## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 14 - Exercises and Problems - Page 264: 72

#### Answer

a) 69 Hz b) 621 meters per second

#### Work Step by Step

a) This is effectively clamped at one end since it is an open pipe, so we know that the frequencies will be a multiple of one quarter of the wave length. We know that the difference between the two given frequencies is 138, so we know that the frequency before these two frequencies is 138 Hz. From this, we see that the frequency before that is 69 Hz. Since 69 cannot be further divided to continue the pattern, we see that the fundamental frequency is $\fbox{69 Hz.}$ b) This allows us to find the sound speed: $v = 4Lf_1 = 4(2.25)(69)=621 \ m/s$

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