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

76 4. Vibrating Stringstherefore differ in phase by π radians at x = 0. With these restrictions Equation(4.7) becomesy = a 1 [sin(ωt − kx) − sin(ωt + kx)] + b 1 [cos(ωt − kx) − cos(ωt + kx)](4.9)Making use of trigonometric transformations for the sine and cosine terms wesimplify Equation (4.9) toy = [−2a 1 cos ωt + 2b 1 sin ωt] sin kxThus y is expressed as a product of a time-dependent term and a coordinatedependentterm.Applying the boundary condition of y(L, t) = 0 (i.e., the displacement is zeroat end point x = L) adds yet another restrictionsin kL = sin nπ where n = 1, 2, 3,...Consequently, the string cannot vibrate freely at any random frequency; it can onlyvibrate with a discrete set of frequencies given byω n = nπc/Lwhere n = 1, 2, 3,..., or, in terms of frequency:f n = nc/(2L) (4.10)4.7 Standing WavesThe boundary conditions at x = 0 and at x = L reduced the general SHM solutionEquation (4.7) to a pattern of standing waves on the string. At the lowest orfundamental frequency, where n = 1, the displacement is given byy 1 = (A 1 cos ωt + B 1 sin ωt) sin k 1 x (4.11)Here k 1 = π/L, and A 1 and B 1 are arbitrary constants of which numerical valuesare established by the initial conditions, i.e., the type of excitation imparted tothe string at t = 0. This fundamental mode of vibration is associated with thefundamental (or first harmonic) frequency f 1 = c/2L. The nth mode of vibrationcorresponding to the nth harmonic frequency is represented byy n = (A n cos ω n t + B n sin ωt) sin k x (4.12)and the frequency is f n = nc/2L, i.e., n times the fundamental frequency. Theconstants A n and B n are determined by the initial excitation.In evaluating the term sin k n x = sin nπ x/L, we recognize that displacementy n = 0 occurs for all values of x when sin nπ x/L = 0, i.e.,nπ x/L = mπ where m = 0, 1, 2, 3,....