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Question on Waves and Interference

  1. Feb 7, 2007 #1
    Please help! I've been stuck on this for a very very long time :confused: .

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

    Two loudspeakers, separated by a distance of d1 = 1.60 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same at the position of a listener, who is d2 = 3.90 m directly in front of one of the speakers.

    See Diagram: http://www.webassign.net/hrw/18_32alt.gif

    (a) For what frequencies in the audible range (20-20,000 Hz) does the listener hear a minimum signal? (Give the lowest, second lowest and highest possible frequencies in the audible range.)

    I JUST FIGURED THIS ONE OUT:(b) For what frequencies is the signal a maximum? (Give the lowest, second lowest and highest possible frequencies in the audible range.)


    KEY: d = delta, f = frequency, v = velocity of air, dL = path length difference, m = nodes I assume?

    If not, what's m in the equation (seen below)?

    2. Relevant equations

    dL = (m+.5)(v/f)
    v = 343 m/s

    3. The attempt at a solution

    d3 = squareroot(1.6^2 + 3.9^2) = 4.22 m

    I was able to derive this: f = (m+.5)(v/dL) from the dL equation given.

    Phase difference at the person is = 2pi(d3-d2)/lamda
    For minimum intensity at the listener, phasediff = (2m+1)pi
    and lamda = 2pi(d3-d2)/phasediff

    freq = v/lamda = [(2m+1)v]/[2(d3-d2)]
    and that leaves me with (2m+1)[v/(2(d3-d2))]

    [v/(2(d3-d2))] = frequency that I get
    For me, it's 543.67Hz.

    And 20000Hz/543.67Hz = 37.3
    2m+1 = 37

    So m needs to be between 1 and 18 I believe...

    ****

    I figured out part b.:rofl:

    lamda = (1/n)(d3-d2)
    so f = v/lamda = nv/(d3-d2) = n(1088.89Hz)

    20,000Hz/1088.89Hz = 18.37

    n = 1 through 18

    Plug in 1, 2, and 18 for frequencies...

    ****

    But I'm still wrong for part a...:mad: :grumpy: :frown: :bugeye: :surprised :devil: :yuck: :cry: :zzz:
     
    Last edited: Feb 7, 2007
  2. jcsd
  3. Feb 7, 2007 #2
    Now here is a prof I like right away. One of my favorite ugrad courses "physics of hi-fi"

    The path length difference between the right and left speakers is
    sqrt(d2^2+d1^2)-d2=Diff. When that difference is a complete wavelength, the two add to produce a maxima, when it is a 1/2 wavelength, you get destructive interference. So the first maxima would occur at a wavelength of .32m and integer multiples of this frequency. Help at all?

    M by the way is the wave number, it starts at zero to capture the first minima, when it is 1, you get the first max
     
    Last edited: Feb 7, 2007
  4. Feb 7, 2007 #3
    You lost me in the last sentence. I understand your response until the sentence ending with "destructive interference." How did you come up with .32m for the wavelength?

    I played around a bit more and here's what I've figured out since posting. Also added this to the original post:

    Phase difference at the person is = 2pi(d3-d2)/lamda
    For minimum intensity at the listener, phasediff = (2m+1)pi
    and lamda = 2pi(d3-d2)/phasediff

    freq = v/lamda = [(2m+1)v]/[2(d3-d2)]
    and that leaves me with (2m+1)[v/(2(d3-d2))]

    [v/(2(d3-d2))] = frequency that I get
    For me, it's 543.67Hz.

    And 20000Hz/543.67Hz = 37.3
    2m+1 = 37

    So m needs to be between 1 and 18 I believe?

    I'm pretty sure I'm close...but somehow the answers I'm getting are still incorrect.
     
  5. Feb 7, 2007 #4
    the .32 is from your math--hypotenuse minus D2. Thats the path length difference, so for constructive interference to occur, the wavelength is equal to that distance or some integer multiple numbers of wavelengths. Forget the profs eqn for a minute as that doesn't add to really understanding the problem. Think about the waves from the two speakers. They can completely add, cancel or mostly do a bit of one or the other depending on their arrival times. so the lowest frequency where they cancel, the diff in pathlengths=1 wavelength=C/f. Help at all?
     
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