Problem 13B

NAME ______________________________________ DATE _______________ CLASS ____________________Holt PhysicsProblem 13BHARMONICSPROBLEMSOLUTIONThe tallest load-bearing columns are part of the Temple of Amun inEgypt, built in 1270 B.C. Find the height of these columns if a standingwave with a frequency of 47.8 Hz is generated in an open pipe that is astall as the columns. The sixth harmonic is generated. The speed of soundin air is 334 m/s.Given: f 6 = 47.8 Hz v = 334 m/sn = number of the harmonic = 6Unknown: L = ?When the pipe is open, the wavelength associated with the first harmonic (fundamentalfrequency) is twice the length of the pipe.f n = n ⎯ 2vL ⎯ n = 1,2,3,...v ( 334m/s)L = n ⎯⎯ = (6) ⎯⎯ =2 f n 2 ( 47.8Hz)21.0 mADDITIONAL PRACTICECopyright © by Holt, Rinehart and Winston. All rights reserved.1. A 47.0 m alphorn was made in Idaho in 1989. An alphorn behaves like apipe with one end closed. If the frequency of the fifteenth harmonic is26.7 Hz, how long is the alphorn? The speed of sound in air is 334 m/s.2. A fully functional acoustic guitar over 8.0 m in length is on display inBristol, England. Suppose the speed of waves on the guitar’s strings is5.00 × 10 2 m/s. If a third harmonic is generated on a string, so that thesound produced in air has a wavelength of 3.47 m, what is the length ofthe string? The speed of sound in air is 334 m/s.3. The unsupported flagpole built for Canada’s Expo 86 has a height of86 m. If a standing wave with a 19th harmonic is produced in an 86 mopen pipe, what is its frequency? The speed of sound in air is 334 m/s.4. A power-plant chimney in Spain is 3.50 × 10 2 m high. If a standing wavewith a frequency of 35.5 Hz is generated in an open pipe with a lengthequal to the chimney’s height and the 75th harmonic is present, what isthe speed of sound?5. The world’s largest organ was completed in 1930 in Atlantic City, NewJersey. Its shortest pipe is 4.7 mm long. If one end of this pipe is closed,what is the number of harmonics created by an ultrasound with a wavelengthof 3.76 mm?Problem 13B 125

Additional Practice 13BGivens1. f 15 = 26.7 Hzv = 334 m/sn = 15Solutions15v15(334 m/s)L = ⎯⎯ = ⎯⎯4 f 1 5 4(26.7 Hz)L = 46.9 m2. l = 3.47 mv s = 5.00 × 10 2 m/sn = 3v a = 334 m/sL = ⎯ l vsn(3.47 m)(5.00 × 10 2 m/s)(3)⎯ = ⎯⎯⎯2va2(334 m/s)L = 7.79 m3. n = 19L = 86 mv = 334 m/snvf 19 = ⎯⎯ = ⎯ 19 ( 334m/s)⎯2 L 2(86m)f 19 = 37 HzII4. L = 3.50 × 10 2 mf 75 = 35.5 Hzn = 75f 75 = ⎯ 7 5 v⎯2Lv = ⎯ 2L f75⎯ =752(3.50 × 10 2 m)(35.5 Hz)⎯⎯⎯75v = 331 m/s5. L = 4.7 × 10 −3 ml = 3.76 × 10 −3 ml n = ⎯ 4 L ⎯n4L 4(4.7 × 10 −3 m)n = ⎯⎯ = ⎯⎯ = 5l n (3.76 × 10 −3 m)Copyright © by Holt, Rinehart and Winston. All rights reserved.II Ch. 13–2Holt Physics Solution Manual