User:EML4507.s13.team4ever.Bonner

Report 7: Problem 2 http://en.wikiversity.org/wiki/Bonnerreport7p2

Report 6: Problem 2 http://en.wikiversity.org/wiki/Bonnerreport6p2

Report 5: Problem 6 http://en.wikiversity.org/wiki/Bonnerreport5p6

Report 5: Problem 1 http://en.wikiversity.org/wiki/Bonnerreport5p1

Report 4: Problem 1 http://en.wikiversity.org/wiki/Bonnerreport4p1

Report 3: Problem 4 http://en.wikiversity.org/wiki/Bonnerreport3p4

Report 3: Problem 8 http://en.wikiversity.org/wiki/Bonnerreport3p8

Problem R2.2
Disccription: We are asked to consider the L2-ODE-CC system referenced in (1) p.53-2 to have the following complex conjugate roots:

$$ \lambda_{1,2} = -0.5 \pm 2i $$

We are asked to find the homogeneous solution in standard form and to find the damping ratio, $$ \zeta $$

We are then to consider the initial conditions

$$ y(0) = 1, y'(0) = 0 $$

We are to plot this solution and then use that plot to find the log decrement and compare it to the log decrement that would be obtained using (6) p.53-5b.

Finally, we're to find the quality factor, Q, and the loss factor, $$ \eta $$

Solution:

From R1.5, we have already obtained that for the above conditions, $$ y(t) = e^{-0.5 t} cos(2t) + (\frac{1}{4}) e^{-0.5 t} sin(2t) $$

with coefficients to the characteristic equation of

$$ a = 1$$ and $$ b = \frac{17}{4}$$

From (1) and (2) p.53-5, we have that

$$ \omega^{2} = b $$

$$ 2 \omega \zeta = a $$

Knowing both a and b we get

$$ \omega = \sqrt{ \frac{17}{4}} = 2.0615 $$

$$ \zeta = \frac{1}{2 \omega} = \frac{1}{2*2.0615} = 0.2425 $$

We then have from equation (1) p.53-5d

$$ \delta = \frac{1}{3}[\delta^{(1)} + \delta^{(2)} + \delta^{(3)}] $$



Using the above graph of the homogeneous solution, accurate estimates were able to be made as to the various values of y at the peaks

Using equation (3) p.53-5c we are able to find

$$ \delta^{(1)} = log(1/0.2075) = 0.68235 $$ $$ \delta^{(2)} = log(0.2075/0.0432) = 0.68216 $$ $$ \delta^{(3)} = log(0.0432/0.009) = 0.68124 $$

We then arrive at

$$ \delta = \frac{1}{3}[0.68235 + 0.68216 + 0.68124] = 0.68192 $$

For comparison, the equation (6) p.53-5b gives

$$ \delta = \frac{2 \pi \zeta}{\sqrt{1 - \zeta^{2}}} = \frac{2 \pi 0.2425}{\sqrt{1 - 0.2425^{2}}} = 1.5705 $$

We now find the quality factor, Q. From (2) p.53-5d, we have

$$ Q = \frac{\delta}{\pi} = \frac{\sqrt{1 - \zeta^{2}}}{2 \zeta} = \frac{\sqrt{1 - 0.2425^{2}}}{2 *0.2425} = 2 $$

From (5) p.53-5d, we know that the loss factor is just the inverse of the quality factor so

$$ \eta = \frac{1}{Q} = \frac{1}{2} = 0.5 $$

Problem R2.4
Description: We are to find an expression for the excitation referenced in (1) p.53-6 and then find the actual solution and the amplification factor, A for the following conditions:

$$ f_{0}/m = r_{0} = 1 $$ $$ \bar{w} = 0.9 w $$

We are then to find the complete solution.

Solution:

$$ y'' + y' +(\frac{17}{4})y = \frac{f_{0}}{m} cos(\bar{w} t) $$

Let us guess the solution is:

$$ y_{p}(t) = A sin(\bar{w} t) + B cos(\bar{w} t) $$

then

$$ y_{p}(t) = A sin(\bar{w} t) + B cos(\bar{w} t) $$ $$ y'_{p}(t) = A \bar{w} cos(\bar{w} t) - B \bar{w} sin(\bar{w} t) $$ $$ y''_{p}(t) = - A \bar{w}^{2} cos(\bar{w} t) - B \bar{w}^{2} sin(\bar{w} t) $$

We then write

$$ [ - A \bar{w}^{2} - B \bar{w} + (\frac{17}{4}) A ] sin(\bar{w} t) + [ - B \bar{w}^{2} + A \bar{w} + (\frac{17}{4}) B] cos(\bar{w} t) = \frac{f_{0}}{m} cos(\bar{w} t) $$

It follows that

$$ [ - A \bar{w}^{2} - B \bar{w} + (\frac{17}{4}) A ] = 0 $$ and that $$ [ - B \bar{w}^{2} + A \bar{w} + (\frac{17}{4}) B] = \frac{f_{0}}{m} $$

We solve for A and B to be

$$ A = \frac{f_{0}/m}{[\bar{w} + (\frac{17}{4})^{2}(\frac{1}{\bar{w}}) - \frac{34 \bar{w}}{4} + \bar{w}^{3} ]} $$ $$ B = \frac{ \frac{f_{0}}{m} [ \frac{17}{4} - \bar{w}^{2}] }{[[\bar{w} + (\frac{17}{4})^{2}(\frac{1}{\bar{w}}) - \frac{34 \bar{w}}{4} + \bar{w}^{3} ] (\bar{w})]} $$

Once given the conditions

$$ f_{0}/m = r_{0} = 1 $$ $$ \bar{w} = 0.9 w $$

we can find

$$ \bar{w} = 0.9 w =  \bar{w} = 0.9 * 2.0615 = 1.855 $$

Plugging these values in yields

$$ A = 0.4528 $$ $$ B = 0.1975 $$

The particular solution is then

$$ y_{p}(t) = 0.4528 sin(1.855 t) + 0.1975 cos(1.855 t) $$

Equations (3) p. 53-8 and (2) p. 53-8 give us

$$ \rho = \frac{\bar{w}}{w} = \frac{1.855}{2.0615} = 0.9$$ $$ A = \frac{1}{[(1 - \rho^{2})^{2} + (2 \zeta \rho)^{2}]^{\frac{1}{2}}} = \frac{1}{[(1 - 0.9^{2})^{2} + (2 *0.2425 *0.9)^{2}]^{\frac{1}{2}}} = 2.628 $$

The following is the graph of the homogeneous solution:



The following is the graph of the particular solution:



The following is the graph of the total solution, $$ y(t) = y_{h}(t) + y_{p}(t) $$:



A graph of all three plots is produced below: