College Physics (4th Edition)

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

Chapter 11 - Problems - Page 431: 63

Answer

The fundamental frequency of this string is $616~Hz$

Work Step by Step

We can find the speed of the wave in the string: $v = \sqrt{\frac{F}{\mu}}$ $v = \sqrt{\frac{mg}{\mu}}$ $v = \sqrt{\frac{(2.20~kg)(9.80~m/s^2)}{3.55\times 10^{-6}~kg/m}}$ $v = 2464~m/s$ We can find the fundamental frequency: $f = \frac{v}{\lambda}$ $f = \frac{v}{2L}$ $f = \frac{2464~m/s}{(2)(2.00~m)}$ $f = 616~Hz$ The fundamental frequency of this string is $616~Hz$.
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