Two in-phase speakers are 2.80 m apart and emit an unknown single frequency of s

Question

Two 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 as shown. The first time the sound intensity is again measured to be a maximum occurs at y = 1.5 m away from the line AB.

What is the frequency of the sound emitted by the speakers in Hz?

(NOTE: The path difference decreases as the detector moves away from the line AB; therefore you can NOT assume that the path difference initially is one wavelength. You must leave "n" as a variable and algebraically eliminate it.)

Explanation / Answer

Let L be wavelength ,
then 2 - 0.8 = n * L = 1.2
At y = 1.5 ,
R1 = distance from source A = sqrt ( 2^2 + 1.5^2) = 2.5 m
R2 = distance from source B = sqrt ( 0.8^2 + 1.5^2) = 1.7 m
so R1 - R2 = (n-1)*L
2.5 - 1.7 = 0.8 = (n-1) *L
0.8 = n*L - L = 1.2 - L
L = 0.4 m
Speed of sound in air = 343 m/s
so frequency = speed/wavelength = 343/0.4 = 857.5 Hz is the answer

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