Question

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.

Answer 1

Answer:

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