# Physics

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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?

• Physics -

Doppler-effect.
At source approaching
f(observed) = [v/(v-u)] •f(source),

f(observed) =
= [343/(343-12)] •395=409.3 Hz
f(beat)= f(observed)=f(source)=
=409.3-395=14.3 Hz

• Physics -

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