In the figure below two in-phase loudspeakers, A and B, are separated by 11.91 m
ID: 1312502 • Letter: I
Question
In the figure below two in-phase loudspeakers, A and B, are separated by 11.91 m. A listener is stationed at point C, which is 2.44 m in front of speaker B. The triangle ABC is a right triangle. Both speakers are playing identical 192-Hz tones, and the speed of sound is 343 m/s. Speaker A is moved further to the left, while ABC remains a right triangle. What is the separation between the speakers when constructive interference occurs again at point C?
http://www.webassign.net/cj8/17-figure-07-alt.gif
Explanation / Answer
The wavelength of the sound is:
Lambda = c/f = 343/192 = 1.786 (m)
Initially:
- distance B-C is 2.44 (m)
- distance A-C is sqrt(2.44^2 + 11.91^2) = 12.16 (m)
- difference between these distances: 12.16 - 2.44 = 9.72 = 5.44*Lambda
So initially, C is NOT a point of constructive interference, because the difference in the distances is not an integral number of wavelengths.
However, when A is moved such that the difference in distances is 6*lambda, it will be.
Therefore:
sqrt(2.44^2 + x^2) - 2.44 = 6*lambda
2.44^2 + x^2 = (6*lambda + 2,44)^2
x = sqrt((6*lambda+ 2.44)^2 - 2.44^2)
= sqrt(36*lambda^2 + 4.88*6*lambda)
= sqrt(lambda*(36*lambda + 4.88*6))
x = 12.928 (m)