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Sound In Air Problem

  1. Feb 22, 2005 #1
    'An ambulance with a siren emitting a whine at 1200 Hz is catching up to a cyclist pedalling a bike in the same direction at 2.80 m/s. Before being passed, the cyclist hears a freqency of 1261 Hz. There is a wind of 10.0 m/s helping the cyclist along. The speed of sound in air is 343 m/s. Calculate the speed of the anbulance.'

    Now i know the ambulance is the source and the cyclist is the listener. So i use the equation:

    FL = FS (v-vL/v-vS)

    FL = 1261 Hz
    Fs = 1200 Hz
    v = 343 + 10
    vL = vs-2.8

    and solve for vS, but i cannot get the right answer. My answer is always turning out to be 297 m/s which is definitly wrong. Can someone tell me where I'm going wrong? any help would be appriciated. Thanks
  2. jcsd
  3. Feb 23, 2005 #2
    hmm I am having a hard time reading you notation for nonrelativistic Doppler would be:
    f_{obs} = f_{source}\frac{1 \pm \frac{v_{obs}}{v_{sound}}}{1 \mp \frac{v_{source}}{v_{sound}}}
    The top signs are uesed if approaching witch is the case so:
    f_{obs} = f_{source}\frac{1 + \frac{v_{obs}}{v_{sound}}}{1 - \frac{v_{source}}{v_{sound}}}
    Now to get rid of the wind. Do not add the wind velocity to the seed of sound, after all it is only going faster in one direction. Rather try subtracting off the wind velocity from the both the ambulance and the biker. Then use the Doppler equation and then add the wind velocity back on.

    By subtracting the winds velocity off the ambulance and biker velocities the air is consider to be still and only the biker and ambulance is moving. This is why it is then valid to use the above Doppler equation with airs normal vsound.

    Does that make sense?
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