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Masja [62]
2 years ago
14

Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre

quency that creates destructive interference at his location is 72.4 Hz. How far away is he from speaker B?
Physics
1 answer:
earnstyle [38]2 years ago
8 0

Answer:

  0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

  λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

  3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

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Answer:

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Divide both sides by 2 to get;

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Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

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d2 - d'2 = 75λ1 - 57λ2

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lbvjy [14]

Answer:

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