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 v = d dt = ȧ(t. (5 ( φ a = ȧ(t. (6 ( φ a = ( φ a 0 + (a a 0 ( 2 φ 2 a 0 + (7 If ω and λ ae epesentative of the fequeny and wave length assoiated with the aousti wave, then (a a 0 ( 2 φ 2 a0 lage{ a 1ω ȧ, a 1 a 0 ȧ}. Hene, if a 1 ω << 1, o equivalently a 1 λ << 1 1

2 , then (a a 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, ( φ 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 (t t 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 ä(t p = ρ 0 a 2 0. (13 It is onvenient to ast (13,14 in tems of the mass flow ate ossing the sphee of adius a 0, m(t = 4πρ 0 a 2 0ȧ. f(t = m 4πρ 0 and (14 φ = m(t 4πρ 0, (15 p = ṁ(t, 4π (16 v = 1 4πρ (ṁ(t + m(t 0 2 (17 This suggests that in the fafield, i.e., >> λ, the aousti pessue and veloity ae in phase and the speifi aousti impedane z = p /v = ρ 0. 2

3 2.1 Hamoni Motion If we have a hamoni motion ȧ = ve iωt, (18 whee v is the amplitude of the pulsation veloity and ω its fequeny. Substituting (18 into (11 and aying out the integation, we get a 0 f(t = a 0 v e iω(t+ a 0 iω. (19 The expessions fo the potential funtion, veloity and the pessue an be eadily obtained by substituting (19 into (9, φ = m 4πρ ω 2 ei(k( a 0+ϕ ωt, (20 v = m 4πρ ω ( ik + 1 e i(k( a0+ϕ ωt, 2 (21 p = iω m 4π 1 + ω 2 ei(k( a 0+ϕ ωt. (22 whee we have intodued ω = ωa 0 /, ϕ = tan 1 ω, k = ω/, and m = 4πa 2 0 vρ 0. Note that the veloity is the sum of two tems. One tems is in-phase with the pessue and at lage distane deays as 1/. The othe tem is out of phase with the pessue and deays at lage distane as 1/ 2. The speifi aousti impedane z = p k = R + ix = ρ 0 (k i, (23 v 1 + k 2 2 whee R is the esistane and X the eatane. In the fafield whee >> λ o, k >> 1, the speifi aousti impedane is dominated by R whih has the same value as a plane wave, z = ρ 0. Howeve, in the nea field the eatane X is ompaable to R fo k = O(1 and X >> R fo k << 1. The expession fo the instantaneous aousti intensity is given by I = p 2 [ Rsin 2 (k( a R 2 + X ϕ ωt + X ] 2 sin[2(k( a 0 + ϕ ωt]. (24 Note R/(R 2 + X 2 = 1/(ρ 0 and X/(R 2 + X 2 = 1/(ρ 0 k, thus (24 simplifies to I = p 2 [ sin 2 (k( a 0 + ϕ ωt 1 ] ρ 0 2k sin[2(k( a 0 + ϕ ωt]. (25 The tem assoiated with the eatane X esults fom oupling the pessue with the outof-phase tem of the veloity. Its time aveage is zeo and hene it does not ontibute to 3

4 the popagation of aousti enegy. Only the tem assoiate with the esistane, whih is the esult of oupling the in-phase veloity with the pessue podues adiated aousti enegy. The aveage aousti intensity ae, Ī = P = m 2 ω 2 32π 2 ρ 0 (1 + ω 2, 2 (26 m 2 ω 2 8πρ 0 (1 + ω 2. (27 Fo a sphee of small adius ompaed to the wave length, i.e., a 0 << λ, the aousti impedane z << ρ 0 in the nea field. Moeove, sine R << X, the impedane is stongly eative and the suounding fluid ats mainly as an inetial mass. The veloity is patially out of phase with the pessue. Thus, a soue of small size is an ineffiient adiato of aousti enegy. 3 The Simple Soue The limit of the pulsating sphee solution as the sphee adius a 0 beomes vey small epesents the simple soue o monopole solution. In this ase, the soue is haateized by the soue mass flow ate m(t = 4πa 2 0ȧ(t, and the exat solution is given by (15. If the soue is loated at the point y, then φ = m(t 4πρ 0, (28 whee = x y. Equation (28 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 veloity and pessue ae given by v = φ = 1 4πρ 0 [ṁ(t p φ = ρ 0 t = ṁ(t 4π ] 2 + m(t (29 ( Hamoni Soues: In this ase m = me iωt (31 φ = m 4πρ 0 e iωτ = m 4πρ 0 ei(k ωt, (32 4

5 whee m epesent the stength of the soue and k = ω/. Similaly, the the veloity expession is given by v = φ = 1 [ṁ(t + m(t ] e ik (33 4πρ 0 2 o Noting that ω = 2π λ, v = v = m 4πρ 0 m 4πρ 0 At lage distane, λ, the aousti intensity is given by [ iω + 1 ] e i(k ωt (34 2 [ i2π λ + 1 ] e i(k t (35 2 p iω m = 4π ei(k ωt. (36 and I = p v = m 2 ω 2 16π 2 ρ o 2 sin2 (ω(t (37 Ī = 1 2 Re(p v = m 2 ω 2 32π 2 ρ o 2 (38 P = m2 ω 2 8ρ o π ( Simple soue distibution: Suppose we have N soues loated at y i with stength m i, then the piniple of supeposition states that: whee i = x y i. φ = N i=1 φ i = 1 4π N i=1 m i (t i i (40 5

6 4 The Dipole: Conside two soues of equal and opposite stength ±m loated at y ± l/2. The potential of the two soues is φ = 1 + m(t [ 0 m(t 0 ], (41 4πρ 0 + whee = x y and ± = x ( y ± l/2. We futhe assume l, then o The pessue is given by o φ = 1 [ m(t l i 4πρ 0 x i φ = 1 4πρ 0 p = 1 4π p = 1 4π ] +... (42 lm(t (43 f(t ( f i(t x i +..., (44, (45 whee f = {f i } = ṁ l. Note that f has the dimension of a foe. The pessue an be moe expliitly expessed as p = 1 4π [ 1 f i (t t + f i(t ] (x i y i, (46 2 We an always assume that dipole is loated at a finite distane, i.e., y is finite. The fa field aousti veloity and pessue ae the leading tems of those fields as the obseve x, o p = p = 1 4π x 1 4π x df(t dt x x. (47 df(t osθ, (48 dt whee θ is the angle between f and x and f = f. Note that fo y << x, has the following expansion = x y x x + O( 1. (49 x 6

7 Note that the geneal expession fo the veloity an be obtained by fom (43. Howeve, in the fafield, the aousti veloity edues to its adial omponent, v, whose expession is simply given in tems of the aousti pessue The aousti intensity has the following expession The aveage aousti intensity is then given by Using (48, we get I = v = p ρ 0. (50 I = p v. (51 I = 1 p p 2 ρ o.. (52 f(t f(t os 2 θ 32π 2 ρ (53 The total powe adiated is obtained by integation of a sphee of adius, this gives P = 1 f(t f(t 24π ρ (54 In many appliations, the foe epesenting the dipole is eated by the inteation of flow nonunifomities and tubulene inteation with stutual omponents suh as wings, fan and ompesso blades. The flutuating pessue along the sufae of these stutue epesent a dipole distibution. The fafield aoustis due to this unsteady pessue an be alulated by summing the ontibution of all the dipoles. As an example, we onside a flat plate aifoil in a nonunifom flow omposed of a unifom flow V in the x 1 dietion and a tansvese gust distubane v = ve i(k 1x 1 ωt, (55 Whee k 1 = ω. The sufae pessue jump is given by V p s = 2ρ 0 V 2 v y 2 1 V + y S(ω e iωt, ( whee ω = ω/(2v is the edued fequeny and S(ω is the Seas funtion. The elementay foe applied on the elementay sufae da = dy 1 dy 2 is df = p sda. The aousti pessue is given by p = 1 4π x 2ρ 0V 2 v V S(ω +b/2 +/2 b/2 /2 2 y y 1 x y iω(t e 0 dy 1 dy 2. (57 7

8 Fo simpliity we use the ompat soue appoximation whih assumes that the body size is small ompaed o the wave length. Hee, it implies ω << 1 o << λ. In this ase f = πρ 0 V 2 v x V S(ω iω( e t 0, (58 is the foe pe unit span b. Using this expession to alulate P, we get 5 The Quadupole: [ π P = (ρ 0 3 0A 6 M 6 ( v V 2 ( ω ] b2 2 SS. (59 2V A The onept of a dipole as a ombination of two monopoles with equal but opposite stength an be genealized to define a quadupole as a ombination of two dipoles of equal but opposite stength sepaated by l = {l i}. Following the same poedue, we obtain If we intodue the tenso p = 1 2 4π l j ( f i(t x i x j. (60 The expession fo the pessue (60 an be ast as p = 1 4π T ij = l jf i. (61 2 ( T ij(t x i x j (62 Note T ij has the dimension of a shea volume. The fa field expession fo a quadupole is p = 1 T 4π 2 ij (t x ix j ( If l is in the same dietion as f, the quadupole is said to be longitudinal, and we have p = 1 d 2 T 4π 2 0 dt 2 os2 θ, (64 whee T = l f. On the othe hand, if l is nomal to f, the quadupole is said to be lateal, and we have p = 1 d 2 T osθsinθ, (65 4π 2 0 dt2 8

9 6 The Geen s Funtion of the Wave Equation Fo Laplae equation, the Geen s funtion G( x, y is solution to the inhomogeneous equation Integate 66 ove a sphee enteed at y of adius = x y, This an be ewitten using the divegene theoem as Σ 2 G = 4πδ( x y.. (66 Σ 2 G d x = Σ 2 G d x = 4π. (67 G n dσ = dg d 4π2 = 4π whih gives the fee-spae Geen s funtion fo Laplae equation as G = 1. (68 The Geen s funtion G( x, y, t, τ fo the wave equation is solution to ( o 2 G = 4πδ( x y δ(t τ. (69 t 2 Beause of the spheial symmety, the geneal solution to 69 is G = f(t τ Fo the ases whee 0, t τ, the funtion satisfies ( o 2 G = 0. t 2., (70 Nea = 0, ( o 2 ( 1 G f(t τ 2 t 2 = 4πδ( x y δ(t τ. Hene, f(t = δ(t and we have the following expession fo the Geen s funtion G = δ ( t τ x y. (71 x y 9

10 6.1 Hamoni Time-Dependene Fo a hamoni time-dependene, we seek solutions of the fom φ( x, t = ϕ( xe iωt. (72 The funtion ϕ satisfies the Helmholtz equation ( 2 + k 2 ϕ = 0, (73 whee k = ω/. The Geen s funtion fo 73 is solution to ( 2 + k 2 g( x, y = 4πδ( x y. (74 It is eadily seen that g( x, y = e±ik. (75 This an be also obtained by taking the Fouie tansfom in time of 71 and noting that e i(ωt±k = 1 + δ (t τ ± e iωτ dτ. (76 o 7 Distibution of Soues and Foes The lineaized Eule equations fo non-visous non-heat-onduting fluid with no-mean motion ae ρ o t v = f p (77 ρ t + ρ o v = q (78 whee f is a foe pe unit volume and q is a mass flow ate pe unit volume. Combining the two equations gives ( o 2 p = q t 2 t + f. (79 In ode to find solutions to the aousti wave equation satisfying 79, we onside the inhomogeneous equation ( p = h( x, t. (80 2 o t 2 Using the Geen s funtion 71, and noting that h( x, t = h( y, τδ( x yδ(t τd ydτ, (81 10

11 we get p = 1 h( y, t d y + p 4π 0, (82 whee p 0 is a solution to the homogeneous equation and aounts fo the effets of boundaies. In what follows we fous on the inhomogeneous solution. 7.1 Soue Distibution In this ase h = q. Hene, we have t p = 1 4π q t ( y, t d y, ( Single Soue Fo a single soue at y 0, q( y, t = m(tδ( y y 0. Substituting into 83, we get bak the pessue expession 30 fo a single soue. whee 0 = x y 0. p = ṁ(t 0 4π 0, ( Foe Distibution Foe a distibution {f i }, we onside the solution to the equation This gives ( o ϕ i = 1 4π Taking the divegene of (86, we get 2 ϕ t 2 i = f i. (85 f i ( y, t p = 1 f i ( y, t 4π x i A moe diet expession fo the pessue is given by The aousti field is given by p = 1 f i [ ( y, t t 4π p = 1 f i [ ( y, t t 4π d y. (86 + f i( y, t 11 d y. (87 ] (x i y i d y. (88 2 ] (x i y i d y. (89

12 7.2.1 Single Foe Fo a single foe at y 0, f i ( y, t = F i (tδ( y y 0. Substituting into (88, we get bak the pessue expession (30 fo a single soue, The aousti field is p = 1 dfi 4π [ (t 0 dt 0 + F i(t ] (x i y 0i 0. (90 p = 1 4π df dt (t 0 osθ, (91 whee F is the magnitude of the foe F and θ is the angle between F and the obseve dietion x. Note that as expeted these esults ae idential to those obtained fo a dipole. This onfims that a dipole epesents the field of a foe. 12