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 - Problems - Page 356: 62

Answer

a) $f_a=1750Hz$ $f_r==1440Hz$ b) $f_a=1880 Hz$ $f_r=1340 Hz$ c) $f_a=1640 Hz$ $f_r=1530 Hz$

Work Step by Step

Given: Siren frequency: $1580Hz$ Speed: $120.0\frac{km}{h}\times\frac{1000m}{1km}\times\frac{1h}{3600s}=33.3\frac{m}{s}$ (page 329):$v_{sound}=343\frac{m}{s}$ Equations: (page 345, 12-2a): $f_a=\frac{f}{1-\frac{v_{source}}{v_{sound}}}$ (page 345, 12-2a): $f_r=\frac{f}{1+\frac{v_{source}}{v_{sound}}}$ (page 345, 12-2a): $f_{both moving}=f\frac{v_{snd}\pm v_{obs}}{v_{snd}\mp v_{source}}$ a) $f_a=\frac{1580Hz}{1-\frac{33.3\frac{m}{s}}{343\frac{m}{s}}}=1750Hz$ $f_r=\frac{1580Hz}{1-\frac{33.3\frac{m}{s}}{343\frac{m}{s}}}=1440Hz$ b) $v_c=90.0\frac{km}{h}\times\frac{1000m}{1km}\times\frac{1h}{3600s}=25\frac{m}{s}$ $f_a=f\frac{343\frac{m}{s}+ 25\frac{m}{s}}{343\frac{m}{s}- 33.3\frac{m}{s}}=1880 Hz$ $f_r=f\frac{343\frac{m}{s}- 25\frac{m}{s}}{343\frac{m}{s}+ 33.3\frac{m}{s}}=1340 Hz$ c) $v_c=80.0\frac{km}{h}\times\frac{1000m}{1km}\times\frac{1h}{3600s}=22.2\frac{m}{s}$ $f_a=f\frac{343\frac{m}{s}- 22.2\frac{m}{s}}{343\frac{m}{s}- 33.3\frac{m}{s}}=1640 Hz$ $f_r=f\frac{343\frac{m}{s}+ 22.2\frac{m}{s}}{343\frac{m}{s}+ 33.3\frac{m}{s}}=1530 Hz$
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