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Two loud speakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from other speaker.

What is the lowest frequency at which destructive interference will occur at this point if the speakers are in phase?

User Marmoy
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1 Answer

3 votes
Answer:

f = 343 Hz

Step by step explanation:

For destructive interference to occur, the sound waves from the two speakers must be out of phase by half a wavelength. This means that the path difference between the two waves must be an odd multiple of half the wavelength.

In this case, the path difference between the two waves is given by:
Δx = d₂ - d₁
where d₂ is the distance from the second speaker to the person, and d₁ is the distance from the first speaker to the person.

Substituting the given values, we get:
Δx = 3.5 m - 3.0 m
Δx = 0.5 m

For destructive interference to occur, the path difference must be an odd multiple of half the wavelength, i.e.:
Δx = (2n + 1)λ/2
where n is an integer.

Solving for the wavelength, we get:
λ = 2Δx/(2n + 1)

The lowest frequency occurs when n is the smallest possible value, i.e. n = 0. Substituting this value, we get:
λ = 2Δx/1
λ = 2(0.5 m)
λ = 1.00 m

The frequency of the sound wave is given by:
f = c/λ
where c is the speed of sound in air (approximately 343 m/s).

Substituting the values, we get:
f = 343 m/s/1.00 m
f = 343 Hz

Therefore, the lowest frequency at which destructive interference will occur at the given point is 343 Hz.
User Ori
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