Two loudspeakers are placed side by side a distance = 4.00 apart. A listener obs

Question

Two loudspeakers are placed side by side a distance = 4.00 apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is = 5.00 . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance = 60.0 from its original position, the listener, who is not moving, observes destructive interference for the first time. (((((Find v, the speed of the sound waves in air. Express your answer in meters per second to three significant figures))))))

Explanation / Answer

The path length difference from the two speakers to the observer L = sqrt(3^2+5^2) - 5 = 0.83095 m. For destructive interference, L should be equal to (an integer i + 0.5, e.g., 0.5, 1.5, etc.) times the wavelength ?, thus ? = L/(i+0.5). Try the series of L/(i+0.5) values until ? is closest to c/500, then compute the frequency as c/L. I get 613.75 Hz (i = 1) when c = 340 m/s.

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