Question

Two in-phase speakers are 2.80 mapart and emit an unknown single frequency of sound in alldirections. At a point 2.00 m away from one speaker on the line ABbetween the two speakers a sound detector records a sound intensitymaximum. The detector is then moved perpendicularly away from theline AB. The first time the sound intensity is again measured to bea 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 theline AB; therefore you can't assume that the path differenceinitially is one wavelength. You must leave n as a variable andalgebraically eliminate it.)

Attempt at answer:

The difference between the 'detector' and points 'A' and 'B'are

S_AD √[(2.00 m)²+(1.01 m)^2] =2.240557966 m

S_BD √[(0.80 m)²+(1.01 m)^2] =1.28844868 m

Beyond this though I'm stuck. I've already tried to divide thespeed of light (v = 343 m/s^s) by (S_AD -S_BD) = 360 Hz

I know this is wrong as the assignment is online.

All I need is a good nudge in the right direction.

Thanks again to anyone who can contribute!