User:Egm4313.s12.team13.rossmeissl/R2work

Problem Statement
Using the spring-dashpot-mass system shown below (from Fig p1.4) and following similar characteristics of (3)p5.5 but using the double real root $$\displaystyle \lambda = -3 $$, find values for the parameters k,c,m.




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$$  \displaystyle {(\lambda -5)}^{2} = {\lambda }^{2}-10\lambda +25 = 0 $$     ((3)p5.5)
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Solution
Rewriting ((3)p5.5) with the double real root $$\displaystyle \lambda = -3 $$ gives the new characteristic equation


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$$  \displaystyle {(\lambda -3)}^{2} = {\lambda }^{2}-6\lambda +9 = 0 $$     (6.0)
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From kinematics:


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$$  \displaystyle y = {y}_{k}+{y}_{c} $$     (6.1)
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From kinetics:


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$$  \displaystyle my''+{f}_{1} = f(t) $$     (6.2)
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$$  \displaystyle {f}_{1} = {f}_{k} = {f}_{c} $$     (6.3)
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Relationships:


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$$  \displaystyle {f}_{k} = k{y}_{k} $$     (6.4)
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$$  \displaystyle {f}_{c} = c{y'}_{c} $$     (6.5)
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From equation (6.1) take the second derivative


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$$  \displaystyle y = {y}_{k} = {y''}_{c} $$     (6.6)
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From equations (6.3), (6.4) & (6.5)


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$$  \displaystyle {f}_{k} = {f}_{c} \Rightarrow k{y}_{k} = c{y'}_{c} \Rightarrow {y'}_{c} = \frac{k}{c}{y}_{k} $$     (6.7)
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From equations (6.6) & (6.7)


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$$  \displaystyle y = {y}_{k}+({y'}_{c})' \Rightarrow {y''}_{k}+(\frac{k}{c}{y}_{k})' $$     (6.8)
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From equations (6.2), (6.3) & (6.8)


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$$  \displaystyle m({y''}_{k}+\frac{k}{c}{y'}_{k})+k{y}_{k} = f(t) $$     (6.9)
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From the characteristic equation (6.0), the corresponding homogeneous L2-ODE-CC is


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$$  \displaystyle y''+6y'+9y = 0 $$     (6.10)
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Now take equation (6.9) and divide through by m


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$$  \displaystyle {y''}_{k}+\frac{k}{cm}{y'}_{k}+\frac{k}{m}{y}_{k} = \frac{f(t)}{m} = 0 $$     (6.11)
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Comparing to equation (6.10) gives


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$$  \displaystyle \frac{k}{cm} = 6 $$     (6.12) &
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$$  \displaystyle \frac{k}{m} = 9 $$     (6.13)
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There are an infinite number of solutions to these two equations, one of these is

$$   \displaystyle k = 9\;\;m = 1\;\;c = \frac{3}{2}$$

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