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 21 - Superposition - Exercises and Problems - Page 621: 12


$\lambda = 1.06\times 10^{-5}~m$ $f = 2.83\times 10^{13}~Hz$

Work Step by Step

We can find the wavelength as: $\lambda = \frac{2L}{m}$ $\lambda = \frac{(2)(0.5300~m)}{100,000}$ $\lambda = 1.06\times 10^{-5}~m$ We can find the frequency as: $f = \frac{v}{\lambda}$ $f = \frac{3.0\times 10^8~m/s}{1.06\times 10^{-5}~m}$ $f = 2.83\times 10^{13}~Hz$
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