Two speakers are driven by the same oscillator with frequency of 172 Hz. They are located 3.72 m apart on a vertical pole. A man really far away walks straight toward the lower speaker in a direction perpendicular to the pole.

How far is he from the pole at the first moment he hears a minimum in sound intensity (short of infinity) if the speed on sound is 330 m/s?

1 answer

  1. First compute the sound wavelength,
    L = Vsound/frequency = 330/172 = 1.919 m

    Require that the difference between the distances to the speakers be one-half wavelength, so that destructive interference occurs.

    If R is the distance from observer to lower speaker and d is the speaker separation,
    sqrt(R^2 + d^2) -R = L/2

    Use the fact that d/R <<1, so that
    sqrt (R^2 + d^2) = R sqrt(1 + (d/R)^2]
    = R (1 + (1/2)(d/R)2 + ...)

    sqrt(R^2 + d^2) -R = (1/2) d^2/R = L/2
    R = d^2/L = 7.2 m

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