Two Speakers - Sound Maximum and Sound Minimum Problem

  1. [Solved] Two Speakers - Sound Maximum and Sound Minimum Problem

    Hi, I am having difficulty solving the following problem:

    1. The problem statement, all variables and given/known data

    Two loudspeakers 5.0 m apart are playing the same frequency. If you stand 13.0 m in front of the plane of the speakers, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 13.0 m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.

    What is the frequency of the sound? Assume a sound speed of 340 m/s.


    2. Relevant equations

    Sound Maximum:
    L1 - L2 = n[tex]\lambda[/tex]

    Sound Minimum:
    L1' - L2 = (n+[tex]\frac{1}{2}[/tex])[tex]\lambda[/tex]

    Frequency:
    f = [tex]\frac{v}{\lambda}[/tex]

    3. The attempt at a solution

    [​IMG]

    Sound Maximum:
    L1 - L2 = n[tex]\lambda[/tex]

    L2 = 13.0 m
    L1 = [tex]\sqrt{13.0^{2}+2.50^{2}}[/tex] = 13.23820229

    L[tex]_{1}[/tex] - L[tex]_{2}[/tex] = n[tex]\lambda[/tex]
    13.23820229 - 13 = n[tex]\lambda[/tex]
    n[tex]\lambda[/tex] = 0.23820229

    Sound Minimum
    L1' - L2 =(n + [tex]\frac{1}{2}[/tex])[tex]\lambda[/tex]

    L2 = 13.0 m
    L1' = [tex]\sqrt{13.0^{2}+5.0^{2}}[/tex] = 13.92838828

    Sub in n[tex]\lambda[/tex]= 0.23820229:

    L1' - L2 = (n + [tex]\frac{1}{2}[/tex])[tex]\lambda[/tex]
    13.92838828 - 13 = n[tex]\lambda[/tex] + [tex]\lambda[/tex]/2
    [tex]\lambda[/tex]/2 = 0.92838828 - 0.23820229
    [tex]\lambda[/tex] = 1.380371974

    Sub in [tex]\lambda[/tex] = 1.380371974:
    f = [tex]\frac{v}{\lambda}[/tex]
    f = [tex]\frac{340}{1.380371974}[/tex]
    f = 246.3104195 Hz

    I'm not sure if my approach is wrong or if I'm interpreting the question incorrectly. Any help would be greatly appreciated!

    Thanks.
     
    Last edited: Jan 31, 2010
  2. jcsd
  3. rl.bhat

    rl.bhat 4,435
    Homework Helper

    In the central position the two speakers are at equal distance. So the path difference is zero. In between the first and the second position, there is neither a maximum nor a minimum. So at the second position ( l1' - l2) = λ/2.
     
  4. Ooh.. no wonder. Thank you very much!
     
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