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 - Exercises and Problems - Page 588: 57

Answer

$-18.8\;\rm m/s,\;0\;m/s,18.8\;m/s$

Work Step by Step

The wave is moving to the right, so from the given graph, it is obvious that point's 1 velocity is downward and point's 3 velocity is upward while point's 2 velocity is zero. The velocity at 1 and 3 are the maximum velocities and their magnitude is given by $$v_{max}=\omega A=2\pi f A$$ where $f=v/\lambda$; $$v_{max}=\omega A=\dfrac{2\pi Av}{\lambda}$$ Plugging the known from the given graph $$v_{max}= \dfrac{2\pi (2\times 10^{-2})(45)}{(30\times 10^{-2})}=\bf 18.8\;\rm m/s$$ Therefore, $$v_1=\color{red}{\bf -18.8}\;\rm m/s$$ $$v_2=\color{red}{\bf 0}\;\rm m/s$$ $$v_3=\color{red}{\bf 18.8}\;\rm m/s$$
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