Problem 21.64 You are standing 2.5 m directly in front of one of the two loudspe

Question

Problem 21.64 You are standing 2.5 m directly in front of one of the two loudspeakers shown in the figure. They are 3.0 m apart and both are playing a 686 Hz tone in phase. Part A As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is 20 degrees C. Express your answer numerically using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas.

Explanation / Answer

Destructive interference occurs when the difference in distance from the in phase speakers is d = ?/2*(2n -1) where n = 1,2,3... If the room temp is 20oC then v = 343m/s so ? = 343/686 = 0.500m So the difference in distances d for minimum sound will be 0.500/2*(2n -1) = 0.250, 0.750, 1.25, 1.75, 2.25, 2.75, 3.25, 3.75, 4.25,4.75.... So we must find the distance such that the hypotenuse of the triangle minus the side = 0.250, 0.750, 1.25, 1.75, 2.25, 2.75, ,.... so sqrt(3^2 + (2.5+x)^2) - (2.5+x) = 0.250, 0.750, 1.25, 1.75, 2.25, 2.75, ..... Therefore 9 + 6.25 + 5x + x^2 = (0.25 + 2.5 + x)^2 = (2.75 + x)^2 = 7.5625 + 5.5x + x^2 So 0.5x = 15.25 - 7.5625 = 7.6875 So x1 = 7.6875/0.5 = 15.375m repeating for d = 0.750 Therefore 9 + 6.25 + 5x + x^2 = (0.75 + 2.5 + x)^2 = (3.25 + x)^2 = 10.5625 + 6.5x + x^2 So 1.5x = 15.25 - 10.5625 = 4.6875 So x2 = 4.6875/1.5 = 3.125m repeating for d = 1.250 Therefore 9 + 6.25 + 5x + x^2 = (1.25 + 2.5 + x)^2 = (3.75 + x)^2 = 14.0625 + 7.5x + x^2 So 2.5x = 15.25 - 14.0625 = 1.1875 So x2 = 1.1875/2.5 = 0.475m repeating for d = 1.750 Therefore 9 + 6.25 + 5x + x^2 = (1.75 + 2.5 + x)^2 = (4.25 + x)^2 = 18.0625 + 9.5x + x^2 So 1.5x = 15.25 - 18.0625 = -2.8125 This is not feasible as you would be closer to the speaker than you started So x = 0.475, 3.125, 15.375 and the distances from the speaker is 2.5 +x = 2.975, 5.625, 17.875

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