This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 16 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 1 = v / f 1 = (343 m/s)/(20 Hz) = 17 m; 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 = = 17 m . ( b ) We find the wavelength from = 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 / ) 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 vD M f . ( a ) For the frequency of 100 Hz, we have 3.0 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 3.0 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 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 ....
View
Full
Document
This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.
 Spring '08
 ROSS

Click to edit the document details