College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 11 - Problems - Page 428: 23


(a) The amplitude is $3.5~cm$ (b) The wavelength is $6.0~cm$

Work Step by Step

(a) In general: $y(x,t) = A~sin(k~x-\omega~t)$ We can see from the equation that $A = 3.5~cm$ The amplitude is $3.5~cm$ (b) We can see from the equation that $k= \frac{\pi~rad}{3.0~cm}$ We can find the wavelength: $\lambda = \frac{2\pi}{k}$ $\lambda = \frac{2\pi}{\pi~rad/3.0~cm}$ $\lambda = 6.0~cm$ The wavelength is $6.0~cm$
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