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Homework Help: Doppler effect car horn question

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A policeman with a very good ear and a good understanding of the Doppler effect stands on the shoulder of a freeway assisting a crew in a 40-mph work zone. He notices a car approaching that is honking its horn. As the car gets closer, the policeman hears the sound of the horn as a distinct B4 tone (494 Hz). The instant the car passes by, he hears the sound as a distinct A4 tone (440 Hz). He immediately jumps on his motorcycle, stops the car, and gives the motorist a speeding ticket. Explain his reasoning.

    3. The attempt at a solution
    f0 = f (V(sound) / (V(sound) - V(source)))
    since the car is moving toward the policemen it is negative
    494Hz = 440Hz * (343 v/s / (343 - V))
    and V is 37.166 m/s...
    which is obviously not..
    i'm not sure where i went wrong probably doppler effect..

    thank you
  2. jcsd
  3. Jul 1, 2011 #2
    Well, I haven't really gone through any of the calculations in your problem... but... just off an initial glance I saw you came up with:

    V = 37.166 m/s

    Which is a decent answer. And you need to prove that this motorist was speeding past the limit of 40 mph... perhaps a conversion? Doing so would put the motorist's speed at ~80mph which is certainly justifiable for the policeman to pull the motorist over. Or maybe I'm misinterpreting what you're looking for here.
    Last edited: Jul 1, 2011
  4. Jul 1, 2011 #3
    i converted to mph from m/s
    and i got 83 mph which still is the wrong answer
    i think it's the formula
  5. Jul 1, 2011 #4
    f(police) = v / (v + v(source)) *f(source)

    440 = 494 (767 / (767 + v))

    v= 94 mph

    does this make sense?
  6. Jul 1, 2011 #5


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    Actually, f(source) = 440 Hz & f(observer) = 494 Hz, so you'll get -94 mph.
  7. Jul 1, 2011 #6
    The correct answer is 83 mph.
  8. Jul 1, 2011 #7


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    Yeah, I didn't look up the Doppler effect formula, but I thought about it for a few minutes and came up with the following for a stationary observer and moving source:

    [tex] f = f_0\left[\frac{1}{1 + v/c_s}\right] [/tex]

    where cs is the sound speed, v is the source speed relative to the observer, and I use the convention that:
    v > 0 if the source moves away from the observer and,
    v < 0 if the source moves toward the observer.

    It makes sense at least to the extent that f > f0 for v < 0 and vice versa.

    Using this formula, with a sound speed of 343.2 m/s (taken from Wikipedia for dry air at 20 C) I get a result of:

    v = -83.9 mi/h
    Last edited: Jul 1, 2011
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