User:Egm4313.s12.team1.wyattling/solutions

Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in Kreyszig 2011 p.85, with an applied force $$r(t)$$ on the ball.

Solution
Kinematics:

$$y=y_k=y_c\!$$

Kinetics:

$$r(t)=my''+{{f}_{k}}+{{f}_{c}}\!$$

$${{f}_{k}}=k{{y}_{k}}\!$$

$${{f}_{c}}=c{{y}_{c}}'\!$$

Rearranging equations,

$$r(t)=my''+k{{y}_{k}}+c{{y}_{c}}'\!$$

$$y=y_k=y_c\!$$

Equation of motion:

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

Author
Solved and typed by - --Egm4313.s12.team1.essenwein 23:06, 25 January 2012 (UTC) Reviewed By -

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

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

Kinetics: $$f(t)=my'' + f_{k}+f_{c}\!$$

Relations:

-Force of spring relation: $$f_{k}=ky_{k}\!$$ -Force of dash-pot relation: $$f_{c}=Cy'_{c}\!$$

Putting kinematics, kinetics, and relations together:

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

Using kinetic relationship $$f(t)=my''_{c}+f_{k}+f_{c}\!$$

Using the kinematics and relations equations: $$f_{k}=ky_{k}=ky_{c}\!$$ $$f(t)=my''_{c}+ky_{c}+f_{c}\!$$ Using dash-pot relation: $$f_{c}=Cy'_{c}\!$$ $$f(t)=my''_{c}+ky_{c}+Cy'_{c}\!$$

'''Final Equation

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

Problem R1.4
Given

Capacitance Equation(1):

$$Q=Cv_{c}\rightarrow \int idt=Cv_{c}\rightarrow i=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

Circuit Equation(2):

$$V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c}$$

First Alternate Circuit Equation (3):

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

Second Alternate Circuit Equation (4):

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

Statement

Using the Circuit equation (2) (see Sec2 Notes), derive equations (3) and (4) from it.

Solution

First we are asked to derive the alternate circuit equation(3) from the original circuit equation(2).

$$V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c}\rightarrow LI''+RI'+\frac{1}{C}I=V'$$

The first step is to take the derivative of the original equation so there is the V' term on one side:

$$V'=LC\frac{\mathrm{d^3}v_{c}}{\mathrm{d} t^3}+RC\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}+\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

L and C are constants and therefore remain unchanged by the derivative.

Next use the capacitance equation which states: $$I=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

Take both the first and second derivative of this equation.

First Derivative: $$I'=C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}$$

Second Derivative: $$I''=C\frac{\mathrm{d^3} v_{c}}{\mathrm{d} t^3}$$

Now substitute in the the I terms into the appropriate places. The equation below groups terms that need to be substituted.

$$V'=L(C\frac{\mathrm{d^3}v_{c}}{\mathrm{d} t^3})+R(C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2})+\frac{({C}\frac{\mathrm{d} v_{c}}{\mathrm{d} t})}{C}$$

The final product after substitution is the alternate circuit equation:

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

We are also asked to derive the second alternate circuit equation $$LQ''+RQ'+\frac{1}{C}Q=V$$ from the original circuit equation.

Once again use the capacitance equation which also states:

$$Q=Cv_{c}\ $$

Take the first and second derivative:

First Derivative: $$Q'=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t}$$

Second Derivative: $$Q''=C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}$$

The C term is constant and does not change when the derivative is taken.

Group the original circuit equation like in the previous problem so that parentheses mark terms that should be substituted for Q terms: $$V=L(C\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2})+R(C\frac{\mathrm{d} v_{c}}{\mathrm{d} t})+\frac{(Cv_{c})}{C}$$

After substituting, the final product is the second alternate circuit equation: $$LQ''+RQ'+\frac{1}{C}Q=V$$

Author
Solved and typed by - Chris Stewart Reviewed By -

Problem R1.5
Find a general solution for the following two problems.

Given

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

$$
 * $$y'' + 4y' + (\pi^2 + 4)y = 0\!$$
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution.

Solution
The characteristic equation of this linear ODE with constant coefficients (L2-ODE-CC) is:


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

$$
 * $$\lambda^2+4\lambda+(\pi^2+4)=0\!$$
 * $$\displaystyle (Eq. 2)

Note: The discriminant of the characteristic equation is negative.
 * }

The roots of this characteristic equation are of the form:


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

$$
 * $$\lambda=-\frac{1}{2}a\pm bi$$
 * $$\displaystyle (Eq. 3)
 * }
 * }

Which yields a general solution of:


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

$$
 * $$y(x)=c_1e^{ax}\cos bx+c_2e^{ax}\sin bx\!$$
 * $$\displaystyle (Eq. 4)
 * }
 * }

By applying the quadratic equation to the characteristic equation, we see that the roots are:


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

$$
 * $$\lambda =-2\pm (4\pi)i$$
 * $$\displaystyle (Eq. 5)
 * }
 * }

This yields a general solution of:


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

$$\displaystyle y(x)=c_1e^{-2x}\cos (4\pi*x)+c_2e^{-2x}\sin (4\pi*x)\!$$ Solved by: Egm4313.s12.team1.durrance 20:23, 25 January 2012 (UTC)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Given

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

$$
 * $$y''=2\pi*y'+\pi^2y\!$$
 * $$\displaystyle (Eq. 1)
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution for this ODE.

Solution
This ODE is a linear ODE with constant coefficients (L2-ODE-CC).

First the equation is rearranged into general form:


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

$$
 * $$y''-(2\pi)y'-(\pi^2)y=0\!$$
 * $$\displaystyle (Eq. 2)
 * $$\displaystyle (Eq. 2)
 * }
 * }

The characteristic equation is:


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

$$
 * $$\lambda^2-2\pi\lambda-\pi^2=0\!$$
 * $$\displaystyle (Eq. 3)
 * $$\displaystyle (Eq. 3)
 * }
 * }

Substituting the characteristic equation coefficients into the quadratic formula, we find:


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

$$
 * $$\lambda_1=\lambda_2=\lambda=\pi\!$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)
 * }
 * }

Therefore, the general form for the general solution is:


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

$$\displaystyle y(x)=c_1e^{\pi*x}+c_2xe^{\pi*x}\! $$ Solved by Egm4313.s12.team1.durrance 20:30, 25 January 2012 (UTC)
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Statement
For each ODE in Fig.2 in Kreyszig 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.

Given

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

$$
 * $$y''=g=const.\!$$
 * $$\displaystyle (Eq. 1)
 * $$\displaystyle (Eq. 1)
 * }
 * }


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

$$
 * $$mv'=mg-bv^2\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }


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

$$
 * $$h'=-k\sqrt{h}\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }


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

$$
 * $$my''+ky=0\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }


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

$$
 * $$y''+w_{0}^2y=coswt, w_{0}=w\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }


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

$$
 * $$LI''+RI'+\frac{C}I=E'\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }


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

$$
 * $$EIy=f(x)\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }


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

$$
 * $$L\Theta''+gsin\Theta=0\!$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Solution
In order to determine the order of any ODE, one must simply find the highest prime term in each equation.

ROUGH ANSWERS IN CASE ANYONE WANTS TO CHECK: (Okay so I think I got the superposition part, but i really want to run it past you all during our meeting before I latex everything etc.) eq. 1 = 2nd order/linear eq. 2 = 1st order/non-linear eq. 3 = 1st order/non-linear eq. 4 = 2nd order/linear eq. 5 = 2nd order/linear eq. 6 = 2nd order/linear eq. 7 = 4th order/linear eq. 8 = 2nd order/non-linear

Author
Solved and typed by - Reviewed By -