Question

The siren of a fire engine that is driving northward at 30.0 m/s emits a sound of frequency 2000 Hz.A truck in front of this fire engine is moving northward at 20.0 m/s.What is the frequency of the siren's sound that the fire engine's driver hears reflected from the back of the truck?

Answer 1

Hey Megan,

doppler effect: apparent freq (n') of actual freq (n)

n ' = n [v-vo] / [v-vs]
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in this formula, all v, (sound spped), vo (observer) and source (vs) are necesarily moving in the same direction (say + x). if one is not in + x, then is will be resolved therein.
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v=343 m/s (+x), vs = 30 m/s (+x), v0 = 20 m/s (+x)
all in same direction
the apparent freq to trucker
n ' = 2000 [343 - 20] / [343 - 30] = 2063.90 Hz
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this n' is reflected and trucker becomes a virtual source
vs = 20 (+x) , vo (engine) = 30 (+x)
but sound is now directed from truck to engine, so resolved as v = - 343 (+x)
n '' (apparent to engine) = n ' [- v - vo] / [- v - vs]
n '' = n ' [v+vo] / [ v+vs]
n '' = 2063.90 [343 +30] / [343 + 20]
n '' = 2120.75 Hz
this is what fire engine driver will hear as reflected freq

Source: http://answers.yahoo.com/question/index?qid=20070503214006AA3oFMT

Hope that helps,

Good luck =)

Answer 2

The apparent frequency of sound due to doppler effect is given by:
\[f_1=\frac{(relative. velocity. of. sound. and. listener)}{(relative. velocity. of. sound. and. source)}f \] where \(f\) is original frequency.