Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 14 - Waves and Sound - Problems and Conceptual Exercises - Page 493: 28

Answer

For note A $27.5Hz;12.5m$ For note C $4.19KHZ;8.18\times 10^{-2}m$

Work Step by Step

We can find the required frequencies and wavelengths as follows: The lowest note on the piano is A and the frequency given in the table=440Hz We are given that note A has octaves below 440 Hz Hence $f=\frac{440}{16}=27.5Hz$ We know that $\lambda=\frac{v}{f}$ We plug in the known values to obtain: $\lambda=\frac{343m/s}{27.5Hz}$ $\lambda=12.5m$ Now the frequency of note C is given as $f^{\prime}=261.7Hz\times 16$ $f^{\prime}=4.19KHz$ and $\lambda^{\prime}=\frac{343}m/s{4.19KHz}=8.18\times 10^{-2}m$
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