**1 Answer**

written 5.0 years ago by |

**When subjected to a The phase difference between the excitation and steady state response is 24.3˚ What is the damping ratio of the isolator and its maximum deflection due to this excitation?**

$X_{st} \ = \ \frac{fo}{k} \ = \ \frac{250}{6 \ \times \ 10^5} \ = \ 5.83 \ \times \ 10^{-4} \ m$

$w_n \ = \ \sqrt{ \frac{k}{m}} \ = \ \sqrt{ \frac{6 \ \times \ 10^5}{30}} \ = \ 141.42$ rad/sec

$tan \ \phi \ = \ \frac{2 \ \xi r}{1 – r^2}$

$r \ = \ \frac{w}{wn} \ = \ \frac{100}{141.42} \ = \ 0.707$

$tan \ 24.3 \ = \ \frac{2 \ \times \ \xi \ \times \ 0.707}{1 \ – \ (0.707)^2}$

$0.4515 \ = \ \frac{1.414 \ \times \ \xi}{0.5}$

$\xi \ = \ 0.1596$

$\therefore$ $\frac{xp}{5.8^3 \ \times \ 10^{-4}} \ = \ \frac{1}{2 \ \times \ 0.1596 \sqrt{ 1 – (0.707)^2}}$

$Xp \ = \ 1.88 \ \times \ 10^{-3} \ m$