Two small loudspeakers are separated by a distance of d = 5.0 cm, as shown in th
ID: 1539427 • Letter: T
Question
Two small loudspeakers are separated by a distance of d = 5.0 cm, as shown in the figure below. The speakers are driven in phase with a sine wave signal of frequency 18 kHz. A small microphone is placed a distance L = 2.00 m away from the speakers on the axis running through the middle of the two speakers, and the microphone is then moved perpendicular to the axis. Where does the microphone record the first minimum and the first maximum of the interference pattern from the speakers? The speed of sound in air is 343/m/s.Explanation / Answer
wavelength lambda = v/f
the microphones moves a distance y towards left
distance of microphone from left speaker
r1 = sqrt(L^2 + (d/2-y)^2)
distance of microphone from right speaker
r2 = sqrt(L^2 + (d/2+y)^2)
path difference = r2 - r1
for 1 st minima path difference = lambda/2
sqrt(L^2 + (d/2+y)^2) - sqrt(L^2+(d/2-y)^2) = 0.5*(v/f)
sqrt(2^2 + (0.05/2+y)^2) - sqrt(2^2+(0.05/2-y)^2) = 0.5*(343/18000)
y = 38.8 cm
for 1 st minima path difference = lambda
sqrt(L^2 + (d/2+y)^2) - sqrt(L^2+(d/2-y)^2) = (v/f)
sqrt(2^2 + (0.05/2+y)^2) - sqrt(2^2+(0.05/2-y)^2) = (343/18000)
y = 82.45 cm
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