College Physics (4th Edition)

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

Chapter 11 - Problems - Page 431: 56

Answer

The fundamental frequency increases by a factor of 1.20 which is an increase of 20%

Work Step by Step

Since we can assume that the tension and mass density of the string do not change, the wave speed also does not change. We can find an expression for the fundamental frequency: $f = \frac{v}{\lambda}$ $f = \frac{v}{2L}$ We can find the new fundamental frequency when the length of the vibrating part of the string is $\frac{5L}{6}$: $f' = \frac{v}{2(\frac{5L}{6})}$ $f' = \frac{6v}{2(5L)}$ $f' = \frac{6}{5}\times \frac{v}{2L}$ $f' = 1.20\times f$ The fundamental frequency increases by a factor of 1.20 which is an increase of 20%
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