A police car emitting a siren wail (600 Hz) is driving at v/25 towards a suspect

Question

A police car emitting a siren wail (600 Hz) is driving at v/25 towards a suspect who is fleeing on foot at v/55, where v is the speed of sound.

a.) Concept:
The distance between the car and suspect  ---Select--((((- increases,,, decreases ,,,,,stays the same overall, ))))))therefore the frequency observed by the suspect will be  ---Select--- ((((higher than, lower than, the same as that emitted by the siren.

b.) Calculate:
What is the frequency the suspect hears? (Make sure your result agrees with part a!)
____ Hz

Explanation / Answer

b) The doppler shift formula which is given as follows:

   Fl = Fs ( v + vL) / (v+vS)

here
Fl is the frequency heard by the listener(600 Hz)
Fs is the frequency heard by the source

v is the velocity of sound  (340m/s)
vL is the velocity of the listener (v/55)
vS is the velocity of the source (-v/25)

Fl = (600 Hz) * (340 m/s - 340/55 m/s)/(340 m/s - 340/25 m/s)
         = (600 Hz) * (333.818 m/s)/(326.4 m/s)
         = 613.64 Hz

a) it makes sense that the suspect would hear a higher pitched sound than the police car would because overall the police car is closing on the suspect so it is similar to the situation in which the suspect is at rest and the police car is closing on them.

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