Use the Doppler effect to find the beat frequency

http://www.flickr.com/photos/53399890@N05/6767463615/in/photostream

Formula:

f ' = f ( v (+/-) vl )/(v (+/ - ) vs )

I don't know why there's a "2vt " .

I think you have to chose the engineer as your reference point. v is the speed of sound. The horn is the source f. The engineer is stationary.



I just need help with part a.

For a person near the wall, the frequency of incoming wave is
(Observer is at rest and the source is approaching him)
n’ = C n /( C- vt)

For the engineer, he now hears a sound wave of wave length = C/n’.
The source is at rest and the observer is moving toward the source.

n” = (C + vt)n’ / C

n” = ( C +vt) {C n /( C- vt) } / C = ( C +vt) n /( C- vt)

Beat frequency

n”- n = ( C +vt) n /( C- vt) – n = n { C +vt - C + vt} / ( C- vt)

= 2vt / ( C- vt)

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The engineer is moving at the same pace as the whistle. So he/she hears the original frequency AS THOUGH stationary. But...
The wall ahead "hears" a frequency that has increased in pitch, and reflects THAT frequency back towards the engineer. However, as the engineer is also moving towards the wall, what he hears back is again increased in pitch by the same increase! That's why the "2".

The stationary man alongside the back of the train hears a direct sound from the whistle that is low in pitch, but the wall "hears" it at the increased pitch as above and reflects that, and because he is stationary that is the reflected pitch he hears.
The beat frequency, of course, is the higher frequency heard minus the lower frequency heard.