Two identical loudspeakers are placed on a wall 3.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speaker

Question

Two identical loudspeakers are placed on a wall 3.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.) (b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

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Thanh Thu 4 years 2021-08-13T17:50:42+00:00 1 Answers 328 views 1

Answers ( )

  1. Tryphena
    0
    2021-08-13T17:51:47+00:00
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    Answer:

    a)  ΔФ = 5.54 rad
    , b)   f = 140 Hz  

    Explanation:

    a) This is a sound interference exercise, which is described by

             Δr /λ = ΔФ / 2π

             ΔФ = Δr 2π /λ

    Let’s find the path difference

           Δr = r₂ -r₁

           r₁ = 4m

           r₂ = √ (x² + y²) = √(3² + 4²) = 5 m

           Δr = 1 m

    To find the wavelength we use the relation of the speed of sound

             v = λ f

             λ = v / f

             λ = 340/300

             λ = 1,133 m

     

    We substitute

            ΔФ = 2π 1 /1.133

             ΔФ = 5.54 rad

    b) to have a minimum intensity the phase difference must be π radians

             λ = Δr 2π /Ф

             λ = 1 2π /π

             λ = 2m

    We look for the frequency

            v = λ f

            f = v /λ

            f = 340/2

            f = 140 Hz  

     

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )