Two loudspeakers, 4.0m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.17 m. What is the frequency of the sound? If the frequency is then increased while you remain 0.17 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Frequency at minimum intensity 0.17 m from center = 500 Hz

Frequency at maximum intensity 0.17 m from center = 1000 Hz

Explanation:

a. Frequency at minimum intensity 0.17 m from center

Since we move 0.17 m away from the center to encounter a minimum, the distance between the first speaker and that point is d₁ =  2 + 0.17 = 2.17 m. The distance between the second speaker and that point is d₂ =  2 - 0.17 = 1.83 m.

The path difference between the two waves is Δd = d₁ - d₂ = 2.17 m - 1.83 m = 0.34 m

Since we have a minimum at this point, there is destructive interference, so Δd = (m + 1/2)λ where λ = wavelength of wave m = 0,1,2...

when m = 0, we have

Δd = (0 + 1/2)λ = 1/2λ

λ = 2Δd = 2 × 0.34 m = 0.68 m.

The frequency of the wave is thus f = v/λ  where v = speed of sound = 340 m/s. f = 340 m/s÷ 0.68 m = 500 Hz

b. Frequency at maximum intensity 0.17 m from center

If the frequency is increased while you remain at 0.17 m a maximum sound intensity is observed when the path difference Δd = mλ     m = 0,1,2...

When m = 1, Δd = 0.34 m = λ

Frequency, f = v/λ = 340m/s ÷ 0.34 m = 1000 Hz = 1 kHz

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