Two loudspeakers in a plane, 6.0 mm apart, are playing the same frequency. If you stand 12.0 mm in front of the plane of the speakers, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 12.0 mm in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers. What is the frequency of the sound?

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nafeesahmed nafeesahmed · Answer 1 · 2020-03-13T23:23:44+01:00

Answer:

Frequency of the sound = 22.97 kHz

Explanation:

Sound waves are traveling waves and they can be modeled as

A(r, t) = A₀(r)sin(kr - ωt + ∅₀)

Where A₀ is the initial amplitude of the wave, r is the distance, ωt is the frequency and ∅₀ is the initial phase shift

First we need to find out the phase difference (Δ∅) between two waves at different distances.

Δ∅ = 2πΔr/λ + Δ∅₀

When you stand centered between the two waves you hear maximum intensity of sound so the the two waves must be in phase

Δ∅ = 2πΔr/λ + 0

λ = 2πΔr/Δ∅

The distance when listening in front of the speakers is given by

Δr = r2 - r1

r1 = 6.0 mm = 0.006 m

r2 = √(0.012²+0.006²) = 0.0134 m

Δr = r2 - r1 = 0.0134 - 0.006 = 0.0074 m

λ = 2π*0.0074/Δ∅

The phase difference Δ∅ = π

λ = 2π*0.0074/π

λ = 0.0148 m

As we know the relation between frequency and wavelength is given by

f = c/λ

Where c = 340 m/s is the speed of light

f = 340/0.0148

f = 22973 Hz

f = 22.97 kHz