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Question

A fire engine has sirens of 500 Hz. What frequency is heard from a stationary observer A) when the engine moves towards her at 12m/s. B) when it moves away from her at 12m/s. (speed of sound is 343m/s).

Answer 1

The apparent (what appears to be) frequency ( n' ) in dopper effect is related to actual frequency ( n ) by

n' = n [v - vo] / [v - vs]

here all v, v0, vs are taken positive in one direction say + x-axis. Also, it is ammuned that wind speed is not impacting the apparant frequency.

given
n = 500 Hz
v = velocity of sound = 343 m/s
vs = velocity of source (engine) = 12 m/s
vo = velocity of observer

case A: n = 500 Hz, vo= 0 stationary, vs = 12 m/s (towards observer) v = 343 m/s (towards observer)

n' = 500 * [343 - 0] / [343 - 12]
n' = 500 * [343] / [331]
n' = 518.126 Hz

the freqency to stationary observer will appear incresingly faster as the distance between them reduces.

freq. shift = n' - n = 518.126 - 500 = 18.126 Hz

base B: when the engine passes her by the direction of v is still towards the observer but vs is not directed away from from the observer ( - ve x). So relative to +v the vs = -12 m/s

n = 500 Hz, vo= 0 , vs = - 12 m/s (towards observer) v = 343 m/s (towards observer)

n' = 500 * [343 - 0] / [343 + 12]
n' = 500 * [343] / [355]
n' = 483.098 Hz

the farthering source will appear to be having lower apparent frequency.

freq. shift = n' - n = 483.098 - 500 = - 16.902 Hz

If wind speed affect the n' then it can be incorporated in the formula by changing v to v+-vw (w - wind)

Answer 2

Find the right formula for moving source for Doppler shift.

fobs = fsource/(1-vsource/vsound) for part a) I think and

fobs = fsource/(1+vsource/vsound) for part b)

Part a) will be higher, of course, by 12/343 roughly
Part b) will be lower by 12/343