Doppler effect ambulance problem

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The discussion revolves around solving a Doppler effect problem involving an ambulance and a cyclist. The ambulance emits a siren at 1300 Hz, and after passing the cyclist, the cyclist hears a frequency of 1280 Hz while traveling at 2.44 m/s. The speed of sound in air is given as 343 m/s. The Doppler equation is set up correctly, but there are concerns about algebraic interpretation and the signs used in the calculations. The key to solving the problem lies in accurately applying the Doppler effect formulas for moving sources and observers.
mikejones2000
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Question: An ambulance with a siren emitting a whine at 1300 Hz overtakes and passes a cyclist pedaling a bike at 2.44 m/s. After being passed, the cyclist hears a frequency of 1280 Hz. Use 343 m/s for the speed of sound in air. How fast is the ambulance moving?

i set up the doppler equation with as follows:

1300 hz =1280 hz((343+(2.44/343))/(343+(vs/343))), I am pretty confidant the signs I choose are correct but must be making a mistake in either interpreting the data or algebra, any help is greatly appreciated..
 
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If you are not sure about the positive direction, I can tell you that the positive direction is aimed towards each other relative to the air.
 
mikejones2000 said:
Question: An ambulance with a siren emitting a whine at 1300 Hz overtakes and passes a cyclist pedaling a bike at 2.44 m/s. After being passed, the cyclist hears a frequency of 1280 Hz. Use 343 m/s for the speed of sound in air. How fast is the ambulance moving?

when both the source and the cyclist move (in this case), then the Dopplereffect is given by:

f_{cyclist} = f_{ambulance}\frac {(v + v_{cyclist})} {(v + v_{ambulance})}

but you can also say that the cyclist stands still and the ambulance passes with a relative velocity v_{rel}. Then use:

f_{cyclist} = f_{ambulance}\frac {v} {(v + v_{rel,amb})} and then add the velocity of the cyclist.
 
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