Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 12 - Sound - Problems - Page 355: 30

Answer

8.6 m to $8.6\times 10^{-3}m$.

Work Step by Step

See figure 12-11. For a pipe open at both ends, the fundamental frequency is $f_1=\frac{v}{2 \mathcal{l}}$. Solve for the length. $$\mathcal{l}=\frac{v}{2f_1}$$ Find the length of the pipe needed for the highest and lowest frequencies. $$\mathcal{l}_{20Hz}=\frac{343m/s}{2(20Hz)}=8.6m$$ $$\mathcal{l}_{20kHz}=\frac{343m/s}{2(20000Hz)}=8.6\times 10^{-3}m$$
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