A proportional controller with gain K is used to control a spring and mass system as shown in the figure below.

If K is adjusted so that the second-order, closed-loop system model is $$(Y(s))/(R(s))=50/(s^2+2s+150)$$, then the system damping ratio is most nearly:

A. 0.8

B. 0.08

C. 0.0067

D. 0

Since we have a second-order, closed-loop system model, we can derive the damping ratio directly from the expression. This information is located on pg. 227 of our FE Reference Handbook:

From the above expression, our damping ratio is contained within the denominator, shown below:

$\frac{Y\left(s\right)}{R\left(s\right)}=\frac{K{{\omega}_{n}}^{2}}{{s}^{2}+2\zeta {\omega}_{n}s+{{\omega}_{n}}^{2}}\phantom{\rule{0ex}{0ex}}\frac{Y\left(s\right)}{R\left(s\right)}=\frac{50}{{s}^{2}+2s+150}\phantom{\rule{0ex}{0ex}}\frac{K{{\omega}_{n}}^{2}}{{s}^{2}+2\zeta {\omega}_{n}s+{{\omega}_{n}}^{2}}=\frac{50}{{s}^{2}+2s+150}$

Lets solve for our natural frequency, ${\omega}_{n}$, first since it is needed in order to solve for our damping ratio:

$\frac{K{{\omega}_{n}}^{2}}{{s}^{2}+2\zeta {\omega}_{n}s+{{\omega}_{n}}^{2}}=\frac{50}{{s}^{2}+2s+150}\phantom{\rule{0ex}{0ex}}{{\omega}_{n}}^{2}=150\phantom{\rule{0ex}{0ex}}{\omega}_{n}=\sqrt{150}\approx 12.25$

Next, solving for the damping ratio, $\zeta $:

$\frac{K{{\omega}_{n}}^{2}}{{s}^{2}+2\zeta {\omega}_{n}s+{{\omega}_{n}}^{2}}=\frac{50}{{s}^{2}+2s+150}\phantom{\rule{0ex}{0ex}}2\zeta {\omega}_{n}s=2s\phantom{\rule{0ex}{0ex}}2(12.25)\left(s\right)\zeta =2s\phantom{\rule{0ex}{0ex}}\zeta =\frac{1}{12.25}\approx .0816$

Therefore, our damping ratio, $\zeta $, is most nearly **.08**, which is answer choice B.

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