University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 15 - Mechanical Waves - Problems - Exercises - Page 497: 15.10

Answer

(a) It takes 1.01 seconds for one complete wave pattern to go past. The wave crest travels a horizontal distance of 15.3 cm in a time of 1.01 seconds. (b) The wave number is 0.410 rad/cm The number of waves that pass each second is 0.987 (c) The speed of the wave crest is 15.1 cm/s The maximum speed of the cork floater is 17.05 cm/s

Work Step by Step

We can write the general equation for a wave equation when the wave is moving in the negative x-direction. $y(x,t) = A~cos(k~x+\omega~t)$ (a) We can find the period. $T = \frac{2\pi}{\omega}$ $T = \frac{2\pi}{6.20~rad/s}$ $T = 1.01~s$ It takes 1.01 seconds for one complete wave pattern to go past. We can find the wavelength. $\lambda = \frac{2\pi}{k}$ $\lambda = \frac{2\pi}{0.410~rad/cm}$ $\lambda = 15.3~cm$ The wave crest travels a horizontal distance of 15.3 cm in a time of 1.01 seconds. (b) The wave number is $k$ which is 0.410 rad/cm We can find the frequency. $f = \frac{\omega}{2\pi}$ $f = \frac{6.20~rad/s}{2\pi}$ $f = 0.987~Hz$ The number of waves that pass each second is 0.987 (c) We can find the speed of the wave crest. $v = \lambda~f$ $v = (15.3~cm)(0.987~Hz)$ $v = 15.1~cm/s$ The speed of the wave crest is 15.1 cm/s We can find the maximum speed of the cork floater. $v_{max} = A~\omega$ $v_{max} = (2.75~cm)(6.20~rad/s)$ $v_{max} = 17.05~cm/s$ The maximum speed of the cork floater is 17.05 cm/s
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.