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

20.2 Modeling Vibration Systems 589Figure 20.3(a) the system is very sluggish and returns to the equilibrium positionvery slowly. In Figure 20.3(b), the system response is rapid, but it overshoots theequilibrium position and eventually the oscillation dies out due to the effect ofdamping. Figure 20.3(c) represents the critically damped system that promptlyreturns to the equilibrium position with no overshoot. This represents the dividingline between the overdamped system (Case 2) and the underdamped system(Case 1). The roots for the underdamped system occur in complex-conjugate pairs,the roots for the overdamped system are real and unequal, and the roots for thecritically damped system are real and equal. Figures 20.3(b) and 20.3(d) show, respectively,the difference between a lightly damped system and a heavily dampedsystem. The response of the more heavily damped system dies out more quickly.If the roots lie on the iω axis as shown in Figure 20.3(d), no damping occursand the system will oscillate forever. If any roots appear on the right half of thes plane, then the response will increase monotonically with time and the systemwill become unstable, as shown in Figure 20.3(f). Thus the iω axis represents theline of demarcation between stability and instability.From Equation (20.4) the natural frequency of the system is√kω n =m(20.7)and the damping ratio isξ =C2 √ km(20.8)When ξ = 1, critical damping occurs. We set C = C c for this value of ξ = 1. Thenfrom which we obtainξ = 1 =C c2 √ kmC c = 2 √ km (20.9)as the critical damping factor. The damping ratio is often written asξ = C C c(20.10)Example Problem 1The system of Figure 20.1 has the following parameters: the weight W of mass mis 28.5 N, C = 0.0650 N s/cm, k = 0.422 N/cm. Determine the undamped naturalfrequency of the system, its damping ratio, and the type of response the systemwould have if the mass is to be initially displaced and released.