College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 12 - Problems - Page 465: 61


(a) The height of the chimney is $5.05~m$ (b) The fundamental frequency is $16.3~Hz$

Work Step by Step

(a) We can find the frequency of a note that is three octaves below middle C: $f = (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})(261.6~Hz) = 32.7~Hz$ We can model the chimney as a pipe that is open at both ends. We can find the height $L$ of the chimney: $\lambda = \frac{v}{f}$ $2L = \frac{v}{f}$ $L = \frac{v}{2f}$ $L = \frac{330~m/s}{(2)(32.7~Hz)}$ $L = 5.05~m$ The height of the chimney is $5.05~m$ (b) We can find the fundamental frequency when the chimney is closed at the bottom: $f = \frac{v}{\lambda}$ $f = \frac{v}{4L}$ $f = \frac{330~m/s}{(4)(5.05~m)}$ $f = 16.3~Hz$ The fundamental frequency is $16.3~Hz$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.