Question

Two loudspeakers S1 and S2, 2.20 m apart, emit the same single-frequency tone in phase at the speakers. A listener L is located directly in front of speaker S1, in other words, the lines LS1 and S1S2 are perpendicular. L notices that the intensity is at a minimum when L is 5.50 m from speaker S1. What is the lowest possible frequency of the emitted tone

Answer 1

Answer:

the lowest possible frequency of the emitted tone is 404.79 Hz

Explanation:

   Given the data in the question;

S₁ ←  5.50 m → L

↑

2.20 m

↓

S₂

We know that, the condition for destructive interference is;

Δr = ( 2m + $\frac{1}{2}$ ) × λ

where m = 0, 1, 2, 3 .......

Path difference between the two sound waves from the two speakers is;

Δr = √( 5.50² + 2.20² ) - 5.50

Δr = 5.92368 - 5.50

Δr = 0.42368 m

v = f × λ

f = ( 2m + $\frac{1}{2}$)v / Δr

m = 0, 1, 2, 3, ....

Now, for the lowest possible frequency, let m be 0

so

f = ( 0 + $\frac{1}{2}$)v / Δr

f = $\frac{1}{2}$(v) / Δr

we know that speed of sound in air v = 343 m/s

so we substitute

f = $\frac{1}{2}$(343) / 0.42368

f = 171.5 / 0.42368

f = 404.79 Hz

Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz