Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 20 - Traveling Waves - Conceptual Questions - Page 585: 5

Answer

$\lambda_1 \gt \lambda_2 \gt \lambda_3$

Work Step by Step

Let $v$ be the speed of sound. Given the frequency $f$, we can write an expression for the wavelength. $\lambda = \frac{v}{f}$ From this equation, we can see that the wavelength decreases as the frequency increases. We know that $f_3 \gt f_2 \gt f_1$. Therefore: $\lambda_1 \gt \lambda_2 \gt \lambda_3$.
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