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Homework Help: Beat Frequency of whistle

  1. Jan 8, 2005 #1
    A Boy is walking away from a wall at a speed of 1m/s in a direction at right angles to the wall. As he walks, he blows a whistle steadily. An observer toward whom the boy is walking hearts 4 beats per second. If the speed of sound is 340 m/s, what is the frequency of the whistle?

    My original approach did not come out anywhere near right, so im just going to ask if this analysis of the situation is right.

    The boy is walking toward someone blowing a whistle. So there are two doppler effects? The one caused from the boy walking toward an observer and the sound wave reflecting off the wall, so I would calculate two doppler effects and combine them? And how can I change beat frequency to frequency, if thats possible even.
  2. jcsd
  3. Jan 8, 2005 #2


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    The observer will hear two different frequencies, as you guessed correctly : one coming directly from the whistler (doppler shifted upwards) and another from the reflection off the wall (doppler shifted downwards). If the whistle frequency is f, you can find f1 (>f) and f2(<f) in terms of f, using the doppler relation. The beat frequency is simply the difference between these two frequencies. So, f1 - f2 = 4 Hz.

    You now have 3 equations in 3 unknowns, which you can solve to find them all.
  4. Jan 8, 2005 #3
    The guy hears two whistle, one is moving toward the guy at 1m/s and the other one is moving away, now you know the frequency of two whistle, and ..........(I don't wanna do the problem for you)
  5. Jan 9, 2005 #4
    Question..this is the doppler effect equation for the source moving towards a stationary observer:

    F'= F / (1-Vsource/V)

    VSource = 1 m/s
    V= 340 m/s for sound of speed

    What goes in for F? 4? THats what I did originally but it didnt work
  6. Jan 9, 2005 #5
    Or will the two doppler effects be equal to each other and thats how you find the F?
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