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

5.5 Mass Concentrated Bars 95symmetrical with respect to the center; otherwise a node in the center correspondsto asymmetrical vibration.As with case of the vibrating string, we can rigidly clamp a bar at any one ofits nodal positions without affecting the modes of vibrations which have a nodeat this position. But the vibrations that do not have a node at this position willbecome suppressed. The nature of the vibration of the free–free bar is such that itis not possible to clamp it at any position that will not eliminate at least some ofthe allowed modes of vibrations.The case of a free-fixed bar also makes for an interesting study. One end remainsfree at x = 0, and the other is rigidly clamped at x = L. The first condition∂ξ/∂x = 0atx = 0 leads again to Equation (5.15), while the second conditionξ = 0atx = L yieldse −ikL + e ikL = 0 or cos kL = 0which means that the allowable frequencies must satisfyork n L = ω nc L = π (2n − 1), n = 1, 2, 3,...2ω n = (2n − 1)π c2L , f n = (2n − 1) c 4lThe fundamental frequency is half that for an otherwise identical free–free bar, andonly odd-numbered harmonic overtones exist. The quality of the sound producedby an oscillating free-fixed bar will thus differ from that of a free–free bar, becauseof the absence of the even harmonics.5.5 Mass Concentrated BarsIn practical situations a vibrating bar is not truly clamped totally nor is it completelyfree to move at its ends. It may incorporate some type of mechanical impedance,most commonly as the result of concentrating a certain amount of mass at a certainlocation. An example is a diaphragm represented as a distributed mass located atone end of a vibrating tube inside a sonar transducer.As an example let us consider a bar that is unfettered at x = 0 and has a loadingconsisting of a mass m concentrated at x = L. The mass is depicted as a pointmass so that it does not move as a unit and thus merely sustain waves propagatingthrough it. The boundary condition ∂ξ/∂x = 0atx = 0 again leads us to Equation(5.17), as the result of A = B. For the boundary condition at x = L we again invokeNewton’s law of motion:( ∂ 2 ξF x (L, t) = m(5.17)∂t)x=L2A positive value of F x , which compresses the bar, will result in acceleration of themass in the positive x-direction. Because the mass is rigidly coupled to the bar,