Chapter 14 - CHAPTER 14 Quick Quizzes 1. (c). Temperature....

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C HAPTER 14 Quick Quizzes 1. (c). Temperature. Although the speed of a wave is equal to the product of its wavelength and frequency, neither one determines the speed of the wave. For example, if the sound from a musical instrument increases in frequency, the wavelength decreases so that v f λ = remains constant. The amplitude of a sound wave determines the size of the oscillations of air molecules, but does not affect the speed of the wave through the air. 2. (a) 10 dB. If we call the intensity of each violin I, the total intensity when all the violins are playing is . Therefore the addition of the nine violins increases the intensity of the sound over that of one violin by a factor of 10. From Equation 14.7 we see that an increase in intensity by a factor of 10 increases the sound level by 10 dB. 91 0 II += I (b) 13 dB. The intensity is now increased by a factor of 20 over that of a single violin. 3. (c). The distance between you and the buzzer is increasing. Therefore, the intensity at your location is decreasing. As the buzzer falls, it moves away from you with increasing speed. This causes the detected frequency to decrease. 4. (b). The speed of sound increases in the warmer air, while the speed of the sound source (the plane) remains constant. Therefore, the ratio of the speed of the source to that of sound (i.e., the Mach number) decreases. 439
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CHAPTER 14 Problem Solutions 14.1 Since , we ignore the time required for the lightning flash to reach the observer in comparison to the transit time for the sound. Then, light sound vv >> ( ) () 3 343 m s 16.2 s 5.56 10 m d ≈= × = 5.56 km 14.2 The speed of sound in seawater at 25°C is 1530 m s . Therefore, the time for the sound to reach the sea floor and return is 2 150 m 2 1530 m s d t v == = 0.196 s 14.3 The speed of the sound wave is ( )( ) 2 0.50 m 700 Hz 3.5 10 m s vf λ . Thus, from ( ) 331 m s 273 K vT = , the temperature of the air is 22 2 3.5 273 K 273 K 305 K 331 m s 331 m s v T ⎛⎞ × = 32 C ° = ⎜⎟ ⎝⎠ 14.4 At , the speed of sound in air is 27 C 300 K T =°= 300 K 331 m s 331 m s 347 m s 273 K 273 K T v === The wavelength of the 20 Hz sound is 347 m s 17 m 20 Hz v f = , and that of the 20 is 000 Hz 2 347 m s 1.7 10 m=1.7 cm 20 000 Hz × . Thus, range of wavelengths of audible sounds at 27°C is 1.7 cm to 17 m . 14.5 10 3 3 2.80 10 N m 1.43 13 600 kg m B v ρ × = 1.43 km s 440
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CHAPTER 14 14.6 The speed of sound at is 10.0 C 283 K T = () 283 K 331 m s 331 m s 337 m s 273 K 273 K a T v === When the plane is at point P , the sound reaching the observer at O originated at point A and has traveled distance == = °° 500 m 539 m cos 22.0 cos 22.0 OP OA The transit time for this sound was 539 m 1.60 s 337 m s s t v == = . In this time interval, the plane traveled distance ( ) ( ) ( ) = ° = tan 22.0 500 m tan 22.0 202 m AP , so the speed of the plane is 202 m 1.60 s p v t = 126 m s O 500 m 22.0° 14.7 From Table 14.1, the speed of sound in the saltwater is 1530 m s w v = . At , the speed of the sound in air is 20 C 293 K T =°= 293 K 331 m s 331 m s 343 m s 273 K 273 K a T v If d is the width of the inlet, the transit time for the sound in the water is w w d t v = , and that for the sound in the air is 4.50 s= aw a d tt v =+ .
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This note was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.

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Chapter 14 - CHAPTER 14 Quick Quizzes 1. (c). Temperature....

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