wo in-phase speakers are 2.80 m apart and emit an unknown single frequency of so

Question

wo in-phase speakers are 2.80 m apart and emit an unknown single frequency of sound in all directions. At a point 2.00 m away from one speaker on the line AB between the two speakers a sound detector records a sound intensity maximum. The detector is then moved perpendicularly away from the line AB. The first time the sound intensity is again measured to be a maximum occurs at y= 1.01 away from the line AB.

What is the frequency of the sound emitted by the speakers? (NOTE: The path difference decreases as the detector moves away from the line AB; therefore you can't assume that the path difference initially is one wavelength. You must leave n as a variable and algebraically eliminate it.)

Explanation / Answer

Try this one first part tell you 2-0.8=(n)* wavelength the second distance,from a to that point is =2.24058m from b to that point was 1.2884487m so the distance difference is 0.9521m 0.9521=(n-1) * wavelength wavelength is equal to 0.24789071m calculate the frequency base on V= wavelength* frequency V= 1384Hz (the wavelength is a bit strange and the speed of sound varies a lot base on temperature , I used 343.26m/s )

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