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:

$$\begin{gathered} \frac{Y\left(s\right)}{R\left(s\right)}=\frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2} \\ \frac{Y\left(s\right)}{R\left(s\right)}=\frac{50}{s^2+2s+150} \\ \frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}=\frac{50}{s^2+2s+150} \end{gathered}$$

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

$$\begin{gathered} \frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}=\frac{50}{s^2+2s+150} \\ {{\omega }_n}^2=150 \\ {\omega }_n=\sqrt{150}\approx 12.25 \end{gathered}$$

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

$$\begin{gathered} \frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}=\frac{50}{s^2+2s+150} \\ 2\zeta {\omega }_{n}s = 2s \\ 2(12.25)\left(s\right)\zeta = 2s \\ \zeta = \frac{1}{12.25}\approx .0816 \end{gathered}$$

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