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In: Physics

At a factory, a noon whistle is sounding with a frequency of 480 Hz. As a...

At a factory, a noon whistle is sounding with a frequency of 480 Hz. As a car traveling at 61.0 km/h approaches the factory, the driver hears the whistle at frequency fi. After driving past the factory, the driver hears frequency ff. What is the change in frequency ff - fi heard by the driver? Assume the speed of sound in air is 343 m/s.

Solutions

Expert Solution

Given
   frequency of whistle(source) f0 = 480 Hz
        speed of car (listner) V0 = 61 kmph = 61*5/18 = 16.94 m/s
   speed of sound V = 343 m/s

   as the car reaches the whistle driver observes the higher frequency than moving away from the factory.

   Doppler effect apparent frequency approaching fi = f0(V+Vo / V)
                                     moving away from whistle ff = f0(V-v0 / V)


           change in frequency = (f0/V)*(2v0)
                                             = (480/343)*(2*16.94)
                                             = 33.88 Hz

change in frequency observed by the car driver is 33.88 Hz

genius_generous answered 2 years ago
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