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Joe unpacks his stereo system. He places the amplifier on his table, one speaker 2.5 m to its left, and the other speaker 2.0 m to its right. Joe then sits in a chair, 2.0 m in front of the amplifier.

**Question 1: **
Draw a picture showing a view from above of Joe, the
amplifier, and both speakers.

**Question 2: **
What is the distance from Joe to the left-hand speaker,
and from Joe to the right-hand speaker?

Answer: distance Joe to left speaker is d(left) = sqrt( [2m]^2 + [2.5m]^2 ) = 3.20 m distance Joe to right speaker is d(right) = sqrt( [2m]^2 + [2m]^2 ) = 2.83 m

Assume that sound travels at v = 343 m/s. The speakers play a note with frequency f = 463.5 Hz.

**Question 3: **
Does Joe hear constructive or destructive interference? Why?

Answer: the wavelength of this sound is v 343 m/s lambda = --- = -------- = 0.740 m f 463.5 Hz The difference in path length from Joe to the two speakers is d(left) - d(right) = 3.20 m - 2.83 m = 0.37 m = lamba / 2 Since the difference in path length is 1/2 of a wavelength, the interference will be destructive.

Joe moves his chair sideways 0.25 m to the right.

**Question 4:**
What sort of interference does he now hear? Why?

Answer: at his new position (dashed figure in diagram), Joe is at a distance d(left) = sqrt( [2m]^2 + [2.75m]^2 ) = 3.40 m d(right) = sqrt( [2m]^2 + [1.75m]^2 ) = 2.66 m So the difference in path length now is d(left) - d(right) = 3.40 m - 2.66 m = 0.74 m Since the difference is exactly 1 wavelength, the two waves interfere constructively.

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This page maintained by Michael Richmond.
Last modified Feb 26, 1997.
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Copyright © Michael Richmond. This work is licensed under a Creative Commons License.