) + c_4 * (6.6.4) = EI(y_{h1} + y_{h2} + y_{h3} + y_{h4} )^{iv} = 0 $$

Therefore $$ \displaystyle y_h = c_1y_{h1} + c_2y_{h2} + c_3y_{h3} + c_4y_{h4} $$

Also $$ \displaystyle y(x) = y_h + y_p = c_1y_{h1} + c_2y_{h2} + c_3y_{h3} + c_4y_{h4} + y_p $$

Superposition applies.

6.7


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

$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Order: $$2^{nd}$$

Linearity: Nonlinear

Superposition: No

Let $$ \displaystyle \Theta_1 $$ and $$ \displaystyle \Theta_2 $$ be solutions to the homogeneous ODE above: $$ \displaystyle L\Theta_1'' + g\sin\Theta_1 = 0 $$ <p style="text-align:right"> (6.7.1)

$$ \displaystyle L\Theta_2'' + g\sin\Theta_2 = 0 $$ <p style="text-align:right"> (6.7.2)

For superposition: Let $$ \displaystyle \bar\Theta = c_1\Theta_1 + c_2\Theta_2 $$

$$ \displaystyle c_1 * (6.7.1) + c_2 * (6.7.2) = L(c_1\Theta_1 + c_2\Theta_2)'' + g(c_1\sin\Theta_1 + c_2\sin\Theta_2) = 0 $$

$$ \displaystyle L\bar\Theta'' + g\sin\bar\Theta = 0 $$

Comparing: $$ \displaystyle (c_1\Theta_1 + c_2\Theta_2) = \bar\Theta $$ BUT $$ \displaystyle \sin\bar\Theta \not= c_1\sin\Theta_1 + c_2\sin\Theta_2 $$

Therefore $$ \displaystyle \bar\Theta $$ is not a solution and superposition cannot be used.