Chapter 16 - 16.1: a) v f ( 344 m s) (100 Hz ) 0 . 344 m ....

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16.1: a) = = = A p p m f v then , 1000 if ) b . 344 . 0 ) Hz 100 ( s) m 344 ( λ 0 0 1000 A Therefore, the amplitude is m. 10 2 . 1 5 - × max max increasing , Since c) p BkA p = while keeping A constant requires decreasing k , and increasing π , by the same factor. Therefore the new wavelength is Hz. 50 m, 9 . 6 ) 20 ( m) 688 . 0 ( m 9 . 6 s m 344 new = = = f 16.2: m. 10 21 . 3 or , 12 Hz) 1000 ( Pa) 9 10 2 . 2 ( 2 ) s m 1480 ( Pa) 2 10 0 . 3 ( 2 max - × - × = × = = A A π πBf v p The much higher bulk modulus increases both the needed pressure amplitude and the speed, but the speed is proportional to the square root of the bulk modulus. The overall effect is that for such a large bulk modulus, large pressure amplitudes are needed to produce a given displacement. 16.3: From Eq. (16.5), . 2 λ 2 max v f πBA BA π BkA p = = = a) Pa. 7.78 s) m (344 Hz) 150 ( m) 10 00 . 2 ( Pa) 10 42 . 1 ( 2 5 5 = × × - π b) Pa. 778 Pa 7.78 100 c) Pa. 77.8 Pa 78 . 7 10 = × = × The amplitude at Hz 1500 exceeds the pain threshold, and at Hz 000 , 15 the sound would be unbearable. 16.4: The values from Example 16.8 are Hz, 1000 Pa, 10 16 . 3 4 = × = f B m. 10 2 . 1 8 - × = A Using Example 16.5, , s m 295 s m 344 293K K 216 = = v so the pressure amplitude of this wave is Pa). 10 16 . 3 ( 2 4 max × = = = A v πf B BkA p Pa. 10 8.1 m) 10 2 . 1 ( s m 295 Hz) (1000 2 3 8 - - × = × π This is 0.27 Pa) 10 (3.0 Pa) 10 1 . 8 ( 2 3 = × × - - times smaller than the pressure amplitude at sea level (Example 16-1), so pressure amplitude decreases with altitude for constant frequency and displacement amplitude. 16.5: a) Using Equation (16.7), [ ] × = = = 2 2 2 s) 400 m)( 8 ( so , ) λ ( B f ρ v B Pa. 10 33 . 1 ) m kg 1300 ( 10 3 × = b) Using Equation (16.8), [ ] 2 4 2 2 s 10 (3.9 m) 5 . 1 ( ) ( - × = = = ρ t L ρ v Y Pa. 10 47 . 9 ) m kg 6400 ( 10 3 × = ×
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16.6: a) The time for the wave to travel to Caracas was s 579 s 39 min 9 = and the speed was s m 10 085 . 1 4 × (keeping an extra figure). Similarly, the time for the wave to travel to Kevo was 680 s for a speed of s, m 10 278 . 1 4 × and the time to travel to Vienna was 767 s for a speed of s. m 10 258 . 1 4 × The average speed for these three measurements is s. m 10 21 . 1 4 × Due to variations in density, or reflections (a subject addressed in later chapters), not all waves travel in straight lines with constant speeds. b) From Eq. (16.7), , 2 ρ v B = and using the given value of 3 3 m kg 10 3 . 3 × = and the speeds found in part (a), the values for the bulk modulus are, respectively, Pa. 10 2 . 5 and Pa 10 5.4 Pa, 10 9 . 3 11 11 11 × × × These are larger, by a factor of 2 or 3, than the largest values in Table (11-1). 16.7:
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 16 - 16.1: a) v f ( 344 m s) (100 Hz ) 0 . 344 m ....

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