THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

590 20. Vibration and Vibration ControlSolutionThe mass is found fromm = W g = 28.5N = 2.91 kg29.807 m/sThen from Equation (20.7)√42.2 N/mω n == 3.81 rad/s2.91 kgFrom Equation (20.8)ξ =C2 √ km = 6.50 N s/m2 √ (42.2 N/m)(2.91 kg) = 0.294Since the damping ratio is less than unity, the response will be underdamped. Thesystem response is that of the form shown in Figure 20.3(b).Example Problem 2For the system of Example Problem 1, find the amount of additional dampingneeded to have the system become critically damped.SolutionFrom Equation (20.9), the condition for a critically damped system isC c = 2 √ km = 2 √ (42.2 N/m)(2.91 kg) = 22.13 N s/m, = 0.2213 N s/cmThe additional damping required isC = 0.2213 − 0.0650 = 0.1563 N s/cmWith this additional damping, Equation (20.2) becomesẍ(t) + 22.13 42.2 ẋ(t) +2.91 2.91 x(t) = 0The characteristic equation in the Laplace transform variable form becomess 2 + 22.132.91 s + 42.22.91 = 0or(s + 3.81) 2 = 0which demonstrates that Case 3 of Figure 20.3 exists.