501_PartUniversity Physics Solution

# 501_PartUniversity Physics Solution - Sound and Hearing...

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Sound and Hearing 16-9 16.36. IDENTIFY: Destructive interference occurs when the path difference is a half integer number of wavelengths. SET UP: ( ) ( ) 344 m s, so λ 344 m/s 172 Hz 2.00 m. vv 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 , BA rr n −=+ where n is any integer. EXECUTE: Requiring ( ) 1 2 λ 0 =++ > gives ( ) ( ) 1 2 8.00 m 2.00 m 4, A nr 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 =+ + = EVALUATE: For 1.00 m, B r = the path difference is 7.00 m. AB rr −= This is 3.5 . 16.37. IDENTIFY: Compare the path difference to the wavelength. SET UP: ( ) 344 m s 860 Hz 0.400 m vf = EXECUTE: The path difference is 13.4 m 12.0 m 1.4 m. = path difference 3.5. = The path difference is a half- integer number of wavelengths, so the interference is destructive. EVALUATE: The interference is destructive at any point where the path difference is a half-integer number of wavelengths. 16.38. IDENTIFY: beat 1 2 . f ff . = SET UP: 344 m/s, v = Let 1 6.50 cm = and 2 6.52 cm. = 21 > so 12 . f f > EXECUTE: 2 22 1 2 11 ( ) ( 3 4 4 m / s ) ( 0 . 0 2 1 0 m ) 16 Hz. (6.50 10 m)(6.52 10 m) v ffv λλ −− ⎛⎞ −× − = = = ⎜⎟ ×× ⎝⎠ There are 16 beats per second. EVALUATE: 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. IDENTIFY: beat . ab f For a stopped pipe, 1 . 4 v f L = SET UP: 344 m/s. v = Let 1.14 m a L = and 1.16 m. b L = ba L L > so 11 . f f > EXECUTE: 2 ) ( 3 4 4 m / s ) ( 2 . 0 0 1 0 m ) 1.3 Hz. 4 4 4(1.14 m)(1.16 m) a b L L LL L L = = There are 1.3 beats per second. EVALUATE: Increasing the length of the pipe increases the wavelength of the fundamental and decreases the frequency. 16.40. IDENTIFY: beat 0 . f . 2 v f L = Changing the tension changes the wave speed and this alters the frequency. SET UP: FL v m = so 1 , 2 F f mL = where 0 . FF F = Let 0 0 1 . 2 F f mL = We can assume that 0 / FF Δ is very small. Increasing the tension increases the frequency, so beat 0 . f EXECUTE: (a) 1/2 0 beat 0 0 0 0 . 2 2 f F F F mL F mL ⎡⎤ Δ =− = + Δ− = + ⎢⎥ ⎣⎦ 00 2 ΔΔ += + when 0 / Δ is small. This gives that beat 0 0 . 2 F F Δ = (b) beat 2 2(1.5 Hz) 0.68%. 440 Hz Ff Δ = EVALUATE: The fractional change in frequency is one-half the fractional change in tension. 16.41. IDENTIFY: Apply the Doppler shift equation L LS S . f f + = + SET UP: The positive direction is from listener to source. S 1200 Hz. f = L 1240 Hz. f = EXECUTE: L 0. v = S 25.0 m/s. v S v f f = + gives SL (2 5 m / s ) ( 1 2 4 0 H z ) 780 m/s. 1200 Hz 1240 Hz v = EVALUATE: f f > since the source is approaching the listener.

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16-10 Chapter 16 16.42. IDENTIFY: Follow the steps of Example 16.19. SET UP: In the first step, S 20.0 m/s v =+ instead of 30.0 m/s. In the second step, L 20.0 m/s v =− instead of 30.0 m/s. + EXECUTE: WS S 340 m/s (300 Hz) 283 Hz. 340 m/s 20.0 m/s v ff vv ⎛⎞ == = ⎜⎟ ++ ⎝⎠ Then L LW 340 m/s 20.0 m/s (283 Hz) 266 Hz. 340 m/s v +− = EVALUATE: When the car is moving toward the reflecting surface, the received frequency back at the source is higher than the emitted frequency. When the car is moving away from the reflecting surface, as is the case here, the received frequency back at the source is lower than the emitted frequency.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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501_PartUniversity Physics Solution - Sound and Hearing...

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