Two loudspeakers are placed above and below each other, as in the figure below,

Question

Two loudspeakers are placed above and below each other, as in the figure below, and driven by the same source at a frequency of 4.00 102 Hz. An observer is in front of the speakers (to the right) at point O, at the same distance from each speaker. If the speed of sound is 345 m/s, what minimum vertical distance upward should the top speaker be moved to create destructive interference at point O?
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Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m

Explanation / Answer

For destructive interference we use f = v/2d where d is the difference in distance between the speakers to the listener So d = v/2f = 345/(2*400) = 0.43125m The distance to the bottom speaker is sqrt(1.85^2 + 8^2) = 8.211m So the distance for the top speaker where destructive interference occurs is 8.211 + 0.43125 = 8.6424m This means the speaker must be moved back x amount where 8.6424 = sqrt((8 +x)^2 + 1.15^2) So 8.6424^2 = (8 + x)^2 + 1.3225 Now x^2 +16x + 64 + 1.3225 - 8.6424^2 = 0 or x^2 + 16x -9.368 = 0 So x = (-16 +-sqrt(16^2 - 4*1*(-9.368))/2 = 0.566m back b) Now if x = 5*0.566 = 2.828m then distance to speaker top = sqrt((8+2.828)^2 + 1.15^2) = 10.888 So the separation distance = 10.888 - 8.211 = 2.677m Now using fd = nv/2d => n = 2d*f/v if n is an integer (1, 3, 5, ...)then destructive interference occurs n = 2*2.677*400/345 = 6.2 not destrcutive So for constructive fc = nv/d so n = f*d/v = 400*2.667/345 = 3 so constructive occurs

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