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Destructive interference and constructive interference

  1. Dec 10, 2014 #1
    • Moved from general physics forum
    Two loudspeakers are placed facing each 4.0 m apart. The speakers emit sound in phase with a frequency of 266 Hz. An observer at the center (2.0 m from each speaker along the line joining them) experiences constructive interference. The speed of sound is 343 m/s

    The lowest frequency at which destructive interference could occur is if 1/2 a wavelength = 1.5 m (the difference between the two)

    The speed of sound can be found with the following formula
    v = 331 m/s + (0.6 m/s/C)•T
    So at 20C you get a speed of 343 m/s.

    We know that (1/2)*lambda = 1.5 m, so lambda = 3.0m
    The wavelength must be 3 m.

    Using the universal wave equation

    v = f*lambda
    f = v/lambda
    f = (343m/s)/(3 m)
    = 114 Hz

    The lowest freq would be 114 Hz

    The next two lowest would be a 3lambda/2 = 1.5 and 5lambda/2 = 1.5

    so lambda = 1m and lambda = 0.6m

    Those give f = 343 Hz and 572 Hz

    i don't know how to find How far toward either speaker must the observer walk to experience destructive interference?
    please help me
    Last edited: Dec 10, 2014
  2. jcsd
  3. Dec 10, 2014 #2


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    Staff: Mentor

    How far what is? (What is "it"?)
  4. Dec 10, 2014 #3
    How far toward either speaker must the observer walk to experience destructive interference?
  5. Dec 10, 2014 #4
    What is the c
    What is the condition for destructive interference? What path difference will give destructive interference?
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