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

20.4 Forced Vibration 595Table 20.1. System Response as a Function of Frequency.Frequency Response Controlling Parameterω 2 ≪ ωn 2 x(ω) = F 0kω 2 ≫ ωn 2 x(ω) = F 0mω 2ω 2 = ωn 2 x(ω) = F 0CωStiffness controlledMass controlledDamping controlledratio ξ affects the magnitude of the oscillation peak at resonance and the sharpnessof this resonant peak.In considering the system response as a function of frequency, we observethat the response varies with the frequency as shown in Table 20.1. The relationshipslisted in the table show that each parameter listed—the stiffness, the mass,and damping—effectively controls the response only within a limited region. Forexample, the damping is primarily effective at resonance. The selection of anyvibratory corrective measure depends on whether the excitation frequency is lessthan, greater than, or equal to the resonant frequency of the system.The effect of the amplitude by stiffness, mass, or damping is exemplified by themagnification factor MF, defined asMF = x(ω)F 0 /k = 1√ (20.27)[1 − (ω/ωn ) 2 ] 2 + [2ξ(ω/w n )] 2At resonance ω = ω n , and therefore(MF) resonance = 1(20.28)2ξA measure of the shape of the resonance peak is given by the bandwidth at thehalf-power points, as shown in Figure 20.5. These points are the two points, one tothe right and one to the left of the peak, which have a magnitude equal to (1/ √ 2)of the value of the peak. The square root occurs because power is proportional tothe square of the magnitude.Let us set h ≡ ω n /ω. At half-power points, Equation (20.27) becomes12 √ 2ξ = 1√(1 − h2 ) 2 + (2ξh) 2Solving the preceding equation algebraically for h 2 results inh 2 = 1 − 2ξ ± 2ξ √ 1 + ξ 2We also assume small values of damping (i.e., ξ ≪ 1) and neglect second-orderterms. Then the following result occurs:h 2 = 1 ± 2ξ