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View Full DocumentCHAPTER 17 SOUND AND OTHER WAVE PHENOMENA ActivPhysics can help with these problems: Activities 10.3, 10.4, 10.5, 10.6, 10.8, 10.9 Sections 17-1 and 17-2: Sound Waves and the Speed of Sound in Gases Problem 1. Show that the quantity p P/ has the units of speed. Solution The units of pressure (force per unit area) divided by density (mass per unit volume) are (N / m 2 ) / (kg / m 3 ) = (N / kg)(m 3 / m 2 ) = (m / s 2 )m = (m / s) 2 , or those of speed squared. Problem 2. Dimensional analysis alone suggests that the sound speed in a gas should be given roughly by p P/. By how much would an estimate based on this simple analysis be in error for a gas with = 7 5 ? Solution The fractional difference between p P/ and p P/ is v/v = ( p P/- p P/ ) / p P/ = p 1 /- 1 . For = 7 / 5 , this is p 5 / 7- 1 =- . 155 - 15% (i.e., the rough value is 15% smaller than Equation 17-1). Problem 3. Find the wavelength, period, angular frequency, and wave number of a 1.0-kHz sound wave in air under the conditions of Example 17-1. Solution The value of the speed of sound in air from Example 17-1 was 343 m/s. Therefore (a) = v/f = (343 m / s) / (1 kHz) = 34 . 3 cm; (b) T = 1 /f = 1 ms; (c) = 2 f = 6 . 28 10 3 s- 1 ; and (d) k = /v = 2 / = 18 . 3 m- 1 . (See Equations 16-1, 3, 4, and 6.) Problem 4. (a) Determine an approximate value for the speed of sound in miles per second. (b) Suppose you see a lightning flash and, 10 s later, hear the thunder. How many miles away did the flash occur? Neglect the travel time for the light (why?) Solution (a) From Example 17-1, v = (343 m / s)(1 mi 1609m) = 0 . 213 mi / s . (b) If it took the sound of the thunder roughly 10 s to reach you, its source was about vt = (0 . 213 mi / s)(10 s) = 2 . 13 mi away. (The speed of light in air, approximately 186,000 mi/s, is so much bigger than the speed of sound that the light travel time is negligible compared to sound travel times.) Problem 5. Timers in sprint races start their watches when they see smoke from the starting gun, not when they hear the sound (Fig. 17-25). Why? How much error would be introduced by timing a 100-m race from the sound of the shot? Solution The sound of the starting gun takes (100 m) (340 m / s) = 0 . 294 s to reach the finish line. An error of this magnitude is significant in short races, where world records are measured in hundredths of a second. (This problem is almost the same as Problem 2, Chapter 2.) Problem 6. The factor for nitrogen dioxide (NO 2 ) is 1.29. Find the sound speed in NO 2 at a pressure of 4 . 8 10 4 N / m 2 and density 0 . 35 kg / m 3 . Solution The given data, substituted into Equation 17-1, gives: v = q 1 . 29(4 . 8 10 4 N / m 2 ) / (0 . 35 kg / m 3 ) = 421 m / s .... View Full Document
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