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Beat frequency of Two in-phase loudspeakers

  1. Nov 30, 2005 #1
    4. Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of 536 Hz. A microphone is moved between the speakers along the line joining the two speakers with a constant speed of 1.60 m/s. What beat frequency is observed? The speed of sound in air is 343 m/s.

    Could anyone help me? I have no idea at all to solve this question since I am quite blur with the concept of beat and beat frequency also. So, could anyone explain about beat and beat frequency to me as well? Thanks.:smile:
     
  2. jcsd
  3. Nov 30, 2005 #2

    Tide

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    The beat frequency will be the sum of the individual frequencies picked up by the microphone but the frequencies will be doppler shifted by an amount dependent on the relative motion of the source and receiver.
     
    Last edited: Dec 1, 2005
  4. Nov 30, 2005 #3
    Thanks. Somebody teach me by using another method.

    First, find out the wavelength and divide by 2. The value is equal to the distance between 2 successive crests. So. The time period to repeat this cycle(from crest to crest) is equal to half of the wavelength divide by 1.6
    And the beat frequency is equal to 1 / time period.

    How if the 2 speakers is not in phase? If so, can this method be applied again?
     
  5. Dec 1, 2005 #4

    Tide

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    According to the problem stated, you need to take into account the fact that the microphone is moving. See my previous post - which I changed to make more accurate.
     
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