501 Chapter 14 Sound Answers to Even Numbered Conceptual Questions 2. The resonant frequency depends on the length of the pipe. Thus, changing the length of the pipe will cause different frequencies to be emphasized in the resulting sound. 8. The speed of light is so high that the arrival of the flash is practically simultaneous with the lightning discharge. Thus, the delay between the flash and the arrival of the sound of thunder is the time sound takes to travel the distance separating the lightning from you. By counting the seconds between the flash and thunder and knowing the approximate speed of sound in air, you have a rough measure of the distance to the lightning bolt. 12. No. Adding two sounds of equal loudness will produce an intensity double that associated with either individual sound. However, the decibel scale is a logarithmic function of intensity, so doubling the intensity only increases the decibel level by 10log 2. Thus, the decibel level with both sounds present will be 53 dB. 14. A beam of radio waves of known frequency is sent toward a speeding car, which reflects the beam back to a detector in the police car. The amount the returning frequency has been shifted depends on the velocity of the oncoming car. 20. A vibrating string is not able to set very much air into motion when vibrated alone. Thus it will not be very loud. If it is placed on the instrument, however, the string's vibration sets the sounding board of the guitar into vibration. A vibrating piece of wood is able to move a lot of air, and the note is louder. Problem Solutions 14.1 Since light s o undvv>>, we ignore the time required for the lightning flash to reach the observer in comparison to the transit time for the sound. Then, ()()3343 m s 16.2 s5.56 10 m5.56 kmd≈=×= 14.9 The decibel level ()010log IIβ=, where 12 201.00 10 W mI−=× .502 CHAPTER 14 (a) If 100β= , then ()0log 10II= giving 10 2 2010 1.00 10 W mII−==× (b) If all three toadfish sound at the same time, the total intensity of the sound produced is 2233.00 10 W mII−== ×′, and the decibel level is ()()()2212 2103.00 10 W m10log1.00 10 W m10log 3.00 10 10 log 3.00 10 105β−−⎛⎞×=′⎜⎟×⎝⎠⎡⎤==+=⎡⎤⎣⎦⎣⎦ 14.10 The sound power incident on the eardrum is IA=à where I is the intensity of the sound and 525.010 mA−=× is the area of the eardrum. (a) At the threshold of hearing, 12 21.010 WmI−=× , and ()()12 2 5 2 171.010 Wm 5.010 m5.010 W−−−=× × =×à (b) At the threshold of pain, 21.0 W mI= , and ()()252 51.0 W m 5.010 m5.010 W−−=×=×à 14.17 (a) The intensity of sound at 10 km from the horn (where β = 50 dB) is ()1012 2 5.072010 1.010 Wm 10 1.010 WmIIβ−−==× =× Thus, from 24Irπ= à , the power emitted by the source is ()()223722441010 m1.010 Wm 1.310 WrIππ−== × × =×à (b) At 50 mr= , the intensity of the sound will be ()232221.310 W4.010 Wm4450 mIrππ−×== =×à and the sound level is ()32912 204.010 Wm10log 10log 10log 4.010 96 dB1.010 WmIIβ−−⎛⎞⎛⎞×== =×=⎜⎟⎜⎟×⎝⎠⎝⎠Sound 503 14.21 When a stationary observer ()0Ov= hears a moving source, the observed frequency is OOS SSSvvvff fvv vv⎛⎞⎛⎞+==⎜⎟⎜⎟−−⎝⎠⎝⎠. (a) When the train is approaching, 40.0 m sSv=+ and ()()345 m s320 Hz 362 Hz345 m s 40.0 m sOa ppr oa c hf⎛⎞==⎜⎟−⎝⎠ After the train passes and is receding, 40.0 m sSv=− and ()()()345 m s320 H z 287 Hz345 m s 40.0 m sOrecedef⎡⎤==⎢⎥−−⎢⎥⎣⎦. Thus, the frequency shift that occurs as the train passes is ()()75.2 HzOO Orecede approachff f∆= − =− , or it is a 75.2 Hz drop (b) As the train approaches, the observed wavelength is ()345 m s0.953 m362 HzOapproachvfλ=== 14.23 Both source and observer are in motion, so OOSSvvffvv⎛⎞+=⎜⎟−⎝⎠. Since each train moves toward the other, 0Ov > and 0Sv > . The speed of the source (train 2) is km 1000 m1 h90.0 25.0 m sh1 km 3600 sSv⎛⎞⎛⎞==⎜⎟⎜⎟⎝⎠⎝⎠ and that of the observer (train 1) is 130 km h 36.1 m sOv== . Thus, the observed frequency is ()345 m s 36.1 m s500 Hz 595 Hz345 m s 25.0 m sOf⎛⎞+==⎜⎟−⎝⎠504 CHAPTER 14 14.31 At point D, the distance of the ship from point A is ()()()222212800 m600 m 800 m 1000 mdd=+ = + = Since destructive interference occurs for the first time when the ship reaches D, it is necessary that 122ddλ−= , or ()()122 2 1000 m600 m 800 mddλ=−= − = 14.35 In the third harmonic, the string forms a standing wave of three loops, each of length 8.00 m2.67 m23λ==. The wavelength of the wave is then 5.33 mλ=. (a) The nodes in this string fixed at each end will occur at distances of 0, 2.67 m, 5.33 m, and 8.00 m from the end. Antinodes occur halfway between each pair of adjacent nodes, or at 1.33 m, 4.00 m, and 6.67 m from the end. (b) The linear density is -3-340.0 10 kg5.00 10 kg m8.00 mmLµ×== = × and the wave speed is -349.0 N99.0 m s5.00 10 kg mFvµ== =× Thus, the frequency is 99.0 m s18.6 Hz5.33 mvfλ== = ABd1Dd2800 mSound 505 14.37 The facing speakers produce a standing wave in the space between them, with the spacing between nodes being ()NN343 m s 10.214 m2 2 2 800 Hzvdfλ⎛⎞== = =⎜⎟⎝⎠ If the speakers vibrate in phase, the point halfway between them is an antinode of pressure, at 1.25 m0.625 m2= from either speaker. Then there is a node at 0.214 m0.625 m0.518 m2−=, a node at 0.518 m0.214 m0.303 m−= , a node at 0.303 m0.214 m0.0891 m−= , a node at 0.518 m0.214 m0.732 m+= , a node at 0.732 m0.214 m0.947 m+= , and a node at 0.947 m0.214 m1.16 m+= from one speaker. 14.41 The speed of transverse waves in the string is 250.000 N70.711 m s1.000 0 10 kg mFvµ−== =× The fundamental wavelength is 121.200 0 mLλ== and its frequency is 1170.711 m s58.926 Hz1.200 0 mvfλ== = The harmonic frequencies are then ()158.926 Hznfnfn== , with n being an integer The largest one under 20 000 Hz is 33919 976 Hz 19.976 kH zf ==506 CHAPTER 14 14.42 The distance between adjacent nodes is one-quarter of the circumference. NN AA20.0 cm5.00 cm24ddλ=== = so 10.0 cm =0.100 mλ= , and 3900 m s9.00 10 Hz 9.00 kH z0.100 mvfλ== = × = The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it. 14.45 Hearing would be best at the fundamental resonance, so