Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 12 - Sound - General Problems - Page 357: 88

Answer

55.3 kHz.

Work Step by Step

Use equation 12–4, because the source and observer are both in motion. There are 2 Doppler shifts in this problem: the first is for the emitted sound with the bat as the source and the moth as the observer $$f’_{moth}=f_{bat}\frac{v_{sound}+v_{moth}}{ v_{sound}-v_{bat}}$$ The second Doppler shift is for the reflected sound, with the moth as the source and the bat as the observer. $$f’’_{bat}=f’_{moth}\frac{v_{sound}+v_{bat}}{ v_{sound}-v_{moth}}$$ $$= f_{bat}\frac{v_{sound}+v_{moth}}{ v_{sound}-v_{bat}}\frac{v_{sound}+v_{bat}}{ v_{sound}-v_{moth}}$$ $$ = (51.35\;kHz)\frac{343+5.0}{343-7.8}\frac{343+7.8}{ 343-5.0}=55.3\;kHz$$
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