# 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 ${\displaystyle \lambda =-3}$, i.e., find the values for the parameters k, c, m.

### Solution

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

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

Dividing the entire equation by m:

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

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

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

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

${\displaystyle y''+6y'+9=0\!}$

Matching the coefficients:

${\displaystyle y''\Rightarrow 1=1\!}$

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

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

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

${\displaystyle k=18\!}$

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

${\displaystyle m=2\!}$

#### Author

Solved and typed by - Egm4313.s12.team4.Lorenzo 20:04, 6 February 2012 (UTC)
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## Problem 7

### Problem Statement

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

• ${\displaystyle e^{t}\!}$
• ${\displaystyle \cos t\!}$
• ${\displaystyle \sin t\!}$

### Solution

Recalling Euler's Formula:

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

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

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

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

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

even powers:

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

odd powers:

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

If we let ${\displaystyle x=it}$:

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

Using the two previous equations:

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

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

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:

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

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

#### Author

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