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501_PartUniversity Physics Solution

501_PartUniversity Physics Solution - Sound and Hearing...

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Sound and Hearing 16-9 16.36. I DENTIFY : Destructive interference occurs when the path difference is a half integer number of wavelengths. S ET U P : ( ) ( ) 344 m s, so λ 344 m/s 172 Hz 2.00 m. v v f = = = = If 8.00 m A r = and B r are the distances of the person from each speaker, the condition for destructive interference is ( ) 1 2 λ , B A r r n = + where n is any integer. E XECUTE : Requiring ( ) 1 2 λ 0 B A r r n = + + > gives ( ) ( ) 1 2 8.00 m 2.00 m 4, A n r λ + > − = − = − so the smallest value of B r occurs when 4, n = − and the closest distance to B is ( ) ( ) 1 2 8.00 m 4 2.00 m 1.00 m. B r = + − + = E VALUATE : For 1.00 m, B r = the path difference is 7.00 m. A B r r = This is 3.5 . λ 16.37. I DENTIFY : Compare the path difference to the wavelength. S ET U P : ( ) ( ) 344 m s 860 Hz 0.400 m v f λ = = = E XECUTE : The path difference is 13.4 m 12.0 m 1.4 m. = path difference 3.5. λ = The path difference is a half- integer number of wavelengths, so the interference is destructive. E VALUATE : The interference is destructive at any point where the path difference is a half-integer number of wavelengths. 16.38. I DENTIFY : beat 1 2 . f f f = . v f λ = S ET U P : 344 m/s, v = Let 1 6.50 cm λ = and 2 6.52 cm. λ = 2 1 λ λ > so 1 2 . f f > E XECUTE : 2 2 1 1 2 2 2 1 2 1 2 1 1 ( ) (344 m/s)(0.02 10 m) 16 Hz. (6.50 10 m)(6.52 10 m) v f f v λ λ λ λ λ λ × = = = = × × There are 16 beats per second. E VALUATE : We could have calculated 1 f and 2 f and subtracted, but doing it this way we would have to be careful to retain enough figures in intermediate calculations to avoid round-off errors. 16.39. I DENTIFY : beat . a b f f f = For a stopped pipe, 1 . 4 v f L = S ET U P : 344 m/s. v = Let 1.14 m a L = and 1.16 m. b L = b a L L > so 1 1 . a b f f > E XECUTE : 2 1 1 1 1 ( ) (344 m/s)(2.00 10 m) 1.3 Hz. 4 4 4(1.14 m)(1.16 m) b a a b a b a b v v L L f f L L L L × = = = = There are 1.3 beats per second. E VALUATE : Increasing the length of the pipe increases the wavelength of the fundamental and decreases the frequency. 16.40. I DENTIFY : beat 0 . f f f = . 2 v f L = Changing the tension changes the wave speed and this alters the frequency. S ET U P : FL v m = so 1 , 2 F f mL = where 0 . F F F = + Δ Let 0 0 1 . 2 F f mL = We can assume that 0 / F F Δ is very small. Increasing the tension increases the frequency, so beat 0 . f f f = E XECUTE : (a) ( ) 1/ 2 0 beat 0 0 0 0 1 1 1 1 . 2 2 F F f f f F F F mL F mL Δ = = + Δ = + 1/ 2 0 0 1 1 2 F F F F Δ Δ + = + when 0 / F F Δ is small. This gives that beat 0 0 . 2 F f f F Δ = (b) beat 0 0 2 2(1.5 Hz) 0.68%. 440 Hz F f F f Δ = = = E VALUATE : The fractional change in frequency is one-half the fractional change in tension. 16.41. I DENTIFY : Apply the Doppler shift equation L L S S . v v f f v v + = + S ET U P : The positive direction is from listener to source. S 1200 Hz. f = L 1240 Hz. f = E XECUTE : L 0. v = S 25.0 m/s. v = − L S S v f f v v = + gives S L S L ( 25 m/s)(1240 Hz) 780 m/s.
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