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Two speakers, one directly behind the other, are each generating a 280-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is 338 m/s.

User IKriKan
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Final answer:

The smallest separation distance between two speakers at 280 Hz for destructive interference is 0.6035 m, calculated using the speed of sound (338 m/s) and the frequency to determine the wavelength and then taking half of that wavelength.

Step-by-step explanation:

To determine the smallest separation distance between two speakers generating a 280-Hz sound wave that will produce destructive interference at a listener standing in front of them, we must consider the concept of path difference leading to destructive interference. Destructive interference occurs when two sound waves are out of phase by ½ of a wavelength, λ/2. Since the frequency, f, is 280 Hz and the speed of sound, v, is given as 338 m/s, we can calculate the wavelength using the formula λ = v/f. Once we have the wavelength, the smallest separation distance that results in destructive interference is λ/2.

Calculating the wavelength: λ = 338 m/s ÷ 280 Hz = 1.207 m. Thus, the smallest separation distance for destructive interference is λ/2 = 1.207 m ÷ 2 = 0.6035 m. This distance represents the smallest separation at which the sound from the rear speaker travels ½ a wavelength further than the sound from the front speaker, causing the two waves to cancel each other out in front of the speakers.

User Megan Speir
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