Constructive Interference of Sound Waves

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Constructive interference of sound waves occurs when the waves from two speakers are in phase, which requires the path length difference to be a multiple of the wavelength. In this scenario, the speakers are separated by three meters, and the listener is four meters from one speaker, creating a path length difference of one meter. To find the lowest frequencies for constructive interference, the wavelength must be calculated using the equation v = fλ, where v is the speed of sound at 346 m/s. The path length difference of one meter indicates that the wavelengths must be 1 m, 0.5 m, and 0.33 m, leading to corresponding frequencies of 346 Hz, 692 Hz, and 1038 Hz. Understanding these principles allows for the determination of the frequencies that produce constructive interference.
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Suppose two loudspeakers are separated by three meters and are being driven by exactly the same sinusoidal signal at a frequency that can be varied. If you are located four meters away directly in front of one of the speakers, determine three lowest frequencies for which you would hear constructive interference. (Assume speed of sound to be 346 m/s) Explain your reasoning carefully.

Through my professor's book, I know that constructive interference happens when waves are in phase. I believe one of the equations I should be using is total phase difference = phase difference of the sources + 2∏Δr/λ. Not sure where to go from here.
 
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It's a 3,4,5 triangle so the two distances are 4 and 5 meters. For constructive interference the difference (eg 1m) must be a multiple of what?

Once you know that apply v=fλ.
 
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