User:Eml4507.s13.team2.cc/R2

Problem Statement
Consider a linear 2nd-order ODE (L2-ODE-CC) with constant coefficients with the following complex conjugate roots:
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$$  \displaystyle \lambda_{1,2}=-0.5 \pm i2 $$ (2.0)
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Find the homogeneous L2-ODE-CC in standard form in terms of $$ \displaystyle y_h (t) $$ with the above roots and find the damping ratio $$ \displaystyle \zeta $$.

Consider the case with no excitation, and find and plot the solution satisfying these initial conditions:
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$$  \displaystyle y(0)=1, \ y'(0)=0 $$     (2.1)
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Using the plot of the solution, find the approximate amplitudes of oscillation in a cycle. Determine the log decrements for $$ \displaystyle n=1,2,3 $$.

Then, find the average log decrement and verify its value.

Finally, find the quality factor Q and the loss factor $$ \displaystyle \eta $$ for the system described by the above roots.

Solution
The form of a homogeneous L2-ODE-CC in standard form in terms of $$ \displaystyle y_h(t) $$ is:
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$$  \displaystyle y_h'' + ay_h' + by_h = 0 $$     (2.2)
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To find the values $$ \displaystyle a $$ and $$ \displaystyle b $$ and derive the L2-ODE-CC, first find the characteristic equation of the form:
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$     (2.3)
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Start by working backward using the quadratic formula and the given roots.
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$$  \displaystyle \lambda_{1,2}=\frac{-a\pm\sqrt{a^2-4b}}{2}=-0.5 \pm i2 $$ (2.4)
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As a result,
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$$  \displaystyle a=1, \ b=4.25 $$     (2.5)
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Then, the characteristic equation is:
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$$  \displaystyle \lambda^2+\lambda+4.25 = 0 $$     (2.6)
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Thus, the homogeneous L2-ODE-CC is:
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$$  \displaystyle y_h'' + y_h' + 4.25y_h = 0 $$     (2.7)
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To find the damping ratio $$ \displaystyle \zeta $$, first find the undamped natural circular frequency $$ \displaystyle \omega $$ of the system.
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$$  \displaystyle \omega^2:=b=4.25 $$     (2.8)
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$$  \displaystyle \omega = \sqrt{4.25}\approx 2.0616 $$     (2.9)
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The damping ratio $$ \displaystyle \zeta $$ can be found with the equation:
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$$  \displaystyle 2\omega \zeta := a = 1 $$     (2.10)
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Solving for $$ \displaystyle \zeta $$ gives:
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$$  \displaystyle \zeta = \frac{a}{2\omega}=\frac{1}{2*2.0616} $$     (2.11)
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$$  \displaystyle \zeta = 0.2425 $$     (2.12)
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The homogeneous solution for the case of two complex conjugate roots is:
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$$  \displaystyle y_h(t)=c_1e^{mt}\cos{nt}+c_2e^{mt}\sin{nt} $$     (2.13)
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The variables $$ \displaystyle m $$ and $$ \displaystyle n $$ are the components of the complex conjugate roots $$ \displaystyle \lambda_{1,2}=m \pm in $$.

Plugging in the given roots results in:
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$$  \displaystyle y_h(t)=c_1e^{-0.5t}\cos{2t}+c_2e^{-0.5t}\sin{2t} $$     (2.14)
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Differentiating gives:
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$$  \displaystyle y'_h(t)=e^{-0.5t}[(-2c_1-0.5c_2) \sin{2t} + (2c_2-0.5c_1) \cos{2t}] $$     (2.15)
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Applying the initial condition $$ \displaystyle y(0)=1 $$ to Eq. 2.14 results in:
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$$   \displaystyle 1=c_1e^{-0.5(0)} \cos(2*0)+c_2e^{-0.5(0)}\sin(2*0) $$      (2.16)
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$$   \displaystyle 1=c_1\cos(0) + c_2 \sin(0) $$      (2.17)
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$$   \displaystyle 1=c_1 $$      (2.18)
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Applying the initial condition $$ \displaystyle y'(0)=0 $$ to Eq. 2.15 results in:
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$$   \displaystyle 0=e^{-0.5t}[(-2c_1-0.5c_2) \sin{(2*0)} + (2c_2-0.5c_1) \cos{(2*0)}] $$      (2.19)
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$$   \displaystyle 0=(2c_2-0.5c_1) $$      (2.20)
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$$   \displaystyle 0.25=c_2 $$      (2.21)
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Then, the overall homogeneous solution is:
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$$  \displaystyle y_h(t)=e^{-0.5t} \cos(2t) + 0.25e^{-0.5t} \sin(2t) $$     (2.22)
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The solution is plotted below.



Using the MATLAB function "ginput" with the above plot, approximate amplitudes of oscillation were found and used to calculate the log decrements for $$ \displaystyle n=1,2,3 $$ from the following:
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$$   \displaystyle \delta^{(n)} = \log \frac{y_h^{(n)}}{y_h^{(n+1)}} $$      (2.23)
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$$   \displaystyle \delta^{1} = \log \frac{0.2080}{0.0433} \approx 1.5694 $$      (2.24)
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$$   \displaystyle \delta^{2} = \log \frac{0.0433}{0.0089} \approx 1.5821 $$      (2.25)
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$$   \displaystyle \delta^{3} = \log \frac{0.0089}{0.0020} \approx 1.4929 $$      (2.26)
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The average approximated log decrement is:
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$$   \displaystyle \delta \approx \frac13 [\delta^{(1)}+\delta^{(2)} + \delta^{(3)}] $$      (2.27)
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$$   \displaystyle \delta \approx \frac13 [(0.6816)+(0.6871 + (0.6484)] $$      (2.28)
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$$   \displaystyle \delta \approx 1.5481 $$      (2.29)
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To verify the value of $$ \displaystyle \delta $$, the following equation is used:
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$$   \displaystyle \delta = \frac{2 \pi \zeta}{\sqrt{1- \zeta^2}} $$      (2.30)
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$$   \displaystyle \delta = \frac{2 \pi 0.2425}{\sqrt{1- (0.2425)^2}} $$      (2.31)
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$$   \displaystyle \delta \approx 1.5708 $$      (2.32)
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This value is close to the approximated average log decrement above.

The quality factor Q is found by:
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$$   \displaystyle Q:= \frac {\pi}{\delta} $$      (2.33)
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$$   \displaystyle Q:= \frac {\pi}{1.5708}=2 $$      (2.34)
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The loss factor $$ \displaystyle \eta $$ is found by:
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$$   \displaystyle \eta:=\frac{1}{Q} $$      (2.35)
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$$   \displaystyle \eta:=\frac{1}{2}=0.5 $$      (2.36)
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