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

96 5. Vibrating Barsthe accelerations of the mass and of the end of the bar should be identical. But ifthe mass had been concentrated at x = 0, a positive (compression) force wouldcorrespond to a reaction force to the left on the mass. The appropriate boundarycondition for this case would be( ∂ 2 ξ−F x (0, t) = m(5.18)∂t 2 )x=0Incorporating the boundary condition Equation (5.18) into Equation (5.16) resultsinwhich rearranges to−E Âe iωt (−ike −ikt + ike ikt ) = mAe iωt (−ω 2 )(e −ikL + e ikL )kEÂ sin kL =−mω 2 cos kLortan kL =− ωmc(5.19)E ÂBecause Equation (5.19) is a transcendental equation, no explicit solution exists.However, if the mass m is very small, m ≈ 0 and hence tan kL ≈ 0 and kL ≈ nω,both of which constitute the allowed conditions for a free–free bar. This is a resultthat should occur, since light loadings render a bar nearly free at both ends. At theother extreme, for very heavy mass loadings, the mass behaves very nearly like arigid support, and the allowed frequencies will approximate those of a free-fixedbar.In the more general case of intermediate mass loading, it is rather cumbersometo solve by hand the transcendental equation (5.19) through graphic means. However,computer programs such as Mathcad r○ , MathLab r○ , Mathematica r○ ,orevenaprofessional-level spread sheet for IBM compatible and Macintosh personal computerscan be used to facilitate solutions. Eliminating Young’s modulus in Equation(5.19) by applying E = ρc 2 from Equation (5.9) and recognizing that the mass ofthe bar is given by m b = ρ ÂL, we can rewrite Equation (5.19) astan kL=− m (5.20)kL m bThe right-hand side of Equation (5.20) is fixed by amount of mass m b in the barand the loading mass m located at x = L. An example of the solution to Equation(5.20) is given in Figure 5.4 for the case of a steel bar with a mass loading m/m bof 20%. The longitudinal velocity of sound propagation of steel is 5050 m/s. Thefundamental frequency f 1 is found from k 1 L through the relationf 1 = k 1L2πand the higher frequencies are similarly established fromf n = k n L2πcLcL