1 Fundamental Solutions to the Wave Equation

Transcription

1 1 Fundamental Solutions to the Wave Equation Physial insight in the sound geneation mehanism an be gained by onsideing simple analytial solutions to the wave equation One example is to onside aousti adiation with spheial symmety about a point y = {y i }, whih without loss of geneality an be taken as the oigin of oodinates If t stands fo time and x = {x i } epesent the obsevation point, suh solutions of the wave equation, ( 2 t 2 2 o 2 φ = 0, (1 will depend only on the = x y It is eadily shown that in this ase (1 an be ast in the fom of a one-dimensional wave equation ( 2 The geneal solution to (2 an be witten as 2 t 2 2 o (φ = 0 (2 2 φ = f(t + g(t + (3 The funtions f and g ae abitay funtions of the single vaiables τ ± = t±, espetively They detemine the patten o the phase vaiation of the wave, while the fato 1/ affets only the wave magnitude and epesents the speading of the wave enegy ove lage sufae as it popagates away fom the soue The funtion f(t epesents an outwadly going wave popagating with the speed The funtion g(t + epesents an inwadly popagating wave popagating with the speed 2 The Pulsating Sphee Conside a sphee enteed at the oigin and having a small pulsating motion so that the equation of its sufae is = a(t = a 0 + a 1 (t, (4 whee a 1 (t << a 0 The fluid veloity at the sphee sufae is At the sufae of the sphee A Taylo expansion of (6 gives u = d dt = ȧ(t (5 a = ȧ(t (6 a = a 0 + (a a 0 ( 2 φ 2 a 0 + (7 1

2 We assume (aa 0 ( 2 φ 2 a0 << ȧ This allows us to lineaize the bounday ondition along the sphee by tansfeing it to the mean position at a 0, a 0 = ȧ(t (8 The veloity potential an be expessed as in (3 Moeove sine the sphee pulsating motion is the soue of aousti waves, the piniple of ausality suggests that g 0 Thus φ = f(t Applying the ondition (8 at the sphee mean loation, Integation of (10 gives φ (9 a0 = f(t o f(t a 0 o = ȧ(t (10 a 2 0 a 0 f(t = a 0 t ȧ(t + a 0 e o (tt a 0 dt (11 Note that if T is a epesentative peiod of the sphee pulsation, T/a 0 = λ/a 0, whee λ is a epesentative of the sound wave length If λ/a 0 >> 1, then most of the ontibution to the integal (11 is when t t Negleting tems of O(a 0 /λ, we get and the aousti field potential funtion is given by The expession fo the aousti pessue is f(t = a 2 0ȧ(t, (12 φ = a2 0ȧ(t (13 ä(t p = ρ 0 a 2 0 (14 It is onvenient to ast (13,14 in tems of the mass flow ate ossing the sphee of adius a 0, m(t = 4πa 2 0ȧρ 0 f(t = ṁ 4πρ 0 and φ = ṁ(t 4πρ 0, (15 p = m(t 4π (16 2

3 21 Hamoni Motion If we have a hamoni motion ȧ = ve iωt, (17 whee v is the amplitude of the pulsation veloity and ω its fequeny Substituting (17 into (11 and aying out the integation, we get a 0 f(t = a 0 v eiω(t+ a 0 + iω (18 The expessions fo the potential funtion and the pessue an be eadily obtained by substituting (18 into (9, φ = m 4πρ ω 2 ei(ωtk(a 0+ϕ, (19 p = iω m 4π 1 + ω 2 ei(ωtk(a 0+ϕ (20 whee we have intodued ω = ωa 0 /, ϕ = tan 1 ω, k = ω/, and m = 4πa 2 0 vρ 0 The aveage aousti intensity and powe an be alulated and we have, 3 The Simple Soue Ī = 1 8π m ω v 1 + ω 2, (21 P = 1 2 m ω 2 0 v 1 + ω 2 (22 The limit of the pulsating sphee solution as the sphee adius vanishes epesents the simple soue o monopole solution In this ase, the soue is haateized by the soue mass flow ate ṁ(t = lim Ra 0 0 4πa2 0u = 4πa 2 0ȧ(t, and the exat solution is the same as fo the low fequeny ase (15 If the soue is loated at the point y, then φ = ṁ(t, (23 4πρ 0 whee = x y Equation (23 states that at the obsevation point x and time t the sound signal eeived was emitted fom the soue point y at the etaded time τ = t The expession fo the pessue is p = m(t 4π 3 (24

4 In the fa field, negleting tems popotional to 1/ 2, we get the following expessions fo the intensity and aousti powe ae The Dipole: I = m2 (t 16π 2 ρ 0 2, (25 P = m2 (t 4πρ 0 (26 Conside two soues of equal and opposite stength ±m i loated at ± l whee ± = x l We futhe assume l x, then φ ± = ṁ(t ± /(4πρ o ±, (27 ± = l x + = l os θ +, (28 whee l = l and θ is the angle between the obsevation veto x and l The potential fo the dipole field is φ = φ + + φ = 1 [ṁ(t + ] ṁ(t 0 (29 4πρ 0 + Using a Taylo seies expansion of(29, we get φ = 1 [ṁ(t 2 4πρ l 0 ] + (30 Intoduing the veto F = 2 m l and letting l = l 0, we obtain the following expession fo the the pessue p = 1 4π F (t (31 The veto F has the dimension of a foe and is alled the dipole stength In the fa field, negleting tems popotional to 1/ 2, we get whih an be witten as p = 1 4π F (t, (32 p = 1 4π F (t osθ, (33 4

5 whee F is the magnitude of F The intensity and aousti powe ae I = P = F 2 (t os 2 θ, 16π 2 ρ (34 F 2 (t 12πρ (35 We define the pessue dietivity by and the intensity dietivity by: D p = p, (36 D I = 2 I (37 Note that the atio of the dipole powe to that of the simple soue is P d = 1 P s 3 ( F 2 (38 m Realling that F = 2l m and assuming a fequeny dependene ω, we get P d P s = 4l2 ω = ( π 3 2 ( 2l λ 2 (39 Sine we have defined the dipole fo l 0, we may onlude that a dipole is a muh less effiient soue of sound than the simple monopole soue 5