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Unless indicated otherwise, assume the speed of | Ch 16 - 41E

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 41E Chapter 16

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 41E

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 cm. Find the beat frequency that they produce when playing together in their fundamentals.

Step-by-Step Solution:
Step 1 of 3

Solution 41E The length of the organ pipes were 1.14 meters. Among these two, one get lengthened by 2 cm. So, it’s length is 1.14 + 0.02 = 1.16 meters. The frequency in the original pipe can be calculated as, Fn= nV / 4l. Here we are dealing with the 1st node where n=1. So, frequency of pipe 1 is, F 1 1×V = 344 = 75.43 4×1.14 4.56 The frequency of the lengthened pipe is, 1×V 344 F 2 4×1.16= 4.64= 74.13 The frequency of the beat will be, F beat= F 1 F =275.43 74.13 = 1.30 Hz . Conclusion: The beat frequency is 1.30 Hz.

Step 2 of 3

Chapter 16, Problem 41E is Solved
Step 3 of 3

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Unless indicated otherwise, assume the speed of | Ch 16 - 41E