1 Fundamental Solutions to the Wave Equation

Transcription

1 1 Fundamental Solutions to the Wave Equation Physical insight in the sound geneation mechanism can be gained by consideing simple analytical solutions to the wave equation. One example is to conside acoustic adiation with spheical symmety about a point y = {y i }, which without loss of geneality can be taken as the oigin of coodinates. If t stands fo time and x = {x i } epesent the obsevation point, such solutions of the wave equation, ( 2 t 2 c2 o 2 φ = 0, (1 will depend only on the = x y. It is eadily shown that in this case (1 can be cast in the fom of a one-dimensional wave equation ( 2 The geneal solution to (2 can be witten as 2 t 2 c2 o (φ = 0. (2 2 φ = f(t + g(t +. (3 The functions f and g ae abitay functions of the single vaiables τ ± = t±, espectively. They detemine the patten o the phase vaiation of the wave, while the facto 1/ affects only the wave magnitude and epesents the speading of the wave enegy ove lage suface as it popagates away fom the souce. The function f(t epesents an outwadly going wave popagating with the speed. The function g(t + epesents an inwadly popagating wave popagating with the speed. 1.1 Acoustic Enegy Conside a fluid blob at est of volume V 0 with pessue p 0 and density ρ 0. As a sound wave eaches the blob of fluid, it acquies a velocity v and its volume, pessue and density change to V, p and ρ, espectively. If we wite V = V 0 + V (4 p = p o + p (5 ρ = ρ o + ρ, (6 the incemental changes V /V 0, p /p 0 and ρ /ρ ae vey small. The blob has the same mass, so ρ o V o = ρv = m. The kinetic enegy of the blob is given by and the acoustic kinetic enegy density is 1 2 ρv2 V = 1 2 ρ ov 2 V o 1 2 ρv2. 1

2 The blob is compessed against the backgound pessue p 0. The compession enegy is given by V W = p dv V V 0 p dρ (7 ρ 0 V o V o If the pocess is assumed to be isentopic, then dp /dρ = c 2 0. Using this elationship and substituting fo dρ in (7 gives W = V 0 ρ 0 c 2 0 V V o p dp = p 2 2ρ 0 c 2 0 o, expessed in tems of the acoustic potential enegy density,. The total enegy density is p 2 2ρ o 2 E = 1 2 ρ ov 2 + p 2 The acoustic enegy flux o acoustic intensity is V 0 (8 2ρ o 2. (9 I = p v. (10 Note that the consevation of enegy implies V E t dv + S I n ds = 0, which using the divegence theoem gives E t + I = 0. (11 Equation (11 can be veified by substituting the expessions fo I and E into the mass and momentum equations, namely ρ t + ρ o v = 0 ρ o v t = p 2

3 2 Plane Waves Plane waves ae solutions to the wave equation (1 of the fom p = pe i( k xωt, (12 whee the wave numbe k = k = ω/. Note that if we define the unit vecto u = k/k, the wave popagates in the diection u and v = p ρ 0 u. (13 Thus fo plane waves the elationship between pessue and velocity is given by p = ρ 0 v. (14 Note also that the enegy of a plane wave is equally divided between kinetic and potential enegy and we have 3 The Pulsating Sphee E = ρ 0 v 2 = p 2 ρ 0 c 2 0 Conside a sphee centeed at the oigin and having a small pulsating motion so that the equation of its suface is = a(t = a 0 + a 1 (t, (16 whee a 1 (t << a 0. The fluid velocity at the sphee suface is At the suface of the sphee A Taylo expansion of (18 gives u = d dt (15 = ȧ(t. (17 ( φ a = ȧ(t. (18 ( φ a = ( φ a 0 + (a a 0 ( 2 φ 2 a 0 + (19 We assume (aa 0 ( 2 φ 2 a0 << ȧ. This allows us to lineaize the bounday condition along the sphee by tansfeing it to the mean position at a 0, ( φ a 0 = ȧ(t. (20 The velocity potential can be expessed as in (3. Moeove since the sphee pulsating motion is the souce of acoustic waves, the pinciple of causality suggests that g 0. Thus φ = f(t 3. (21

4 Applying the condition (20 at the sphee mean location, Integation of (22 gives φ a0 c = f(t o f(t a 0 c o = ȧ(t (22 a 2 0 a 0 f(t = a 0 t ȧ(t + a 0 e co (tt a 0 dt. (23 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 contibution to the integal (23 is when t t. Neglecting tems of O(a 0 /λ, we get and the acoustic field potential function is given by The expession fo the acoustic pessue is f(t = a 2 0ȧ(t, (24 φ = a2 0ȧ(t. (25 ä(t p = ρ 0 a 2 0 (26 It is convenient to cast (25, 26 in tems of the mass flow ate cossing the sphee of adius a 0, m(t = 4πa 2 0ȧρ 0. f(t = m and 3.1 Hamonic Motion If we have a hamonic motion φ = m(t, (27 p = ṁ(t. 4π (28 ȧ = ve iωt, (29 whee v is the amplitude of the pulsation velocity and ω its fequency. Substituting (29 into (23 and caying out the integation, we get a 0 f(t = a 0 v eiω(t+ a 0 + iω. (30 The expessions fo the potential function and the pessue can be eadily obtained by substituting (30 into (21, 4

5 φ = m 1 + ω 2 ei(ωtk(a 0+ϕ, (31 p = iω m 4π 1 + ω 2 ei(ωtk(a 0+ϕ. (32 whee we have intoduced ω = ωa 0 /, ϕ = tan 1 ω, k = ω/, and m = 4πa 2 0 vρ 0. The aveage acoustic intensity and powe can be calculated and we have, 4 The Simple Souce Ī = 1 8π mc ω v 1 + ω 2, (33 P = 1 2 mc ω 2 0 v 1 + ω. 2 (34 The limit of the pulsating sphee solution as the sphee adius vanishes epesents the simple souce o monopole solution. In this case, the souce is chaacteized by the souce mass flow ate m(t = lim Ra 0 0 4πa2 0u = 4πa 2 0ȧ(t, and the exact solution is the same as fo the low fequency case (27. If the souce is located at the point y, then φ = m(t, (35 whee = x y. Equation (35 states that at the obsevation point x and time t the sound signal eceived was emitted fom the souce point y at the etaded time τ = t. The velocity and pessue ae given by u = φ = 1 [ṁ(t 2 + m(t (36 Hamonic souces: In this case p φ = ρ 0 t = ṁ(t 4π (37 m = me iωt (38 φ = m eiω(t (39 5

6 Noting that ω = 2π λ, u = m u = m p = [ iω + 1 e iω(t 2 [ i2π λ + 1 e iω(t 2 iω m 4π ei(t At lage distance, λ, the acoustic intensity is given by (40 (41 (42 and I = p u = m 2 ω 2 16π 2 2 ρ o sin 2 (ωt (43 Ī = p u = m 2 ω 2 32π 2 2 ρ o (44 P = m2 ω 2 8π ρ o (45 Simple souce distibution: Suppose we have N souces located at y i with stength m i, then the pinciple 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 (46 The Dipole: Conside two souces of equal and opposite stength ±m i located at ± l. We futhe assume l x, then φ ± = ±m(t ± /(4πρ o ± (47 ± = l x +... = l cos θ +... (48 φ = φ + + φ = 1 m(t ± ± [ m(t + + = m(t [ṁ(t c ± l cos θ 0 m(t 2 + m(t (49 (50 6

7 φ = 2l cos θ [ṁ(t 2 + m(t If we assume l to be small and conside the field at a distance >> λ, then let µ = 2lṁ, (51 φ = 2lṁ ( t cos θ (52 p = 1 µ (t cos θ (53 4π c0 µ is called the stength of the dipole. The dipole stength has the dimension of a foce. We define the pessue diectivity by: (p = 1 µ (t cos θ (54 4π c0 7