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Phase difference equation.

  1. Apr 28, 2003 #1
    Hi i posted here before and got great help, and I ran into another problem maybe you can help me this is the question.

    Two loud speakers are placed on a wall 2.00 m apart. A listener stands 3.00m from the wall direclty infront of one of the speakers. A single ocillator is driving the speakers in phase at a frequency of 300Hz.
    A. What is the phase difference between tehe two waves when they reach the observer.
    B. What is the frequency closest to 300 hz. to which the oscillator may be adjusted such that the observer hears minimal sound?

    This is what I did..

    r1= square root of (3.00)^2 = 3
    r2= square root of ((3.00)^2 + (2.00)^2) = 3.61

    The difference is = r2-r1 = .61

    Now this is where i got stuck I wasnt sure if i had to use .61/2pie to find the phase difference.
    Also what forumla should I use for part B.
    Thank you in advance.
     
  2. jcsd
  3. Apr 28, 2003 #2
    Hi Jenn_Lane2001,
    I think you got the difference in distance correctly. Next you should find out what the wavelength is for a 300Hz sonic wave.
     
  4. Apr 28, 2003 #3
    R u sure..?

    I didnt think you need to know the wavelength. more the velocity but that is what i wasnt sure calculating. I am not sure of the equation to use. Thanks
     
  5. Apr 28, 2003 #4
    [lamb] = v/f,
    where
    [lamb] is wavelength
    v is velocity
    f is frequency
    I think, for sonic waves, v = 340 m/s or something.
     
  6. Apr 28, 2003 #5
    Oh ok so sonic is the speed of sound

    the speed of sound is 343v and then you will divide that by frequency which is 300 hz which equals wavelength= 1.143. now that i have the wavelenght where would i put taht into use as to determine the answers. Thanks alot for the help.
     
  7. Apr 29, 2003 #6
    Right. Since the progress in phase is 2[pi] for each wavelength travelled, your phase difference is
    [psi] = .61m / 1.143m * 2[pi].
    That's the answer to A.

    Part B is more difficult. You know two waves cancel out each other when [psi] = [pi]. But that's only true if amplitudes are equal. You know the amplitude decreases (sound becomes softer) as you move away from a speaker. So the speaker that is more distant from the observer has to be turned up a bit if you want waves to cancel out. Anyway, as no info about amplitudes is given in the problem, it will probably do to solve for [psi] = [pi].
     
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