16_InstructorSolutions

# 16_InstructorSolutions - SOUND AND HEARING 16 16.1....

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16-1 S OUND AND H EARING 16.1. IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) / (344 m/s)/1000 Hz 0.344 m vf λ == = (b) max p BkA = and Bk is constant gives max1 1 max2 2 // p Ap A = 85 max2 21 2 max1 30 Pa 1.2 10 m 3.0 10 Pa p AA p −− ⎛⎞ × = × ⎜⎟ × ⎝⎠ (c) max 2/ p BkA BA π max 2 constant pB A λπ so max1 1 max2 2 pp = and 2 max1 3 max2 (0.344 m) 6.9 m 1.5 10 Pa p p λλ × = × / (344 m/s)/6.9 m 50 Hz fv = EVALUATE: The pressure amplitude and displacement amplitude are directly proportional. For the same displacement amplitude, the pressure amplitude decreases when the frequency decreases and the wavelength increases. 16.2. IDENTIFY: Apply max p BkA = and solve for A . SET UP: 2 k = and , = so 2 f k v = and 2 . fBA p v = EXECUTE: 2 12 max 9 (3.0 10 Pa) (1480 m s) 3.21 10 m. 2 2 (2.2 10 Pa) (1000 Hz) pv A π Bf π × = × × EVALUATE: Both v and B are larger, but B is larger by a much greater factor, so / vB is a lot smaller and therefore A is a lot smaller. 16.3. IDENTIFY: Use Eq.(16.5) to relate the pressure and displacement amplitudes. SET UP: As stated in Example 16.1 the adiabatic bulk modulus for air is 5 1.42 10 Pa. B Use Eq.(15.1) to calculate from f , and then 2 / . k πλ = EXECUTE: (a) 150 Hz f = Need to calculate k : / = and 2 / k = so 2 / (2 rad)(150 Hz)/344 m/s 2.74 rad/m. kf v = Then 3 max (1.42 10 Pa)(2.74 rad/m)(0.0200 10 m) 7.78 Pa. k A 5− × × = This is below the pain threshold of 30 Pa. (b) f is larger by a factor of 10 so 2 / v = is larger by a factor of 10, and max p BkA = is larger by a factor of 10. max 77.8 Pa, p = above the pain threshold. (c) There is again an increase in f , k , and max p of a factor of 10, so max 778 Pa, p = far above the pain threshold. EVALUATE: When f increases, decreases so k increases and the pressure amplitude increases. 16.4. IDENTIFY: Apply max . p BkA = 22 , f k v so max 2 . p v = SET UP: 344 m/s v = EXECUTE: 3 max 56 (344 m/s)(10.0 Pa) 3.86 10 Hz 2 2 (1.42 10 Pa)(1.00 10 m) vp f BA ππ = × ×× EVALUATE: Audible frequencies range from about 20 Hz to about 20,000 Hz, so this frequency is audible. 16.5. IDENTIFY: . = Apply Eq.(16.7) for the waves in the liquid and Eq.(16.8) for the waves in the metal bar. SET UP: In part (b) the wave speed is 4 1.50 m 3.90 10 s d v t × 16

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16-2 Chapter 16 EXECUTE: (a) Using Eq.(16.7), 22 ( λ ), Bv ρ f ρ == so [ ] 2 31 0 (8 m)(400 Hz) (1300 kg m ) 1.33 10 Pa. B × (b) Using Eq.(16.8), 2 4 3 1 0 ( ) (1.50 m) (3.90 10 s) (6400 kg m ) 9.47 10 Pa. Yv ρ Lt ρ ⎡⎤ = × = × ⎣⎦ EVALUATE: In the liquid, 3200 m/s v = and in the metal, 3850 m/s. v = Both these speeds are much greater than the speed of sound in air. 16.6. IDENTIFY: /. vd t = Apply Eq.(16.7) to calculate B . SET UP: 33 3.3 10 kg/m EXECUTE: (a) The time for the wave to travel to Caracas was 9 min 39 s 579 s = and the speed was 4 1.08 10 m/s. × Similarly, the time for the wave to travel to Kevo was 680 s for a speed of 4 1.28 10 m/s, × and the time to travel to Vienna was 767 s for a speed of 4 1.26 10 m/s. × The average for these three measurements is 4 1.21 10 m/s.
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## This note was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at University of Arizona- Tucson.

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16_InstructorSolutions - SOUND AND HEARING 16 16.1....

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