# Ch16Word - Chapter 16 Page 1 CHAPTER 16 – Sound 1 Because...

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Unformatted text preview: Chapter 16 Page 1 CHAPTER 16 – Sound 1. Because the sound travels both ways across the lake, we have L = ! vt = ! (343 m/s)(1.5 s) = 2.6 × 10 2 m . 2. ( a ) We find the extreme wavelengths from λ 1 = v / f 1 = (343 m/s)/(20 Hz) = 17 m; λ 2 = v / f 2 = (343 m/s)/(20,000 Hz) = 1.7 × 10 –2 m = 1.7 cm. The range of wavelengths is 1.7 cm = λ = 17 m . ( b ) We find the wavelength from λ = v / f = (343 m/s)/(10 × 10 6 Hz) = 3.4 × 10 –5 m . 3. The speed in the concrete is determined by the elastic modulus: v concrete = ( E / ρ ) 1/2 = [(20 × 10 9 N/m 2 )/(2.3 × 10 3 kg/m 3 )] 1/2 = 2.95 × 10 3 m/s. For the time interval we have ? t = ( d / v air ) – ( d / v concrete ); 1.4 s = d {[1/(343 m/s)] – [1/(2.95 × 10 3 m/s)]}, which gives d = 5.4 × 10 2 m . 4. ( a ) For the time interval in sea water we have ? t = d / v water = (1.0 × 10 3 m)/(1560 m/s) = 0.64 s . ( b ) For the time interval in the air we have ? t = d / v air = (1.0 × 10 3 m)/(343 m/s) = 2.9 s . 5. If we let L 1 represent the thickness of the top layer, the total transit time is t = ( L 1 / v 1 ) + [( L – L 1 )/ v 2 )]; 4.5 s = [ L 1 /(331 m/s)] + [(1500 m – L 1 )/(343 m/s)], which gives L 1 = 1200 m . Thus the bottom layer is 1500 m – 1200 m = 300 m . 6. Because the distance is d = vt , the change in distance from the change in velocity is ? d = t ? v ; so the percentage change is (? d / d )100 = (? v / v )100 = (343 m/s – 331 m/s)(100)/(343 m/s) = 3.5% . 7. We find the displacement amplitude from ? P M = 2p ρ vD M f . ( a ) For the frequency of 100 Hz, we have 3 . × 10 –3 Pa = 2p(1.29 kg/m 3 )(331 m/s) D M (100 Hz), which gives D M = 1.1 × 10 –8 m . ( b ) For the frequency of 10 kHz, we have 3 . × 10 –3 Pa = 2p(1.29 kg/m 3 )(331 m/s) D M (10 × 10 3 Hz), which gives D M = 1.1 × 10 –10 m . 8. We find the pressure amplitude from ? P M = 2p ρ vD M f . ( a ) For the frequency of 50 Hz, we have ? P M = 2p(1.29 kg/m 3 )(331 m/s)(3 × 10 –10 m)(50 Hz) = 4 × 10 –5 Pa . ( b ) For the frequency of 5.0 kHz, we have ? P M = 2p(1.29 kg/m 3 )(331 m/s)(3 × 10 –10 m)(5.0 × 10 3 Hz) = 4 × 10 –3 Pa . Chapter 16 Page 2 9. The pressure variation is ? P = ? P M sin ( kx – ω t ). ( a ) For the frequency of 50 Hz, we have ω = 2p f = 2p(50 Hz) = 315 s –1 ; k = ω / v = (315 s –1 )/(331 m/s) = 0.949 m –1 . The pressure variation is ? P = (4 × 10 –5 Pa) sin [(0.949 m –1 ) x – (315 s –1 ) t ] . ( b ) For the frequency of 5.0 kHz, we have ω = 2p f = 2p(5.0 × 10 3 Hz) = 3.15 × 10 4 s –1 ; k = ω / v = (3.15 × 10 4 s –1 )/(331 m/s) = 94.9 m –1 . The pressure variation is ? P = (4 × 10 –3 Pa) sin [(94.9 m –1 ) x – (3.15 × 10 4 s –1 ) t ] ....
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Ch16Word - Chapter 16 Page 1 CHAPTER 16 – Sound 1 Because...

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