College Physics (4th Edition)

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

Chapter 11 - Problems - Page 433: 77


The new fundamental frequency in each wire is $933~Hz$

Work Step by Step

We can find an expression for the original fundamental frequency: $f = \frac{v}{\lambda}$ $f = \frac{\sqrt{\frac{mg}{\mu}}}{2L}$ $f = \frac{1}{2}\sqrt{\frac{mg}{\mu~L^2}}$ When the wire is cut in half, the length of each wire is half of the original length but the mass per unit length is still the same. Each wire has half the tension since each wire supports half the weight of the sign. We can find the new fundamental frequency: $f' = \frac{v}{\lambda}$ $f' = \frac{\sqrt{\frac{mg/2}{\mu}}}{2(L/2)}$ $f' = \frac{2}{\sqrt{2}}\times \frac{\sqrt{\frac{mg}{\mu}}}{2L}$ $f' = \sqrt{2}\times \frac{1}{2}\sqrt{\frac{mg}{\mu~L^2}}$ $f' = \sqrt{2}\times f$ $f' = \sqrt{2}\times 660~Hz$ $f' = 933~Hz$ The new fundamental frequency in each wire is $933~Hz$.
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