Two speakers separated by distance d1 = 2.00 m emit sound waves that are in phas

Question

Two speakers separated by distance d1 = 2.00 m emit sound waves that are in phase. A listener, L, is located at a distance d2 = 3.75 m directly in front of one speaker (i.e., a line connecting the listener and the speaker directly in front of her makes a right angle with the line connecting the speakers). Consider the full audible range for normal hearing, 20 Hz to 20 kHz. The speed of sound here is 343 m/s.

(a) What is the lowest frequency min, 1 f , that gives minimum signal (destructive interference) at the listener

Explanation / Answer

speakers as A and B, and the observer as O, thenAO= 2.37 m, AB = 2.69 m, so BO (listener's distance from the "other" speaker) = sqrt (2.37^2 + 2.69^2) = 3.585 m.

Now that is the path length BO, and you already know the path length AO, so you can take the difference for the path length difference BO - AO = 3.75 - 2. = 1.75 m.

The lowest frequency for max constructive interference should have the peak of an AO wave exactly match the peak of a BO wave, so BO - AO = one wavelength. So you want the frequency where one wavelength = 1.75 m. Since speed of sound was given as 3.43 * 10^2 m/s, f = v/w = 3.43*10^2 / 1.75 .

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