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Phase Difference - Homework Help Needed!

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Two sound fonts make plain armonic waves with the same amplitude and frequency. If f=200 Hz and V=340 m/s, the phase difference in a point situated 8m away from one font and 25 m away from the other one will be...

    2. Relevant equations

    Y: Wave longitude
    Phase Difference=kr2-kr1 (Found it online and is so far the only one I know about)

    3. The attempt at a solution

    Phase Difference=k(r2-r1)

    Nut in the answer sheet it says the answer is: The waves are in phase. Please explain why?
  2. jcsd
  3. Apr 13, 2007 #2
    Here is how I would reason about the problem: Let A be the source of sound 8 meters away and B the one 25 meters away. The time it takes for you to hear A is 8/340 seconds. The time it takes for you to hear B is 25/340 seconds which is roughly 3*(8/340) seconds. Hence, the phase difference would be very small.
  4. Apr 13, 2007 #3


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    Science Advisor
    Homework Helper

    340/200 is NOT 17/100. Please correct that, ok? You have a phase difference of 100*pi (it's not really that, but you'll correct it). 2*pi is one whole wave length. A wave shifted by 2*pi in phase is the same wave again since the wave pattern repeats every 2*pi. So a shift of 2*pi is the same as no shift at all. So if your phase shift is a multiple of 2*pi then the two sounds are 'in phase'. If you get a phase shift bigger than 2*pi then just take the remainder after dividing by 2*pi to get the phase shift.
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