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Doppler Effect & Beat Frequency

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Horseshoe bats use the Doppler Effect to determine their
    location. A horseshoe bat flies toward a wall at a speed of 15 m/s while emitting a sound of
    frequency 35 kHz. What is the beat frequency between the emission frequency and the
    echo? Assume that the air temperature is 20C.


    2. Relevant equations
    Doppler effect
    f0=[(1-v0/v)/(1-vs/v)]fs
    Beat Frequency
    fbeat = Δf


    3. The attempt at a solution

    f0=[(1-(-343m/s)/(15m/s))/(1-(343m/s)/(15m/s))]*35kHz
    f0=-38.20121951kHz

    fbeat = Δf = fs-f0
    fbeat = 35kHz-(-38.20121951kHz)
    fbeat = 73.2kHz

    I'm not sure that I have solved this correctly. It seems strange to me that the echo has more than double the frequency of the original sound. edit; or should I say, that the beat frequency is more than double that of the original sound.
     
    Last edited: Feb 25, 2009
  2. jcsd
  3. Feb 25, 2009 #2
    Also, I assumed v0 was negative because the sound is traveling in the opposite direction of the source at that point. Is this correct?
     
  4. Mar 12, 2010 #3
    Well, the answering 'tutor's' fomulas may be approximately OK, especially if lifted from some random handbook, but his understanding of beat frequency is rubbish.

    Instead mechanically using the signs that come out of these arbitrary formulas, you should find the difference between absoulte values (modulus), ie compare both frequencies without the minus sign. When you do this you will have computed approx 3kHz, which could be a more reasonable answer.

    Frequency here is a scalar, not a vector. You complare them just like you compare the lengths of two sticks: you subtract from the longer one the length of the shorter, no matter in which direction the sticks may be pointing.
     
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