College Physics (4th Edition)

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

Chapter 12 - Problems - Page 464: 44

Answer

Mandy hears a beat frequency of $~6.4~Hz$

Work Step by Step

We can use the equation for the Doppler effect when the observer is approaching: $f_1 = \left(\frac{v+v_o}{v}\right)~f_s$ $f_1 = \left(\frac{343~m/s+1.56~m/s}{343~m/s}\right)~(698~Hz)$ $f_1 = 701.2~Hz$ We can use the equation for the Doppler effect when the observer is moving away: $f_2 = \left(\frac{v-v_o}{v}\right)~f_s$ $f_2 = \left(\frac{343~m/s-1.56~m/s}{343~m/s}\right)~(698~Hz)$ $f_2 = 694.8~Hz$ The beat frequency is the difference between the two frequencies that Mandy hears from the sirens. We can find the difference of the two frequencies: $f_1-f_2 = 701.2~Hz-694.8~Hz = 6.4~Hz$ Mandy hears a beat frequency of $~6.4~Hz$.
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