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Phase difference of speakers=?

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Two loudspeakers emit sound waves along the x-axis. a listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x=0.50m. If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching anohter maximum when speaker 1 is as x=0.90m

    A) (solved already) What is the frequency of he sound? Assume V(sound)=340 m/s

    B) What is the phase difference between the speakers??

    2. Relevant equations

    v=λf

    Δ∅= [(2∏Δd)/λ] + Δ∅o

    3. The attempt at a solution

    Well for A) I used v=λf. With a λ of 0.40m, I got the right answer of a frequency of 850 Hz.

    For B) I am unsure as to the initial phase difference, Δ∅o.



    For the record the answer is -(∏/2)
     
    Last edited: Jan 25, 2013
  2. jcsd
  3. Jan 25, 2013 #2

    Simon Bridge

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    If you have trouble with remembering what the equations are saying, you need to go back to the physics.

    You know the relative phase of the waves where the listener is standing, for example, and you also know the equation for the wave at any particular time. So work it backwards.
     
  4. Jan 26, 2013 #3

    rude man

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    You can let speaker 1 output = cos(ωt), then speaker 2 output = cos(ωt - kx + ψ) at position of speaker 1, where x = distance between speakers and k = 2π/λ = ω/v, v = 340 m/s.

    The argument of cos() must then be the same at distance a = 0.5m, and also (the same + 2π) at distance b = 0.9m. So just solve for ω (which you already & correctly did) and also for ψ. Reduce the computed ψ in magnitude by integer multiples of 2π as needed to drive the value to -π < ψ < +π & you will get the posted answer.
     
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