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

Published by Pearson

# Chapter 21 - Superposition - Exercises and Problems - Page 621: 13

#### Answer

(a) $\lambda_1 = 2.42~m$ $\lambda_2 = 1.21~m$ $\lambda_3 = 0.807~m$ (b) $\lambda_1 = 4.84~m$ $\lambda_3 = 1.61~m$ $\lambda_5 = 0.968~m$

#### Work Step by Step

(a) For a tube that is open at both ends, the wavelengths are $\lambda_m = \frac{2L}{m}$ We can find the three longest wavelengths: $\lambda_1 = \frac{2L}{1} = \frac{(2)(1.21~m)}{1} = 2.42~m$ $\lambda_2 = \frac{2L}{2} = \frac{(2)(1.21~m)}{2} = 1.21~m$ $\lambda_3 = \frac{2L}{3} = \frac{(2)(1.21~m)}{3} = 0.807~m$ (b) For a tube that is closed at one end, the wavelengths are $\lambda_m = \frac{4L}{m}$, where $m = 1, 3, 5,...$ We can find the three longest wavelengths: $\lambda_1 = \frac{4L}{1} = \frac{(4)(1.21~m)}{1} = 4.84~m$ $\lambda_3 = \frac{4L}{3} = \frac{(4)(1.21~m)}{3} = 1.61~m$ $\lambda_5 = \frac{4L}{5} = \frac{(4)(1.21~m)}{5} = 0.968~m$

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