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Two loudspeakers, 4.5 m 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 0.31 m . Assume the speed of sound is 340 m/s. What is the frequency of the sound? If the frequency is then increased while you remain 0.31 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Respuesta :

Answer:

f = 16.03 kHz

f = 32.07 kHz

Explanation:

For maximum intensity of sound, the two waves must be in phase.

The phase difference (Δ∅) between two waves is given by

Δ∅ = 2πΔr/λ

λ = 2πΔr/Δ∅

where Δr is given by

Δr = r2 - r1

Δr = √(4.5²+0.31²) - 4.5

Δr = 0.0106 m

λ = 2π*0.00106/Δ∅

For cancellation, the phase difference must be Δ∅ = π

λ = 2π*0.0106/π

λ = 2Δr

λ = 0.0212 m

Finally the frequency can be found using

f = c/λ

f = 340/0.0212

f = 16038 Hz

f = 16.03 kHz

Now if the frequency is then increased while you remain 0.31 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

The λ will be reduced to half

λ = 2Δr/2

λ = Δr

λ = 0.0106 m

f = c/λ

f = 340/0.0106

f = 32075 Hz

f = 32.07 kHz

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