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 - Problems - Page 324: 41

Answer

The tension in the cord is 4.8 N

Work Step by Step

We can find the speed $v$ of the wave as: $v = \frac{L}{t}$ $v = \frac{8.7~m}{0.85~s}$ $v = 10.24~m/s$ We can find the tension $F_T$ in the cord as: $v = \sqrt{\frac{F_T}{\mu}}$ $F_T = v^2~\mu$ $F_T = \frac{v^2~m}{L}$ $F_T = \frac{(10.24~m/s)^2(0.40~kg)}{8.7~m}$ $F_T = 4.8~N$ The tension in the cord is 4.8 N.
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