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

The lowest possible frequency of the emitted tone is approximately 31.2 Hz.

Step-by-step explanation:

The minimum intensity occurs when the waves from both speakers interfere destructively. This happens when the path difference between the two waves is equal to an odd multiple of half the wavelength.

In this case, the path difference between L and S1 is equal to LS2 - S1S2 = 5.50 m. Since the lines LS1 and S1S2 are perpendicular, this path difference is equal to half a wavelength, so:

λ/2 = 5.50 m (where λ is the wavelength)

The shortest possible wavelength corresponds to the highest frequency, so:

f = v/λ = v/(2(5.50 m))

where v is the speed of sound, which is approximately 343.00 m/s.

Substituting the values:

f = 343.00 m/s / (2(5.50 m))

f ≈ 31.2 Hz

Answer 2

the lowest possible frequency of the emitted tone is 404.79 Hz

Step-by-step 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 +
`(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 +
`(1)/(2)`)v / Δr

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

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

so

f = ( 0 +
`(1)/(2)`)v / Δr

f =
`(1)/(2)`(v) / Δr

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

so we substitute

f =
`(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