College Physics (4th Edition)

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

Chapter 11 - Problems - Page 432: 66


The three largest weights are $~1.02~N~$, $~0.256~N~$, and $~0.114~N$

Work Step by Step

We can write an expression for the wave speed in the string: $v = \lambda_n~f = \frac{(2L)(f)}{n}$, for some integer $n$ The tension in the string will be equal to the weight $W$. We can find an expression for the weight: $\sqrt{\frac{F}{\mu}} = v$ $\sqrt{\frac{W}{\mu}} = \frac{2L~f}{n}$ $\frac{W}{\mu} = (\frac{2L~f}{n})^2$ $W = (\frac{2L~f}{n})^2~(\mu)$ $W = \left(\frac{(2)(0.42~m)(110~Hz)}{n}\right)^2~(0.120\times 10^{-3}~kg/m)$ $W = \frac{1.0245~N}{n^2}$ We can find the three largest weights: $W_1 = \frac{1.0245~N}{1^2} = 1.02~N$ $W_2 = \frac{1.0245~N}{2^2} = 0.256~N$ $W_3 = \frac{1.0245~N}{3^2} = 0.114~N$ The three largest weights are $~1.02~N~$, $~0.256~N~$, and $~0.114~N$
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