Aousti Waves in a Dut 1 One-Dimensional Waves The one-dimensional wave approximation is valid when the wavelength λ is muh larger than the diameter of the dut D, λ D. The aousti pressure disturbane p is then governed by 1 2 p 2 t 2 p = 0, 1) 2 x2 where is the speed of sound. In air at 293 o K, = 343m/s. The general solution of 1) is of the form p x, t) = f t x ) + g t + x ) 2) For harmoni waves, the general solution is of the form p x, t) = Ae i t x ) + Be it+ x ). 3) In order to determine a partiular solution to 1), we need to speify the initial and boundary onditions assoiated with a real problem. For example, let us onsider a piston loated at x = 0 in a semi-infinite dut as shown in Figure 1. Figure 1: One-Dimensional Wave Propagation in a Dut At time t = 0, the piston begins to osillate about its mean position with a speed Ut). Sine the dut extends from 0 to, the physial priniple of ausality tells us that all waves must propagate from the soure of sound outward, i.e., from the piston to the right. This is equivalent to a homogeneous boundary ondition imposed at infinity. As a result, g t + x ) 0. The fluid veloity u is related to the aousti pressure p by the momentum equation This gives u t = 1 p ρ x 1
ux, t) = p x, t) ρ We now apply the boundary ondition at x = 0, = 1 ρ f t x ) 4) ut, 0) = Ut) = 1 ft) for t 0. ρ o and solving for f gives f t x ) = ρ U t x ) for t x. Note that the initial ondition u0, x) = 0 for x > 0 implies that the sound produed by the osillation of the piston will not reah loations x > t. This is equivalent to taking Ut) 0 for t < 0. Substituting the expression of f into Equation 4) gives p = ρ U t x ). Note that if the piston has been osillating for a very long time, we an take the initial time at instead of 0. In this ase Ut) an be defined for all t. This is the ase for harmoni osillations where Ut) = Ae it. The solution takes the form p = Aρe i t x ), where A is the wave amplitude. Suh a wave is alled a plane wave sine its phase is onstant in a plane perpendiular to the x-axis. The physial solution is the real part or the imaginary part) of this solution, p = Aρ os [ )] t x 5) Note that v = p /ρ). The aousti intensity I = p v = A 2 ρos 2 [t x )]. 6) The average intensity Ī and aousti power P an then readily alulated where S is the dut ross setion. Ī = 1 2 A2 ρ 7) P = 1 2 A2 ρs, 8) 2
The following parameters are usually used to desribe harmoni waves. Cirular Frequeny: = 2πf [Radians per seond] 9) Frequeny: f = 2π [Hertz=yles per seond] 10) Period: Wavelength: Wave Number: T = 1 f = 2π λ = 2π k = = 2π λ [Seonds] 11) 12) 13) Phase Speed: = k. 14) 2 Retangular Dut with Rigid Boundaries 15) Figure 2: Wave Propagation in a Retangular Dut A shemati of the dut is shown in Figure 2. The governing equation for the aousti pressure is similar to that of the one-dimensional ase, 1 2 2 p t 2 2 p = 0. 16) The rigid walls assumption gives the boundary ondition p = 0. We use the method of n separation of variables to find a time-harmoni solution to this problem. Thus we assume a solution of the form p x, t) = fx 1 ) gx 2 ) hx 3 ) T t) 17) Substituting 17) into 16) gives If the osillation frequeny is, then 1 T 2 T f f g g h h T T = 2 = T t) = e it. 3 = 0. 18)
We ould have onsidered a solution of the form e it. This would give the omplex onjugate of the solution obtained with e it. This, however,does not affet the physial solution whih is the real part of the mathematial solution. Substituting the expression for T into 18) gives If we assign g g h and h f f + g g + h h = 2 2 to be onstant, we obtain g g = α2 = gx 2 ) = A 1 osαx 2 ) + B 1 sinαx 2 ) h h = β2 = hx 3 ) = C 1 osβx 3 ) + D 1 sinβx 3 ) Applying the boundary ondition at x 2 = {0, a} and x 3 = {0, b} gives: B 1 = 0, αa = mπ, D 1 = 0, βb = nπ, and ) ) mπ nπ gx 2 ) = A 1 os a x 2, hx 3 ) = C 1 os b x 3, where m and n are integers. Finally, we have giving, f ) mπ 2 nπ f = 2 + + 2 a b ) 2 fx 1 ) = E 1 e ikmnx 1 + F 1 e ikmnx 1, where k mn is defined as ) kmn 2 = 2 mπ 2 ) nπ 2 2 a b Sine we have assumed > o, and we are onsidering aousti waves propagating to the right, if we take 2 ) mπ 2 ) nπ 2, k mn = 19) 2 a b then E 1 = 0. Equation 19) whih gives the wave number k mn in terms of the frequeny is known as the dispersion equation. From the solutions for T, f, g, and h, and taking F 1 = 1, we have the solution ) p mπx2 mn x, t) = os os a nπx2 b ) e ik mnx 1 t). 20) 4
p mn is referred to as the m, n) mode. The veloity in the x 1 diretion is where the impedane Z mn = The general solution is of the form v 1mn = p mn Z mn, 21) ρ. 22) 1 mπ a )2 nπ )2 b p x, t) = m= m=0 n= n=0 a mn p mn x, t) 23) where a mn are onstants to be determined from the boundary onditions at x 1 = 0. If for example the aousti waves are aused by the osillations of a membrane desribed by U 1 x 2, x 3 )exp it), with the following Fourier ewxpansion then U 1 x 2, x 3 ) = m= m=0 n= n=0 mπx2 b mn os a ) os nπx2 b ), 24) a mn = b mn Z mn. 25) When < [ mπ a )2 + nπ b )2 ] k mn is purely imaginary and the amplitude of the wave would either inrease or derease exponentially. Sine growing waves are unaeptable physially, the waves will deay exponentially. These waves are alled evanesent waves. As inreases a new {m, n} mode begins to propagate or is said to ut on. Thus at a given frequeny, one the transients represented by the evanesent modes deay, the solution will have only a finite number of propagating modes. Note that the dispersion equation is not linear, i.e., k mn is not a linear funtion of. The axial phase speed, p = k mn > and depends on. Waves with different frequenies have different phase speeds and as a result the waves disperse as they propagate. Suh waves are alled dispersive waves. The energy of dispersive waves propagate with the group veloity g defined as g = d dk = k 2 mn = 2 p <. If we assume for simpliity that a = b. Then 2 k mn = π2 2 a 2 m2 + n 2 ) 26) Only waves where > π a m2 + n 2 ) 1 2 will propagate. Introduing the wave length λ = 2π, gives the following ondition for a wave to propagate, 5
a > λ 2 m 2 + n 2) 1 2. If a < λ, only the 0,0) mode orresponding to a plane wave propagates in the dut. The 2 higher order modes m > 0, n > 0) are ut-off. Thus if the dut diameter is muh smaller than the wavelength the problem is redued to that of the one-dimensional wave approximation. This result justifies the one-dimensional approximation for duts with a diameter small ompared to the wavelength. 6