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Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator, whose frequency range is 1399 Hz to 1812 Hz. A point P is located 5.1 m from oneloudspeaker and 3.6 m from the other. The speed of sound is 344 m/s. The frequency produced by the oscillator, for which constructive interference of sound occurs atpoint P, in SI units, is closest to:
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Answer #1
For constructive interference, n*lambda = 5.1m-3.6m = 1.5m, where n = 0, 1, 2,...
So lambda = 1.5/n = 3/2n. And f = v/lambda = 344n/3*2 = 229.33n.
When n = 7, f = 1605Hz.
answered by: Me!
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Answer #2
Path difference between two waves reaching point PΔx=5.1-3.6=1.50mFor constructive interference tooccurΔx=n λ=nv/ fWherespeed of sound in airv=344m/sFrequency of sound waves=fThen1.50=n * 344 / ff=n* 344 / 1.50=n * 229.33Since minimum frequency is 1399 Hz, value of n should be so chosen that f 1399 HzHenceforn=7,f=7 * 229.33=1605Hz
answered by: jward1970
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