12_InstSolManual_PDF_Part8

12_InstSolManual_PDF_Part8 - Eq. (12.7) applies to a pipe...

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(b) closed and with 3, 12.28. Set Up: There is a displacement node at each closed end of the pipe. The standing wave pattern for the first three normal modes are shown in Figure 12.28. Figure 12.28 Solve: (a) For the fundamental and since the node to node distance is Each successive over- tone adds another along the length of the pipe, so 2, and 2, (b) This result is the same for a pipe open at both ends. 12.29. Set Up: For a stopped pipe, Solve: This result is a reasonable value for the mouth to diaphragm distance for a typical adult. Reflect: 1244 Hz is not an integer multiple of the fundamental frequency of 220 Hz, it is 5.65 times the fundamen- tal. The production of sung notes is more complicated than harmonics of an air column of fixed length. 12.30. Set Up: There are displacement nodes at the closed end of an air column. Problem 12.28 shows that
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Unformatted text preview: Eq. (12.7) applies to a pipe closed at both ends as well as one open at both ends. Solve: (a) There are displacement nodes at the floor and ceiling. L 5 v 4 f 1 5 344 m / s 4 1 220 Hz 2 5 39.1 cm. f 1 5 220 Hz f 1 5 v 4 L . 3, c . n 5 1, l n 5 2 L n , 3, c . n 5 1, n l n 2 5 L , l / 2 l / 2. l 5 2 L , l / 2 5 L N N A Fundamental A A N N N 1st overtone A N A A N N N 2nd overtone L L L f 7 5 7 f 1 5 7770 Hz f 5 5 5 f 1 5 5550 Hz, f 3 5 3 f 1 5 3330 Hz, f 1 5 v 4 L 5 354 m / s 4 1 0.080 m 2 5 1110 Hz, l 7 5 4.6 cm l 5 5 6.4 cm, l 3 5 10.7 cm, l 1 5 32.0 cm, 5, c . n 5 1, f n 5 n v 4 L , l n 5 4 L n f 4 5 4 f 1 5 8840 Hz f 3 5 3 f 1 5 6630 Hz, f 2 5 2 f 1 5 4420 Hz, f 1 5 v 2 L 5 354 m / s 2 1 0.080 m 2 5 2210 Hz, 12-8 Chapter 12...
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