University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 16 - Sound and Hearing - Problems - Exercises - Page 539: 16.26


a) $$ L = 0.328 m$$ b) $$λ = 1.312 m$$ c) $$ f = 262.2 Hz$$

Work Step by Step

a) The formula for the frequency in a open pipe is: $f=\frac{nv}{2L}$ The question specifies fundamental frequency so n=1 (i.e. the first harmonic). $Length = \frac{nv}{2f}$ $Length = \frac{1\times344 m/s}{2\times524 Hz}$ $$Length = 0.328 m$$ b) In a closed pipe, the formula for the frequency is $f=\frac{nv}{4L}$. Since $f=\frac{v}{λ}$, we can rearrange for the wavelength and substitute the equation for the frequency to get: $ λ = \frac{4L}{n}$ Therefore, $$λ = \frac{4\times\times\times0.328 m}{1} = 1.312 m$$ c) $f=\frac{nv}{4L}$. $$ f = \frac{1\times344 m/s}{4\times0.328 m} = 262.2 Hz$$
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