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: 43

Answer

The length of the cord is 21 meters.

Work Step by Step

We can write an expression for the speed $v$ of the vibrations. Let $L$ be the length of the cord. $v = \frac{L}{t}$ We can write another expression for the speed of the vibrations. $v = \frac{F_T}{\mu} = \sqrt{\frac{F_T}{m/L}}$ To find the length $L$, we can equate the two expressions for the speed $v$. $\frac{L}{t} = \sqrt{\frac{F_T}{m/L}}$ $\frac{L^2}{t^2} = \frac{F_T~L}{m}$ $L = \frac{F_T~t^2}{m}$ $L = \frac{(35~N)(0.55~s)^2}{0.50~kg}$ $L = 21~m$ The length of the cord is 21 meters.
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