Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands

Question

Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 5.0 m away from one speaker and 6.2 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 330 m/ s. A) 183 Hz B) 275 Hz C) 413 Hz D) 137 Hz E) 550 Hz

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Bình An 4 years 2021-09-03T01:55:08+00:00 1 Answers 4 views 0

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  1. thongdat2
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    2021-09-03T01:57:00+00:00
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    Answer:

    C) 413 Hz

    Explanation:

    For destructive interference, the path difference ΔL = (n + 1/2)λ where ΔL = L₂ – L₁ where L₁ = person’s distance from one speaker (the closer one) = 5.0m and L₂ = person’s distance from other speaker (the farther one) = 6.2 m and λ = wavelength = v/f where v = speed of sound = 330 m/s and f = frequency

    So, ΔL = (n + 1/2)λ

    L₂ – L₁  = (n + 1/2)v/f

    f = (n + 1/2)v/(L₂ – L₁)

    At the second lowest frequency that results in destructive interference at the point where the person is standing, n = 1.

    So,

    f = (1 + 1/2)v/(L₂ – L₁)

    f = 3v/2(L₂ – L₁)

    Substituting the values of the variables into the equation, we have

    f = 3v/2(L₂ – L₁)

    f = 3(330 m/s)/2(6.2 m – 5.0 m)

    f = 3(330 m/s)/2(1.2 m)

    f = 990 m/s ÷ 2.4 m)

    f = 412.5 Hz

    f ≅ 413 Hz

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