User:EML4507.s13.team4ever.Bonner/Bonnerreport3p8

Problem R3.8
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

We are asked to redo 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

$$ \rho = \frac{\bar{w}}{w} = \frac{1.855}{2.0615} = 0.9$$ $$ A = \frac{1}{[(1 - \rho^{2})^{2} + (2 \zeta \rho)^{2}]^{\frac{1}{2}}} = \frac{1}{[(1 - 0.9^{2})^{2} + (2 *0.2425 *0.9)^{2}]^{\frac{1}{2}}} = 2.1006 $$

We then need to find the total solution given the boundary conditions from R2.4; http://en.wikiversity.org/wiki/Fead-s13-team4-R2#Solution

We have for the homogeneous solution from R1.5:

$$ y_h = Ae^{-0.5t}cos(2t) + Be^{-0.5t}sin(2t) $$

The total solution can then be found from

$$ y(t) = y_h + y_p $$

We have then for the total solution:

$$ y(t) = Ae^{-0.5t}cos(2t) + Be^{-0.5t}sin(2t) + 0.4528sin(1.855t) + 0.1975cos(1.855t) $$

We now need to find the coefficients A and B using the initial conditions:

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

Taking the derivative of y(t), we obtain:

$$ y'(t) = -0.5Ae^{-1.5t}cos(2t) - 2Ae^{-0.5t}sin(2t) - 0.5Be^{-1.5t}sin(2t) + 2Be^{-0.5t}cos(2t) + 0.8399cos(1.855t) - 0.3663sin(1.855t) $$

Plugging in t = 0 for both the equation of y(t) and y'(t) allows us to find the values of A and B.

From the equation for y(t):

$$ A(1) + 0.1975 = 1 $$

$$ A = 0.8025 $$

From the equation for y'(t):

$$ -0.5A + 2B + 0.8399 = 0 $$

$$ 2B = -0.8399 + 0.5*0.8024 = -0.4386 $$

$$ B = -0.2193 $$

Hence, the total solution is:

$$ y(t) = 0.8025e^{-0.5t}cos(2t) - 0.2193e^{-0.5t}sin(2t) + 0.4528sin(1.855t) + 0.1975cos(1.855t) $$

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: