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 21 - Superposition - Exercises and Problems - Page 621: 17

Answer

$T = 2180~N$

Work Step by Step

We can find the speed of the wave along the wire. $v = f~\lambda$ $v = (f)~(2L_{vib})$ $v = (27.5~Hz)(2)(1.90~m)$ $v = 104.5~m/s$ We can find the tension in the wire. $v = \sqrt{\frac{T}{\mu}}$ $v = \sqrt{\frac{T}{m/L}}$ $T = \frac{v^2~m}{L}$ $T = \frac{(104.5~m/s)^2(0.400~kg)}{2.00~m}$ $T = 2180~N$
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