MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fal2005. (Massachusetts Institute of Technology: MIT OpenCourseWare)http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit:http://ocw.mit.edu/terms l .� � � � 6.013 - Electromagnetics and Applications Fall 2005 Lecture 23 - Acoustic Resonators Prof. Markus Zahn December 13, 2005 I. Governing Acoustic Equations ∂v¯ρ0 ∂t = −�p ∂p 2 ∂t = −ρ0cs� · v¯1 ∂2p2 2⇒ � p = c ∂t2 s II. Acoustic Resonator - Closed Box A. Boundary Conditions vx(x = 0) = vx(x = a) = 0 vy(y = 0) = vy(y = b) = 0 vz(z = 0) = vz(z = d) = 0 B. Solutions to Acoustic Equations 1 ∂2p2 2� p = c ∂t2 s p(x, y, z, t) = Re ˆj(ωt−k¯·r¯)p e= Re ˆj(ωt−kxx−ky y−kz z)p e∂2p ∂2p ∂2p 1 ∂2p 2�2 p = ∂x2 + ∂y2 + ∂z2 = cs ∂t2 1 Image by MIT OpenCourseWare.� � ˆp � kx 2 + ky 2 + kz 2� pω2 ⇒ −ˆ = − c2 s ω2 k2 + k2 + k2 = x y z 2cs ∂v¯ρ0 ∂t = −�ρ ∂vx ∂p ρ0 = −∂t ∂x ∂vy ∂p ρ0 ∂t = −∂y ∂vz ∂p ρ0 = −∂t ∂z vx ∝ sin(kxx), vx(x = 0) = 0 ⇒ p ∝ cos(kxx) vy ∝ sin(kyy), vy(y = 0) = 0 ⇒ p ∝ cos(kyy) vz ∝ sin(kzz), vz(z = 0) = 0 ⇒ p ∝ cos(kzz) p(x, y, z, t) = Re pˆcos(kxx) cos(kyy) cos(kzz)ejωtρ0jωvˆx = ˆ sin(kxx) cos(kyy) cos(kzz)pkx ˆpkx ⇒ vˆx = ρ0jω sin(kxx) cos(kyy) cos(kzz) ˆvˆy = pky cos(kxx) sin(kyy) cos(kzz)ρ0jω ˆvˆz = pkz cos(kxx) cos(kyy) sin(kzz)ρ0jω mπ vˆx(x = a) = 0 kxa = mπ kx = ⇒ ⇒ a nπ vˆy(y = b) = 0 kyb = nπ ky = ⇒ ⇒ b vˆz(z = d) = 0 kzd = pπ kz = pπ ⇒ ⇒ d �� m �2 � n �2 � p �2 �1/2 ω π + + = a b d cs cs �� m �2 � n �2 � p �2 �1/2 fmnp = + + 2 a b d cs ≈ 330 m/s, a = .203 m (8 in), b = .203 m (8 in), d = .305m (12 in) (The value for sp e ed of sound and the determined frequencies are for waves in air.) 2� � � � m n p fmnp(Hz) fmeasured(Hz) 0 0 1 541 550 0 1 0 813 0 0 2 1082 1 1 0 1150 1150 1 1 1 1270 1 1 2 1579 0 0 3 1623 0 2 0 1626 0 1 3 1815 1 2 2 2115 C. Comparison of Theory to Experiment Air - T = 0◦ C, p0 = 1 atm = 1.01 105 Nt/m2 (1 Nt/m2 = 1 Pascal) · ρ0 = 1.293 g/L = 1.293 kg/m3 cs = γp0 = 331 m/s ρ0 Helium -T = 0◦ C, p0 = 1 atm = 1.01 105 Pascals, γ = 1.67· ρ0 = 0.178 g/L = 0.178 kg/m3 cs = γp0 = 973 m/s ρ0 Helium/air mixture -f001 = cs = 950 Hz was measured 2d cs = 2d(950) = 2(.305)950 = 580 m/s cs,air < cs,air/He mixture < cs,He 331 < 580 < 973m/s III. Acoustic Resonator - z = 0 side open, other 5 s ides c losed A. Boundary Conditions vx(x = 0) = vx(x = a) = 0 vy(y = 0) = vy(y = b) = 0 vz(z = d) = 0, p(z = 0) = 0 B. Solutions p(x, y, z, t) = Re pˆcos(kxx) cos(kyy) sin(kzz)ejωtˆvˆx = pkx sin(kxx) cos(kyy) sin(kzz)ρ0jω ˆvˆy = pky cos(kxx) sin(kyy) sin(kzz)ρ0jω pkz vˆz = −ρˆ0jω cos(kxx) cos(kyy) cos(kzz)ejωt 3vˆx(x = 0) = 0 mπ vˆx(x = a) = 0 kxa = mπ kx = ⇒ ⇒ a vˆy(y = 0) = 0 nπ vˆy(y = b) = 0 kyb = nπ ky = ⇒ ⇒ b π (2p + 1)π vˆz(z = d) = 0 kzd = (2p + 1) kz = ⇒ 2 ⇒ 2d pˆ(z = 0) = 0 ω2 �� m �2 � n �2 �2p + 1 �2 � = π2 + + c2 a b 2ds cs �� m �2 � n �2 �2p + 1 �2 �1/2 fmnp = + +2 a b 2d (2p + 1)cs �� f000 = � , cs = 330m/s (air), d = .305 m f000 = 270.5 Hz 4d p=0 ⇒ �� �2 � �2 �1/2 cs 1 1 f100 = + = 856.6 Hz 2 a 2d
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