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:

$m(y^{\prime }{\text{'}}_k+\frac{k}{c}y{\text{'}}_k)+k{y}_k=f(t)$

Dividing the entire equation by m:

$y^{\prime }{\text{'}}_k+\frac{k}{cm}y{\text{'}}_k+\frac{k}{m}{y}_k=f(t)$

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:

$y^{″}+6y^{\prime }+9=0$

Matching the coefficients:

$y^{″}\Rightarrow 1=1$

$y^{\prime }\Rightarrow \frac{k}{cm}=6$

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

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)

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

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

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

even powers:

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

odd powers:

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

If we let $x=it$:

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

Using the two previous equations:

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

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

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:

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

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

Author

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

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