Posted by **josh** on Sunday, July 4, 2010 at 5:25am.

Two identical loudspeakers are 2 m apart. A person stands 4.6 m from one speaker and 3.4 m from the other. What is the lowest frequency at which destructive interference will occur at this point?

- physics -
**drwls**, Sunday, July 4, 2010 at 5:49am
Destructibe interference will occur if the difference between the distances to the two speakers is n + 1/2, where n is an integer zero or higher. When n = 0, you get the longest wavelength and lowest frequency with destructive interference.

Therefore 4.6 - 3.4 = 1.2 = (wavelength)/2

wavelngth = 2.4 m

freq

- physics -
**drwls**, Sunday, July 4, 2010 at 5:52am
Destructive interference will occur if the difference between the distances to the two speakers is n + 1/2, where n is an integer equal to zero, or higher. When n = 0, you get the longest wavelength and lowest frequency with destructive interference.

Therefore 4.6 - 3.4 = 1.2 = (wavelength)/2

wavelength = 2.4 m

frequency = (speed of sound)/2.4 m

= 140 Hz (approx.)

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