Physics
posted by hayley on .
Two cars have identical horns, each emitting a frequency of f= 395 hz. One of the cars is moving with a speed of 12 m/s toward a bystander waiting at a corner, and the other car is parked. The speed of sound is 343 m/s. what is the beat frequency heard by the bystander?

Dopplereffect.
At source approaching
f(observed) = [v/(vu)] •f(source),
f(observed) =
= [343/(34312)] •395=409.3 Hz
f(beat)= f(observed)=f(source)=
=409.3395=14.3 Hz 
The given are (a) the speed of sound, which is 395 m/s, (b) the velocity of the moving source, which is 12 m/s, and (c) the frequency of the sound both cars are emitting, which is 395 Hz.
Let's assume that f{o} is the frequency the observer can hear and f{s} is the frequency the source emits, Vs is the velocity of the moving source, and v is the speed of sound.
The equation for a source moving towards an observer is f{o} = f{s} x (1 (Vs÷v))
f{o} = 395 Hz x (1  (12 m/s ÷ 343 m/s))
f{o} = 381.180758 Hz
To compute for the beat frequency, let's assume that f{beat} is the beat frequency:
f{beat} = f{a}  f{b}; where f{a} is the larger of the two.
f{beat} = f{a}  f{b}
f{beat} = 395 Hz  381.180758 Hz