Two loudspeakers are placed on a stage 5.4 m apart and are radiating the same fr

Question

Two loudspeakers are placed on a stage 5.4 m apart and are radiating the same frequency, in phase. An observer walks along a line parallel to the loudspeakers and 12.7 m from them. Starting from the central point of the line (equidistant from the 2 loudspeakers) he observes that as he walks, the sound intensity first decreases and then increase to a maximum again when he has travelled 1.4 m.

A.Explain this observation.

B.Calculate the frequency of the sound, given that the speed of sound is 343 m/s.

Please provide a full detailed explanation for the solution (especially for part B).

Explanation / Answer

A. The path length difference ?d between the two speakers and the observer starts = 0. As the observer moves, ?d becomes 1/2 wavelength where observer hears a minimum, then 1 wavelength where observer hears a maximum.

B. Wavelength ? = path length diff. = L1-L2 = sqrt(12.7^2+(2.7+1.4)^2) - sqrt(12.7^2+(2.7-1.4)^2) = 0.579049 m.
Frequency = v/? = 592.35 Hz.

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