In the figure, two speakers separated by distance d1 = 2.00 m are in phase. Assu

Question

In the figure, two speakers separated by distance d1 = 2.00 m are in phase. Assume the amplitudes of the sound waves from the speakers are approximately the same at the listener's ear at distance d2 = 3.40 m directly in front of one speaker. Consider the full audible range for normal hearing, 20 Hz to 20 kHz and use 343 m/s for the speed of sound in air.

(a) What is the lowest frequency that gives minimum signal (destructive interference) at the listener's ear? ___ Hz (b) What is the second lowest frequency that gives minimum signal? ___ Hz (c) What is the third lowest frequency that gives minimum signal? ___ Hz (d) What is the lowest frequency that gives maximum signal (constructive interference) at the listener's ear? ___ Hz (e) What is the second lowest frequency that gives maximum signal? ___ Hz (f) What is the third lowest frequency that gives maximum signal? ___ Hz

Explanation / Answer

Use the pythagorean theorem to calculate the distance from the other speaker, which is sqrt[2^2+3.4^2]=3.488 m.
The difference in distance to the two speakers is 3.488-3.10=0.388 m. The wavelength of the lowest audible frequency is 343/20=17.15 m.
(a) The distance 0.388 m must equal half a wavelength for this question, asuming it to fall within the audible range. Thus wavelength=lambda=0.776m, which is within the audible range. The frequency is f=v/lambda=343/0.776=442 Hz.
(b) The next destructive interference occurs when 0.388 m equals 1 1/2 wavelengths, which is three times that of (a). The frequency will thus be a third of the previous answer, or 147.3 Hz.
(d)For this, 0.388m equals one full wavelength, yielding 221Hz frequency.
(e) For this, 0.388m equals two full wavelengths, giving 110.5 Hz for the frequency.

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