University of Florida/Egm4313/s12.team4.Lorenzo/R2

Problem 6

Problem Statement

Realize spring-dashpot-mass systems in series as shown in Fig. p.1-4 with the similar characteristic as in (3) p.5-5, but with double real root $\lambda =-3$, i.e., find the values for the parameters k, c, m.

Solution

Recall the equation of motion for the spring dashpot mass system:

$m(y''_{k}+{\frac {k}{c}}y'_{k})+ky_{k}=f(t)\!$

Dividing the entire equation by m:

$y''_{k}+{\frac {k}{cm}}y'_{k}+{\frac {k}{m}}y_{k}=f(t)\!$

The characteristic equation for the double root :$\lambda =-3\!$ is:

$(\lambda +3)^{2}=\lambda ^{2}+6\lambda +9=0\!$

The corresponding L2-ODE-CC (with excitation) is:

$y''+6y'+9=0\!$

Matching the coefficients:

$y''\Rightarrow 1=1\!$

$y'\Rightarrow {\frac {k}{cm}}=6\!$

$y\Rightarrow {\frac {k}{m}}=9\!$

After algebraic manipulation it is found that the following are the possible values for k, c, and m:

$k=18\!$

$c={\frac {3}{2}}\!$

$m=2\!$

Author

Solved and typed by - Egm4313.s12.team4.Lorenzo 20:04, 6 February 2012 (UTC)
Reviewed By -
Edited by -

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Problem 7

Problem Statement

Develop the MacLaurin series (Taylor series at t=0) for:

• $e^{t}\!$
• $\cos t\!$
• $\sin t\!$

Solution

Recalling Euler's Formula:

$e^{i\omega x}=\cos \omega x+i\sin \omega x\!$

Recall the Taylor Series for $e^{x}$ at :$x=0$ (also called the MacLaurin series)

$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\!$

By replacing x with t, the Taylor series for $e^{t}$ can be found:

$e^{t}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\!$

even powers:

$i^{2k}=(i^{2})^{k}=(-1)^{k}\!$

odd powers:

$i^{2k+1}=(i^{2})^{k}i=(-1)^{k}i\!$

If we let $x=it$:

$e^{it}=\sum _{n=0}^{\infty }{\frac {i^{n}t^{n}}{n!}}=\sum _{k=0}^{\infty }{\frac {i^{2k}t^{2k}}{(2k)!}}+\sum _{k=0}^{\infty }{\frac {i^{2k+1}t^{2k+1}}{(2k+1)!}}\!$

Using the two previous equations:

$e^{it}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k}}{(2k)!}}+\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}\!$

$\Rightarrow e^{it}=\cos t+i\sin t\!$

Therefore, the first part of the equation is equal to the Taylor series for cosine, and the second part is equal to the Taylor series for sine as follows:

$\cos t=\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k}}{(2k)!}}\!$

$\sin t=\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}\!$

Author

Solved and typed by - Egm4313.s12.team4.Lorenzo 20:05, 6 February 2012 (UTC)
Reviewed By -
Edited by -

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