Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 14 - Waves and Sound - Problems and Conceptual Exercises - Page 495: 71

Answer

(a) $3.6KHz;9.6cm$ (b) $11KHz;3.2cm$ (c) greater than

Work Step by Step

(a) We can find the required fundamental frequency and wavelength as follows: $f_1=\frac{v}{4L}$ We plug in the known values to obtain: $f_1=\frac{343m/s}{4\times 2.4\times 10^{-2}m}$ $f_1=3.57\times 10^3Hz=3.6KHz$ and $\lambda_1=4L$ $\lambda_1=4(2.4cm)=9.6cm$ (b) We can find the required frequency and wavelength as follows: $f_3=3f_1$ $f_3=3(3.6KHz)=11KHz$ and $\lambda_3=\frac{\lambda_1}{3}$ $\lambda_3=\frac{9.6cm}{3}=3.2cm$ (c) We know that a shorter wavelength has a greater frequency; thus, if the person's ear canal is shorter than $L=2.4cm$, then the fundamental frequency will be greater.
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