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Doppler effect ambulance problem

  1. Aug 5, 2006 #1
    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..
  2. jcsd
  3. Aug 5, 2006 #2
    If you are not sure about the positive direction, I can tell you that the positive direction is aimed towards eachother relative to the air.
  4. Aug 6, 2006 #3
    when both the source and the cyclist move (in this case), then the Dopplereffect is given by:

    [tex]f_{cyclist} = f_{ambulance}\frac {(v + v_{cyclist})} {(v + v_{ambulance})}[/tex]

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

    [tex]f_{cyclist} = f_{ambulance}\frac {v} {(v + v_{rel,amb})}[/tex] and then add the velocity of the cyclist.
    Last edited: Aug 6, 2006
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