Answer:
1270.44 Hz
Explanation:
[tex]v_{L}[/tex] = velocity of the our car = 35.0 m/s
[tex]v_{P}[/tex] = velocity of the police car = ?
[tex]v_{S}[/tex] = velocity of the sound = 343 m/s
[tex]f_{app}[/tex] = frequency observed as police car approach = 1310 Hz
[tex]f_{rec}[/tex] = frequency observed as police car go away = 1240 Hz
[tex]f[/tex] = actual frequency of police siren
Frequency observed as police car approach is given as
[tex]f_{app}= \frac{(v_{s}-v_{L})f}{v_{s} -v_{P} }[/tex]
inserting the values
[tex]1310 = \frac{(343 - 35)f}{343 -v_{P} }[/tex] eq-1
Frequency observed as police car goes away is given as
[tex]f_{rec}= \frac{(v_{s} + v_{L})f}{v_{s} + v_{P} }[/tex]
inserting the values
[tex]1240 = \frac{(343 + 35)f}{343 + v_{P} }[/tex] eq-2
Dividing eq-1 by eq-2
[tex]\frac{1310}{1240} = \left ( \frac{343 - 35}{343 - v_{P} } \right )\frac{(343 + v_{P})}{343 + 35 }\\[/tex]
[tex]v_{P}[/tex] = 44.3 m/s
Using eq-1
[tex]1310 = \frac{(343 - 35)f}{343 - 44.3 }[/tex]
f = 1270.44 Hz
