[SOLVED] Interference of Sound Waves Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s. What is the closest you can be to speaker B and be at a point of destructive interference? __________ I am having the hardest time trying to visualize the problem. I know that destructive interference occurs when the difference in path lengths traveled by sound waves is a half integer number of wavelengths. So I need to know the wavelength of the sound which is just 2m. I also know that in general if d_a and d_b are paths traveled by two waves of equal frequency that are originally emitted in phase, the condition for destructive interference is d_a-d_b=n(wavelength)/2 where wavelength is what I calculated it to be (2m) and n=any nonzero odd integer. I think I need to know what the value of n is that corresponds to the shortest distance d_b to solve my prob. (is d_a=8m? then what is d_b?) I'm going around in circles and getting nowhere. Please help!
Usually, it is important to know the distance between the two speakers. Is this not given? Then you can move B wherever you like. If you are 8 m from A, and wavelength is 2 m, the sound form A has traveled 4 full integer multiples of the wavelength. You must place B at a location such that the sound will arrive 1/2 wavelength out of phase. One full wavelength? or less? how much less?
apparently n=1 is not the right answer. all info has been given. gahh. i need help. Yes, sound from A has traveled 4X the wavelength of sound emitted by the loud speakers. Sound from B must be moved so that the sound arrives 1/2 wavelength out of phase (d_a-d_b=n(wavelength)/2) so are you telling me the same thing that Vijay told me, that n=1?
Re: [SOLVED] Interference of Sound Waves The correct answer is 1.00m. Hint: The closest you can be to B. (lamda/2)