17 - CHAPTER E xercises 1. 6. 17 = v /f = 3.4 mm v = (B/ )...

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Exercises 1. λ = v/f = 3.4 mm 6. v = (B/ ρ ) 1/2 : (a) 314 m/s; (b) 972 m/s 11. ∂s/∂t = – ω s o cos(kx – ω t). Using v = ω /k and P o = Bks o’ , we see that (B/v) ∂s/∂t = P (See Eq. 17.14) 12. P = 2 sin(5.3x + 1800t) 18. f 2 /f 1 = (T 2 /T 1 ) 1/2 , f 2 =983 Hz 22. (a) (340/310)(1200) = 1320 Hz; (b) (340/370)(1200) = 1100 Hz 26. v = 18.1 m/s. f’ = 1800(340)/(340 + 18.1) = 1710 Hz, 1900 Hz 27. (a) f 1 ’ = 400(325/300) = 433.3 Hz, f 2 ’ = 400(355/380) = 373.7 Hz. Thus Δf’ = 59.6 Hz; (b) f 1 ’ = 400(355/300) = 473.3 Hz, f 2 ’ = 400(325/380) = 342.1 Hz. Thus Δf’ = 131 Hz 31. 640 = 600(340)/(340 – v s ), thus v s = 21.25 m/s. f’ = 600(340)/(361.25) = 565 Hz 32. 50 = 10 log(I 1 /I o ), thus I 1 = 10 5 I o = 10 –7 W/m 2 β = 10 log (2 × 10 4 I 1 /I o ) = 10 log (2 ×  10 9 ) = 93 dB 38. 80 = 10 log(I 2 /I 1 ), thus I 2 = 10 8 I 1 42. (a) s o /d = v p /v s’ is the particle velocity and v s is the speed of sound. So v p = (10 –6 )(340)/(5 ×  10 –2 ) = 6.8 mm/s (b) s o /2d = v p /v s’ , thus v p = –3.4 mm/s. Problems 1. (a) I = P/4πr 2 = 4.97 × 10 –6 W/m 2 ; 3. (a) dv/dT = 1/2 (20)T
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This note was uploaded on 05/01/2011 for the course ELECTRIC MAT351 taught by Professor Ahmettantuni during the Spring '10 term at Abant İzzet Baysal University.

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17 - CHAPTER E xercises 1. 6. 17 = v /f = 3.4 mm v = (B/ )...

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