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16_InstructorSolutions

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

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16-1 S OUND AND H EARING 16.1. I DENTIFY and S ET U P : Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. E XECUTE : (a) / (344 m/s)/1000 Hz 0.344 m v f λ = = = (b) max p BkA = and Bk is constant gives max1 1 max2 2 / / p A p A = 8 5 max2 2 1 2 max1 30 Pa 1.2 10 m 1.2 10 m 3.0 10 Pa p A A p = = × = × × (c) max 2 / p BkA BA π λ = = max 2 constant p BA λ π = = so max1 1 max2 2 p p λ λ = and 2 max1 2 1 3 max2 3.0 10 Pa (0.344 m) 6.9 m 1.5 10 Pa p p λ λ × = = = × / (344 m/s)/6.9 m 50 Hz f v λ = = = E VALUATE : 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. I DENTIFY : Apply max p BkA = and solve for A . S ET U P : 2 k π λ = and , v f λ = so 2 f k v π = and 2 . fBA p v π = E XECUTE : 2 12 max 9 (3.0 10 Pa) (1480 m s) 3.21 10 m. 2 2 (2.2 10 Pa) (1000 Hz) p v A π Bf π × = = = × × E VALUATE : Both v and B are larger, but B is larger by a much greater factor, so / v B is a lot smaller and therefore A is a lot smaller. 16.3. I DENTIFY : Use Eq.(16.5) to relate the pressure and displacement amplitudes. S ET U P : 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 λ from f , and then 2 / . k π λ = E XECUTE : (a) 150 Hz f = Need to calculate k : / v f λ = and 2 / k π λ = so 2 / (2 rad)(150 Hz)/344 m/s 2.74 rad/m. k f v π π = = = Then 3 max (1.42 10 Pa)(2.74 rad/m)(0.0200 10 m) 7.78 Pa. p BkA 5 = = × × = This is below the pain threshold of 30 Pa. (b) f is larger by a factor of 10 so 2 / k f v π = 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. E VALUATE : When f increases, λ decreases so k increases and the pressure amplitude increases. 16.4. I DENTIFY : Apply max . p BkA = 2 2 , f k v π π λ = = so max 2 . fBA p v π = S ET U P : 344 m/s v = E XECUTE : 3 max 5 6 (344 m/s)(10.0 Pa) 3.86 10 Hz 2 2 (1.42 10 Pa)(1.00 10 m) vp f BA π π = = = × × × E VALUATE : Audible frequencies range from about 20 Hz to about 20,000 Hz, so this frequency is audible. 16.5. I DENTIFY : . v f λ = Apply Eq.(16.7) for the waves in the liquid and Eq.(16.8) for the waves in the metal bar. S ET U P : 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 E XECUTE : (a) Using Eq.(16.7), 2 2 ( λ ) , B v ρ f ρ = = so [ ] 2 3 10 (8 m)(400 Hz) (1300 kg m ) 1.33 10 Pa. B = = × (b) Using Eq.(16.8), 2 2 2 4 3 10 ( ) (1.50 m) (3.90 10 s) (6400 kg m ) 9.47 10 Pa. Y v ρ L t ρ = = = × = × E VALUATE : 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. I DENTIFY : / . v d t = Apply Eq.(16.7) to calculate B . S ET U P : 3 3 3.3 10 kg/m ρ = × E XECUTE : (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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16_InstructorSolutions - SOUND AND HEARING 16 16.1 IDENTIFY...

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