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 two loudspeakers) he/she observes that, as he/she walks, the sound intensity first decreases and then increases to a maximum again when he/she 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.

Explanation / Answer

A) Explain this observation.

The path length difference i.e. d between the two speakers and the observer starts = 0.

As the observer moves, the path difference d becomes half of the wavelength, where observer hears a minimum,

then equal to wavelength where observer hears a maximum.

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

Wavelength = path length difference = L1-L2

L1 = sqrt(12.7^2+(2.7+1.4)^2)

L2 = sqrt(12.7^2+(2.7-1.4)^2)

then =0.579049 m.

Now since frequency = v/

and v = 343 m/s,

f = 592.35 Hz.

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