University of Florida/Egm4313/s12 Report 2, Problem 2.6

Problem 2.6

Solved by: Andrea Vargas

Problem Statement

For the following spring-dashpot-mass system (in series) find the values for the parameters ${\displaystyle k,c,m\!}$ knowing that the system has the double real root ${\displaystyle \lambda =-3\!}$

Solution

Previously, we have derived the following equation for such a system:
(From Sec 1 (d), (3) p.1-5)
${\displaystyle m\left(y_{k}''+{\frac {k}{c}}y_{k}'\right)+ky_{k}=f(t)\!}$
We can write this equation in standard form by diving through by ${\displaystyle m\!}$:
${\displaystyle y_{k}''+{\frac {k}{c}}y_{k}'+{\frac {k}{m}}y_{k}={\frac {f(t)}{m}}\!}$
Here, we can take the coefficients of ${\displaystyle y_{k}'\!}$ and ${\displaystyle y_{k}\!}$ as ${\displaystyle a\!}$ and ${\displaystyle b\!}$:

${\displaystyle y_{k}''+\underbrace {\frac {k}{c}} _{a}y_{k}'+\underbrace {\frac {k}{m}} _{b}y_{k}={\frac {f(t)}{m}}\!}$

Next,considering the double real root:
${\displaystyle \lambda =-3\!}$
We can find the characteristic equation to be:
${\displaystyle \left(\lambda +3\right)^{2}=\lambda ^{2}+6\lambda +9=0\!}$
Which is in the form:
${\displaystyle \lambda ^{2}+a\lambda +b=0\!}$

Then, we know that ${\displaystyle a=6\!}$ and ${\displaystyle b=9\!}$:

Setting ${\displaystyle a\!}$ and ${\displaystyle b\!}$ from the first equation equal to these, we obtain:

                                                             ${\displaystyle {\frac {k}{c}}=6\;and\;{\frac {k}{m}}=9\!}$


Clearly, there is an infinite amount of solutions to this problem because we have 2 equations but 3 unknowns. This can be solved by fixing one of the values and finding the other two.

Example of Solution

An example of fixing one of the constants to find the other two is provided here. By solving the simple equations above, we can illustrate how to find ${\displaystyle k,c,m\!}$ . We had:
${\displaystyle {\frac {k}{c}}=6\;and\;{\frac {k}{m}}=9\!}$

If we fix the mass to ${\displaystyle m=10\;kg\!}$. We find:
${\displaystyle {\frac {k}{10}}=9\!}$

${\displaystyle k=90\!}$

Then,
${\displaystyle {\frac {90}{c}}=6\!}$

${\displaystyle c=15\!}$

Finally, we obtain:

                                                             ${\displaystyle m=10,\;k=90,\;and\;c=15\!}$


--Andrea Vargas 21:44, 7 February 2012 (UTC)