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Unformatted text preview: homework 24 FIERRO, JEFFREY Due: Apr 27 2008, 11:00 pm 1 Question 1, chap 16, sect 3. part 1 of 1 10 points Consider two organ pipes. The first pipe is open at both ends and its 1 . 165 m long. The second pipe is open at one end and closed at the other end; its 1 . 714 m long. When both pipes are played together, the first overtone the lowest harmonic above the fundamental frequency of the second pipe produces beats agains the fundamental harmonic of the first pipe. What is the fre quency of these beats? Take the sound speed in air to be 343 m / s. Correct answer: 2 . 87721 Hz (tolerance 1 %). Explanation: The first pipe is open at both ends, so a resonating standing wave in it has pressure nodes at both ends. This implies L o = 2 , 2 2 , 3 2 , 4 2 , . . . and therefore f = v 2 L o , 2 v 2 L o , 3 v 2 L o , 4 v 2 L o , . . . For the fundamental harmonic of the pipe L o = 2 = f fund o = v 2 L o . Numerically, f fund o = 343 m / s 2(1 . 165 m) = 147 . 21 Hz . The second pipe is open at one and and closed at the other. A resonating standing wave in this pipe has a pressure node at the open end and a displacement node at the closed end. A displacement node is an antin ode of pressure and lies quarterwavelength from the nearest pressure node. Hence, the distance between the two ends of the string must be L co = 4 + 2 an integer = 4 , 3 4 , 5 4 , 7 4 , . . . and therefore f = v 4 L co , 3 v 4 L co , 5 v 4 L co , 7 v 4 L co , . . . Note that all the harmonics are odd multi plets of the fundamental harmonic. For the fundamental harmonic of this pipe we have L co = 4 = f fund co = v 4 L co , while the first overtone corresponds to L co = 3 4 = f 1st over co = 3 v 4 L co = 3 f fund co . Numerically, f 1st over co = 3 343 m / s 4(1 . 714 m) = 150 . 088 Hz . The first overtone of the second pipe is close to the fundamental harmonic of the first pipe, but the two frequencies are not exactly equal. So when they sound together, they beat with frequency f beat =  f 1st over co f fund o  . Numerically, f beat =  150 . 088 Hz 147 . 21 Hz  = 2 . 87721 Hz . Question 2, chap 16, sect 3. part 1 of 1 10 points The waves from a radio station can reach a home receiver by two paths. One is a straight line path from transmitter to home, a distance of 43 . 7 km. The second path is by reflection from the ionosphere (a layer of ionized air molecules near the top of the atmosphere). Assume: This reflection takes place at a point midway between receiver and transmit ter. Assume: No phase changes on reflection. If the wavelength broadcast by the radio station is 465 m, find the minimum height homework 24 FIERRO, JEFFREY Due: Apr 27 2008, 11:00 pm 2 h of the ionospheric layer that produces de structive interference between the direct and reflected beams....
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This note was uploaded on 05/11/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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