T2a2s - Physics 102{1 Assignment 2 solutions(Spring Term 14.33 The wavelength of the sound is v 345 m s \u03bb= = = 0.500 m f 690 Hz Speaker 2 d1 0.700

T2a2s - Physics 102{1 Assignment 2 solutions(Spring...

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Physics 102 Assignment 2 solutions (Spring Term){1} 14.33The wavelength of the sound is 345 m s0.500 m690 Hzvfλ===. (a) At the first relative maximum (constructive interference), 1220.500 mdddλ=+=+Using the Pythagorean theorem, , giving ()(222220.500 m0.700 mdd+=+)2d=0.240 mSpeaker 2Speaker 1Observer0.700 md1d2(b) At the first relative minimum (destructive interference), 12220.250dddmλ=+=+Therefore, the Pythagorean theorem yields , or ()()222220.250 m0.700 mdd+=+2d=0.855 m{2} 14.44 (a) In the fundamental resonant mode of a pipe open at both ends, the distance between antinodes is AA2dLλ==. Thus, ()220.320 m0.640 mLλ===and 340 m s0.640 mvfλ===531 Hz(b) AA340 m s110.0425 m2224 000 Hzvdfλ=====4.25 cm{3} 14.47(a) The speed of sound is 331 m sat 0 °C, so the fundamental wavelength of the pipe open at both ends is 112vLfλ==giving ()1331 m s22300 HzvLf===0.552 m
(b) At , 30 C303 KT=°=()()303 K331 m s331 m s349 m s273 K273 KTv===and ()11349 m s220.552 mvvfLλ====316 Hz{4} 14.35

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