University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 15 - Mechanical Waves - Problems - Exercises - Page 501: 15.58


(a) $f_1 = 156~Hz$ $\lambda_1 = 1.60~m$ (b) F = 467 N

Work Step by Step

(a) Let $f_1$ be the fundamental frequency. $n~f_1 = 624~Hz$, for some whole number $n$. $(n+1)~f_1 = 780~Hz$, for some whole number $n$. We can subtract the first equation from the second equation. $f_1 = 780~Hz-624~Hz = 156~Hz$ We can find the wavelength $\lambda_1$. $\lambda_1 = 2L$ $\lambda_1 = (2)(0.800~m)$ $\lambda_1 = 1.60~m$ (b) We can find the speed of the wave along the string. $v = f_1~\lambda_1$ $v = (156~Hz)(1.60~m)$ $v = 249.6~m/s$ We can find the tension $F$. $\sqrt{\frac{F}{\mu}} = v$ $F = v^2~\mu$ $F = (249.6~m/s)^2(0.00750~kg/m)$ $F = 467~N$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.