Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - General Problems - Page 327: 84

Answer

1.59 cm.

Work Step by Step

The wave speed is given by $v=\lambda f$. The maximum speed of particles on the cord is given by equation 11–7, $v_{max}=2\pi Af$. Set the two expressions equal to one another. $$\lambda f = 2 \pi A f$$ $$A=\frac{\lambda}{2 \pi}=\frac{10.0\;cm}{2 \pi}=1.59\;cm$$
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