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### Assign13

Course: PHYSICS 104, Fall 2010
School: Rutgers
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Word Count: 301

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2130 PHY Homework solutions Assignment 13 14.4 At T = 27C = 300 K , the speed of sound in air is v = ( 331 m s ) T = ( 331 m s ) 273 K 300 K = 347 m s 273 K The wavelength of the 20 Hz sound is = 20 000 Hz is = v 347 m s = = 17 m , and that of the f 20 Hz 347 m s = 1.7 10-2 m=1.7 cm . Thus, range of wavelengths of 20 000 Hz 1.7 cm to 17 m audible sounds at 27C is 14.25 . With the train...

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2130 PHY Homework solutions Assignment 13 14.4 At T = 27C = 300 K , the speed of sound in air is v = ( 331 m s ) T = ( 331 m s ) 273 K 300 K = 347 m s 273 K The wavelength of the 20 Hz sound is = 20 000 Hz is = v 347 m s = = 17 m , and that of the f 20 Hz 347 m s = 1.7 10-2 m=1.7 cm . Thus, range of wavelengths of 20 000 Hz 1.7 cm to 17 m audible sounds at 27C is 14.25 . With the train approaching at speed v t , the observed frequency is 345 m s + 0 345 m s 442 Hz = f = f 345 m s - v t 345 m s - v t As the train recedes, the observed frequency is 345 m s + 0 441 H z = f = 345 m s - ( -v t ) Dividing equation (1) by (2) gives 345 m s f 345 m s + v t (2) (1) 442 345 m s + v t = , 441 345 m s - v t 0.391 m s and solving for the speed of the train yields v t = 14.36 The mass per unit length of the wire is = m 0.300 103 kg = = 4.29 kg 104 m , 2 L 70.0 10 m and the speed of transverse waves is v= F 600 N = = 1.18 103 m s 4.29 104 kg m The fundamental or first harmonic of the wire has a wavelength of 1 = 2L = 1.40 m , and v 1.18 103 m s f1 = = = 845 H z . frequency 1 1.40 m The frequency of the second harmonic is f 2 = 2 f1 = harmonic is f 3 = 3 f1 = 2.54 10 H z 3 1.69 3 H z 10 and that of the third . 14.45 Hearing would be best at the fundamental resonance, so = 4L = 4 ( 2.8 cm ) and f= 340 m s 100 cm v 3 = = 3.0 10 Hz = 4 ( 2.8 cm ) 1 m 3.0 kH z 14.47 (a) The speed of sound is 331 m s at 0 C, so the fundamental wavelength of the pipe open at both ends is 1 = 2 L = v 331 m s v = = giving L = f1 2 f1 2 ( 300 Hz ) 0.552 m (b) At T = 30 C = 303 K , v = ( 331 m s ) f1 = T 303 K = ( 331 m s ) = 349 m s 273 K 273 K 316 H z and 349 m s v v = = = 1 2 L 2 ( 0.552 m )
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Rutgers - PHYSICS - 104
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