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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
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
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Physics
3 years
2021-08-18T22:50:19+00:00
2021-08-18T22:50:19+00:00 1 Answers
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Answers ( )
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 +
) × 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 +
)v / Δr
m = 0, 1, 2, 3, ….
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 +
)v / Δr
f =
(v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f =
(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