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Sagot :
Answer:
a) 0, ±1.65 b) ± 0.825m
Explanation:
This is a sound interference exercise, it can be described by the path difference between the two waves
for the case of constructive interference
Δr = 2n λ/ 2 n = 0, 1, 2 ...
for the case of destructive interference
Δr = (2n + 1) λ/ 2
the speed of sound is related to the wavelength
v = λ f
λ = v / f
λ= 340/206
λ = 1.65 m
let's set a reference system in the center between the two speakers
a) let's find the distances for constructive interference
Δr = 2n 1.65 / 2
Δr = 1.65 n
* the first interference occurs at n = 0
Δr = 0
therefore the interference in the center is maximum
* n = 1
Dr = 1.65 m
the second inference occurs at 1.65 m from the center, therefore there is a right wing and a left wing,
We do not have any more interference between the speakers because
n = 2 Δr = 3.3m this distance can be from the speaker
b) let's look for the destructive interference points
Δr = (2n + 1) 1.65 / 2
Δr = (2n + 1) 0.825
m = 0 Δr = 0.825m
m = 1 Δr = 2,475m
We can see that we only have the first destructive interference, one on each side.
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