Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.

Question

Ryder Fox

Physics

21 February, 20:40

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.

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Answers (1)

Know the Answer?

Moises Berger · Answer 1

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2) λ/2 where n = 0,1,2 ...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √ (2.00² + 5.50²) = √ (4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2) λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2) λ/2 = (0 + 1/2) λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340 / (4 * 0.35) = 242.85 Hz