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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?

Respuesta :

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