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

104 5. Vibrating Barsthe light beam travel with different speeds thereby causing dispersion. A vibratingbar thus acts as a dispersive medium for transverse waves.The real part of Equation (5.33) constitutes the actual solution of Equation(5.29). We make use of the following hyperbolic and trigonometric identities:sin y = (e iy − e −iy )/2, cos y = (e iy + e −iy )/2sin(iy) = i sinh y, sinh(iy) = i sin ycos(iy) = cosh y, cosh(iy) = cos yto recast Equation (5.33) asy = cos(ωt + φ)(A cosh ωx ωx ωx+ B sinh + C cosv v vωx)+ D sinv(5.34)Here A, B, C, D are real constants that occur from the rearrangement of the originalcomplex constants A, B, C, D. The intricate relationships between the real set ofconstants and the set of complex constants are not really of much concern to us,because it is the application of the initial and boundary conditions that provides theevaluation of these constants. However, there are twice as many arbitrary constantsin the transverse equation (5.34) as in the longitudinal wave equation (5.7), due tothe fact that the former is of the fourth-differential order rather than the seconddifferentialorder. Therefore, twice as many boundary conditions are required, andthis can be satisfied by specifying pairs of boundary conditions at the ends of thebars. The nature of the supports establishes the boundary conditions that generallyfall into the categories of free and clamped ends.5.8 Boundary Conditions for Transverse Vibrations1. If the bar is rigidly clamped at one end, the both the displacement and the slopemust be zero at that end at all times, and the boundary conditions are expressed as:y = 0, ∂y/∂x = 0 (5.35)2. On the other hand, neither an externally applied moment nor a shear force mayexist at a free end of a vibrating bar. But the displacement and the slope of the barat a free end are not constrained, excepting for the mathematical stipulation theyremain small. From Equations (5.26) and (5.28) the boundary conditions become∂ 2 y∂x 2 = 0,Case 1: Bar Clamped at One End∂ 3 y∂x 3 = 0 (5.36)Consider a bar of length L that is rigidly clamped at x = 0 but is free at x = L.At x = 0, the two conditions of Equation (5.35) apply, so A =−C and B =−D.The general solution (5.35) reduces toy = cos(ωt + φ)[A(cosh ωxv− cosωxv) (+ B sinh ωxvωx)]− sinv