B. Suppose the speakers are reconnected so that the 472-Hz sounds they emit are
ID: 1325713 • Letter: B
Question
B. Suppose the speakers are reconnected so that the 472-Hz sounds they emit are exactly out of phase.
At what positions are the intensity maximum and minimum now? Assume
xmidpoint=0.
Two loudspeakers are placed 3.00 m apart, as shown in the figure below. They emit 472-Hz sounds, in phase. A microphone is placed 3.20 m distant from a point midway between the two speakers, where an intensity maximum is recorded. How far must the microphone be moved to the right to find the first intensity minimum? B. Suppose the speakers are reconnected so that the 472-Hz sounds they emit are exactly out of phase. At what positions are the intensity maximum and minimum now? Assume xmidpoint=0.Explanation / Answer
lambda = 343 / 472 = 0.727 m
The difference in paths s1 and s2 must yield
S2 - S1 = 1/2 * lambda
S2 = sqrt [(L/2 + x)2 + (d2)]
S1 = sqrt [(L/2 - x)2 + (d2)]
1/2 * lambda = [sqrt [(L/2 + x)2 + (d2)]] - [sqrt [(L/2 - x)2 + (d2)]]
(4 L2 - lambda2) x2 = lambda2L2 / 4 + lambda2 d2 - lambda4/16
L = 3.00 m, d = 3.2 m, lambda = 0.727
(4 32 - 0.7272) x2 = 0.7272 * 32 / 4 + 0.7272 ?* 3.22 - lambda2/16
x = 0.431 m
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The position of maxima and minima will be interchanged
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