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Sound, phase difference and number of minima

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data
    http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

    number 5b

    The speed of sound is 330ms-1

    The frequency of sound from S1 and S2 is increased. Determine the number of minima that will be detected at M as the frequency is increased from 1k to 4k Hz.


    2. Relevant equations
    v=fλ
    (n+1)λ


    3. The attempt at a solution
    Okay, first have to find the path difference, which first we have to find the hypotenuse, 128cm.
    and then, 128-100=28 cm, that is the path difference
    330=(1000)λ
    330=(4000)λ
    λ is found to be 33 cm and 8.25 cm respectively.

    After that, Im not too sure what to do. ..
     
  2. jcsd
  3. May 26, 2012 #2

    Doc Al

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    Staff: Mentor

    How many wavelengths does that path difference correspond to? As the wavelength is slowly decreased, at what points will there be destructive interference?
     
  4. May 26, 2012 #3
    28=1/2λ, so λ is 56cm.


    So its like 5/2λ=28, 7/2λ=28 and so on?
    So path difference is constant? for all frequencies?
    minima will be detected at 18.7 and 11.2 cm. So 2 minima are detected
     
  5. May 26, 2012 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You already found the range of wavelengths. Now make use of that by figuring out how many wavelengths fit into that path difference.

    If the path difference ends up being a whole number of wavelengths, what does that mean? Under what conditions will there be destructive interference?
     
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