College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 11 - Problems - Page 432: 64


$\mu = 0.45~g/m$

Work Step by Step

We can find the wave speed in the string: $v = \lambda~f$ $v = 2L~f$ $v = (2)(0.65~m)(329.63~Hz)$ $v = 428.5~m/s$ We can find the mass per unit length of the string: $v = \sqrt{\frac{F}{\mu}}$ $v^2 = \frac{F}{\mu}$ $\mu = \frac{F}{v^2}$ $\mu = \frac{82~N}{(428.5~m/s)^2}$ $\mu = 4.5\times 10^{-4}~kg/m$ $\mu = 0.45~g/m$
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