Two speakers separated by a distance of 4.40 m emit sound. the speakers have opposite phase. a person listens from a location 3.00 m directly in front of one of the speakers.

what is the lowest frequency that gives destructive interference in this case?

f = 147.21 Hz

Explanation:

In order to have a destructive interference, as the source emit in opposite phases, the path difference between the distance to the person, measured in a straight line from the speakers, must be equal to an integer number of wavelengths.

We need to know the distance from the listener to the other speaker, located 4.4 m from the one which is directly in front of him, which we can find using Pythagorean theorem, as follows:

l₂ = √(3)²+(4.4)² = 5.33 m

The difference in path will be, then:

d = l₂-l₁ = 5.33 m - 3.00 m = 2.33 m

For the lowest frequency that gives destructive interference, the wavelength will be highest possible, which happens when the distance is just one wavelength.

⇒ d = λ = 2.33 m

In any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

v = λ*f Κ  ⇒ f = v/λ

Taking the speed of sound as 343 m/s, and solving for f, we get:

f= 343 m/s / 2.33 m = 147.21 Hz

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