Physics: Principles with Applications (7th Edition)

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

Chapter 11 - Oscillations and Waves - General Problems - Page 326: 82

Answer

$\lambda_n=\frac{4l}{2n-1}$

Work Step by Step

Here, we have one node where the string can't move and one antinode where the string can freely move. The possible combinations of wavelength from length include: $\lambda_1=4l$, $\lambda_3=\frac{4l}{3}$, and $\lambda_5=\frac{4l}{5}$ $\lambda_n=\frac{4l}{2n-1}$
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