A pair of speakers separated by 0.700 m is driven by the same oscillator at a fr

Question

A pair of speakers separated by 0.700 m is driven by the same oscillator at a frequency of 655 Hz. An observer, originally positioned at one of the speakers, begins to walk along a line perpendicular to the line joining the speakers. (Assume the speed of sound is 345 m/s.)

(a) How far must the observer walk before reaching a relative maximum in intensity?


(b) How far will the observer be from the speaker when the first relative minimum is detected in the intensity?

Explanation / Answer

wavelength = 345/675 = 0.5111 m If L is the hypoteneuse of thetriangle and D is the distance between the speakers andx is the distance from the person to the speaker closer tothem then m ? = L - x for constructive interference Or m? + x = L If we square both sides m2?2 + x2 + 2m?x = L2 Using pythag theorem L2 = D2 + x2 m2?2 + x2 + 2m?x = D2 + x2 m2?2 + 2m?x = D2 m2 *0.51112 + 2 * m * 0.5111 *x = 0.7002 0.26122m2 + 1.0222 m x = 0.49 x = ( 0.49 - 0.26122m2 ) / 1.0222 m m is any integer, but x must be a positive number. So we cansee that 0.49 - 0.26122 m2 > 0 1.75 > m so wemust have m = 1 x = ( 0.49 - 0.26122 * 12 ) / 1.0222 * 1 = = 0.223 meters

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