HW9_solution473

# HW9_solution473 - ECE 473/TAM 413 Homework Assignment#9...

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Homework Assignment #9 Solutions - 1 ECE 473/TAM 413 Homework Assignment #9 Solutions For your information Hour Exam 2: Wednesday, April 23, 2008, 7:00-8:30 pm Hour exams are open book (the text for the course) with one sheet of notes (you can use both sides). Your class notes and lecture notes are not permitted. You can use calculators. 1. A baffled circular piston source with a radius of 10 centimeters operates in air (20˚C). a) For an operating frequency of 10 kHz, what is the distance to the far field? b) For an operating frequency of 1 kHz, what is the distance to the far field? c) For an operating frequency of 4 kHz, what is the zero-to-zero beamwidth? d) For an operating frequency of 8 kHz, what is the – 6 dB beamwidth? e) For an operating frequency of 10 kHz, what is the radiation impedance of the air acting on the piston? The answer should be nonzero values for both the real and imaginary parts. Express answer in rectangular form. f) For the piston source in part (e), what is the radiated acoustic power? The peak amplitude surface displacement is 100 micrometers. g) For the piston source in part (e), what is the mass loading? For air at 20˚C, ! o = 1.21kg / m 3 c = 343 m / s ! o c = 415 Pa " s/ m a) ! = c f = 343m / s 10 kHz = 3.43 cm , a ! = 10 cm 3.43cm = 2.9 From Eq. 7.4.9, r 1 = a 2 ! " ! 4 = 10 cm ( ) 2 3.43 cm " 3.43 cm 4 = 29.16 cm " 0.86 cm = 28.30cm b) ! = c f = 343m / s 1kHz = 34.3 cm , a ! = 10 cm 34.3cm = 0.29 r 1 = a 2 ! " ! 4 = 10 cm ( ) 2 34.3 cm " 34.3 cm 4 = 2.92 cm " 8.58 cm < 0 , no near field, far field starts at the transducer surface. c) H ! ( ) = 2J 1 kaSin ! ( ) kaSin ! , ka = 2 ! fa c = 2 ! 4 kHz ( ) 0.1m ( ) 343 m / s = 7.327 First zero in H ! ( ) occurs when kaSin ! = 3.83 " ! = Sin # 1 3.83 7.327 \$ % ( ) = 31.5 o Thus, the zero-to-zero beamwidth = 2 31.5˚ ( ) = 63.0 o . d) ka = 2 ! fa c = 2 ! 8 kHz ( ) 0.1m ( ) 343m / s = 14.655 20 ! log 10 H " ( ) = 20 ! log 10 2J 1 kaSin " ( ) kaSin " = # 6 dB 2J 1 kaSin ! ( ) kaSin ! = 10 " 6 20 = 0.501 when kaSin ! " 2.2 , ! = Sin " 1 2.2 14.655 # \$ % ( = 8.63 o Thus, the -6 dB beamwidth = 2 8.63 o ( ) = 17.26 o .

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Homework Assignment #9 Solutions - 2 e) ka = 2 ! fa c = 2 ! 10 kHz ( ) 0.1m ( ) 343 m / s = 18.32 From Eq. 7.5.11, Z r = ! a 2 " o c R 1 2ka ( ) + jX 1 2ka ( ) { } = ! 0.1m ( ) 2 415 Pa " s/ m ( ) R 1 36.64 ( ) + jX 1 36.64 ( ) { } For 2ka >> 1, R 1 2ka ( ) = 1 ! 2J 1 2ka ( ) 2ka " 1 ! 8 # Cos 2ka ! 3 # 4 \$ % ( ) 2ka ( ) 3 " 1 X 1 2ka ( ) = 2H 1 2ka ( ) 2ka ! 4 " 2ka ( ) # 8 " Sin 2ka # 3 " 4 \$ % ( ) 2ka ( ) 3 ! 4 " 2ka ( ) = 4 " 36.64 ( ) = 0.0348 Z r = ! 0.1m ( ) 2 415Pa " s/ m ( ) 1 + j0.0348 { } = 13.04 + j0.454 Pa " s " m U o = !" o = 2 # \$ 10 4 / s ( ) 1 \$ 10 % 4 m ( ) = 6.283m / s f) From Eq. 7.5.6, ! = 1 2 U o 2 R r = 1 2 6.283m / s ( ) 2 13.04 Pa " s " m ( ) = 257 W g) m r = X r !
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HW9_solution473 - ECE 473/TAM 413 Homework Assignment#9...

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