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Interference - Frequency

  1. Jun 22, 2010 #1
    Interference -- Frequency

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

    Two loudspeakers at an outdoor rock concert are located 3.5 meters apart. You are standing 16.1 meters from one of the speakers and 19 from the other. During a sound check, the technician sends the exact same frequency to both speakers while you listen. The technician starts at 20Hz and slowly increases it to 30,000Hz.
    --------------------------------------------------------------------------------
    a) What is the lowest frequency where you will hear a minimum signal ?
    f = Hz
    --------------------------------------------------------------------------------
    b) What is the second lowest frequency where you will hear a minimum signal ?
    f = Hz
    --------------------------------------------------------------------------------
    c) What is the lowest frequency where you will hear a maximum signal ?
    f = Hz
    --------------------------------------------------------------------------------
    d) What is the second lowest frequency where you will hear a maximum signal ?
    f = Hz


    2. Relevant equations

    [tex]\omega[/tex]=2[tex]\pi[/tex]f
    v=[tex]\sqrt{T/\mu}[/tex]

    3. The attempt at a solution

    I'm not sure where to start!
     
  2. jcsd
  3. Jun 22, 2010 #2
    Re: Interference -- Frequency

    What is the general condition at which the net amplitude is max/min when two waves of the same frequency (and the same vibrating direction) superimpose? Hint: Something about phase difference.
     
  4. Jun 22, 2010 #3
    Re: Interference -- Frequency

    When there is no phase difference or the phase difference is divisible by pi?
     
  5. Jun 22, 2010 #4
    Re: Interference -- Frequency

    Okay so what I have done so far is found the path length difference, ([tex]\Delta[/tex]L = L1 - L2).
    I know that [tex]\Delta[/tex]L/[tex]\lambda[/tex] = [tex]\Phi[/tex] / 2[tex]\Pi[/tex], but is this the right direction?

    Where do I go from here?
     
  6. Jun 23, 2010 #5
    Re: Interference -- Frequency

    The problem says nothing about the initial phase difference, so I assume that the initial signals coming out of the loudspeakers are in phase.
    [tex]\Phi[/tex] is the phase difference, right? So you're on the right track ;)
    1 - Now what would [tex]\Phi[/tex] be if it's maximum? And if it's minimum?
    2 - Let's take the sound speed v=340m/s. You have [tex]\Delta L[/tex]. So from the above equation you've just pointed out:[tex]f = \frac{v}{\lambda} = \frac{\Phi}{2\pi \Delta L}v[/tex]
    Subtitute [tex]\Phi[/tex] for each case (max/min), you will get f.
     
  7. Jun 23, 2010 #6
    Re: Interference -- Frequency

    Got it! Thank you so so much!
     
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