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1 Fundamental Solutions to the Wave Equation 2 the Pulsating Sphere

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1 Fundamental Solutions to the Wave Equation 2 the Pulsating Sphere 1 Fundamental Solutions to the Wave Equation Physical insight in the sound generation mechanism can be gained by considering simple analytical solutions to the wave equation. One example is to consider acoustic radiation with spherical symmetry about a point ~y = fyig, which without loss of generality can be taken as the origin of coordinates. If t stands for time and ~x = fxig represent the observation point, such solutions of the wave equation, @2 ( − c2r2)φ = 0; (1) @t2 o will depend only on the r = j~x − ~yj. It is readily shown that in this case (1) can be cast in the form of a one-dimensional wave equation @2 @2 ( − c2 )(rφ) = 0: (2) @t2 o @r2 The general solution to (2) can be written as f(t − r ) g(t + r ) φ = co + co : (3) r r The functions f and g are arbitrary functions of the single variables τ = t± r , respectively. ± co They determine the pattern or the phase variation of the wave, while the factor 1=r affects only the wave magnitude and represents the spreading of the wave energy over larger surface as it propagates away from the source. The function f(t − r ) represents an outwardly co going wave propagating with the speed c . The function g(t + r ) represents an inwardly o co propagating wave propagating with the speed co. 2 The Pulsating Sphere Consider a sphere centered at the origin and having a small pulsating motion so that the equation of its surface is r = a(t) = a0 + a1(t); (4) where ja1(t)j << a0. The fluid velocity at the sphere surface is dr v = =a _(t): (5) r dt At the surface of the sphere @φ ( ) =a _(t): (6) @r a A Taylor expansion of (6) gives @φ @φ @2φ ( ) = ( ) + (a − a )( ) + ··· (7) @r a @r a0 0 @r2 a0 If ! and λ are representative of the frequency and wave length associated with the acoustic 2 @ φ a1! a1 ja1j! a1 wave, then (a−a0)( 2 )a ≈ largerf a;_ a_g. Hence, if << 1, or equivalently << 1 @r 0 c0 a0 c0 λ 1 @2φ , then j(a − a0)( @r2 )a0 j << ja_j. This allows us to linearize the boundary condition along the sphere by transferring it to the mean position at a0, @φ ( ) =a _(t): (8) @r a0 The velocity potential can be expressed as in (3). Moreover since the sphere pulsating motion is the source of acoustic waves, the principle of causality suggests that g ≡ 0. Thus f(t − r ) φ = co : (9) r Applying the condition (8) at the sphere mean location, f(t − a0 ) f_(t − a0 ) @φ co co = − 2 − =a _(t) (10) @r a0 a0co Integration of (10) gives Z t a co 0 0 0 − a (t−t ) 0 f(t) = −a0co a_(t + )e 0 dt : (11) −∞ c0 Note that if T is a representative period of the sphere pulsation, coT=a0 = λ/a0, where λ is a representative of the sound wave length. If λ/a0 >> 1, then most of the contribution to 0 the integral (11) is when t ≈ t. Neglecting terms of O(a0/λ), we get 2 f(t) = −a0a_(t); (12) and the acoustic field potential function is given by a2a_(t − r ) φ = − 0 co : (13) r The expression for the acoustic pressure is a¨(t − r ) p0 = ρ a2 co (14) 0 0 r It is convenient to cast (13,14) in terms of the mass flow rate crossing the sphere of radius 2 m a0, m(t) = 4πρ0a a_. f(t) = − and 0 4πρ0 m(t − r ) φ = − co ; (15) 4πρ0r m_ (t − r ) p0 = co ; (16) 4πr m_ (t − r ) m(t − r ) 1 c0 c0 vr = ( + 2 ) (17) 4πρ0 rc0 r This suggests that in the farfield, i.e., r >> λ, the acoustic pressure and velocity are in 0 phase and the specific acoustic impedance z = p =vr = ρ0c0. 2 2.1 Harmonic Motion If we have a harmonic motion a_ =ve ¯ −i!t; (18) wherev ¯ is the amplitude of the pulsation velocity and ! its frequency. Substituting (18) into (11) and carrying out the integration, we get −i!(t+ a0 ) e c0 f(t) = −a0c0v¯ : (19) c0 − i! a0 The expressions for the potential function, velocity and the pressure can be readily obtained by substituting (19) into (9), m¯ φ = − p ei(k(r−a0)+'−!t); (20) 2 4πρ0r 1 +! ~ m¯ −ikr + 1 v = p ( ) ei(k(r−a0)+'−!t); (21) r 2 4πρ0r 1 +! ~ r i!m¯ p0 = − p ei(k(r−a0)+'−!t): (22) 4πr 1 +! ~2 −1 2 where we have introduced! ~ = !a0=c0, ' = tan !~, k = !=c0, andm ¯ = 4πa0vρ¯ 0. Note that the velocity is the sum of two terms. One terms is in-phase with the pressure and at large distance decays as 1=r. The other term is out of phase with the pressure and decays at large distance as 1=r2. The specific acoustic impedance p0 kr z = = R + iX = ρ0c0 2 2 (kr − i); (23) vr 1 + k r where R is the resistance and X the reactance. In the farfield where r >> λ or, kr >> 1, the specific acoustic impedance is dominated by R which has the same value as a plane wave, z = ρ0c0. However, in the near field the reactance jXj is comparable to R for kr = O(1) and jXj >> R for kr << 1. The expression for the instantaneous acoustic intensity is given by jp0j2 X I = Rsin2(k(r − a ) + ' − !t) + sin[2(k(r − a ) + ' − !t))] : (24) R2 + X2 0 2 0 2 2 2 2 Note R=(R + X ) = 1=(ρ0c0) and X=(R + X ) = −1=(ρ0c0kr), thus (24) simplifies to 0 2 jp j 2 1 I = sin (k(r − a0) + ' − !t) − sin[2(k(r − a0) + ' − !t))] : (25) ρ0c0 2kr The term associated with the reactance X results from coupling the pressure with the out- of-phase term of the velocity. Its time average is zero and hence it does not contribute to 3 the propagation of acoustic energy. Only the term associate with the resistance, which is the result of coupling the in-phase velocity with the pressure produces radiated acoustic energy. The average acoustic intensity are, 2 2 ¯ m¯ ! I = 2 2 2 ; (26) 32π ρ0c0(1 +! ~ )r m¯ 2!2 P = 2 : (27) 8πρ0c0(1 +! ~ ) For a sphere of small radius compared to the wave length, i.e., a0 << λ, the acoustic impedance jzj << ρ0c0 in the near field. Moreover, since R << X, the impedance is strongly reactive and the surrounding fluid acts mainly as an inertial mass. The velocity is practically out of phase with the pressure. Thus, a source of small size is an inefficient radiator of acoustic energy. 3 The Simple Source The limit of the pulsating sphere solution as the sphere radius a0 becomes very small rep- resents the simple source or monopole solution. In this case, the source is characterized by the source mass flow rate 2 m(t) = 4πa0a_(t); and the exact solution is given by (15). If the source is located at the point j~yj, then m(t − r ) φ = − co ; (28) 4πρ0r where r = j~x − ~yj. Equation (28) states that, at the observation point ~x and time t, the sound signal received was emitted from the source point ~y at the retarded time τ = t − r . co The velocity and pressure are given by "m_ (t − r ) m(t − r )# @φ 1 c0 c0 vr = = + 2 (29) @r 4πρ0 rc0 r @φ m_ (t − r ) p0 = −ρ = c0 (30) 0 @t 4πr 3.1 Harmonic Sources: In this case m =me ¯ −i!t (31) m¯ m¯ φ = − e−i!τ = − ei(kr−!t); (32) 4πρ0r 4πρ0r 4 wherem ¯ represent the strength of the source and k = !=c0. Similarly, the the velocity expression is given by " # @φ 1 m_ (t) m(t) ikr vr = = + 2 e (33) @r 4πρ0 rc0 r or m¯ −i! 1 i(kr−!t) vr = + 2 e (34) 4πρ0 rc0 r Noting that ! = 2π , c0 λ m¯ −i2π 1 i(kr−t) vr = + 2 e (35) 4πρ0 rλ r i!m¯ p0 = − ei(kr−!t): (36) 4πr At large distance, r λ, the acoustic intensity is given by 2 2 0 m¯ ! 2 r I = p vr = 2 2 sin (!(t − )) (37) 16π ρoc0r c0 and 2 2 ¯ 1 0 m¯ ! I = Re(p v¯r) = 2 2 (38) 2 32π ρoc0r m¯ 2!2 P¯ = (39) 8ρoπc0 3.2 Simple source distribution: Suppose we have N sources located at ~yi with strength mi, then the principle of superposition states that: N N ri mi(t − ) X 1 X c0 φ = φi = − (40) i=1 4π i=1 ri where ri = j~x − ~yij. 5 4 The Dipole: Consider two sources of equal and opposite strength ±m located at ~y ± ~`=2. The potential of the two sources is 1 m(t − r+ ) m(t − r− ) φ = − [ c0 − c0 ]; (41) 4πρ0 r+ r− ~ ~ where r = j~x − ~yj and r± = j~x − (~y ± `=2)j. We further assume j`j r, then "m(t − r )# 1 @ c0 φ = `i + ::: (42) 4πρ0 @xi r or 2~ r 3 1 `m(t − c ) φ = r · 4 0 5 + :::: (43) 4πρ0 r The pressure is given by 2 ~ r 3 0 1 f(t − c ) p = − r · 4 0 5 + :::; (44) 4π r or r 1 @ fi(t − ) p0 = − ( c0 ); (45) 4π @xi r ~ ~ ~ where f = ffig =m _ `. Note that f has the dimension of a force.
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