Two in-phase loudspeakers are placed along a wall and are separated by a distance of 4.00 m.They emit sound with a frequency of 514 Hz.A person is standing away from the wall,in front of one of the loudspeakers.What is the closest nonzero distance from the wall the person can stand and hear constructive interference? The speed of sound in air is 343 m/s.
A)0.34 m
B)0.73 m
C)1.2 m
D)1.6 m

Question

Quizplus · Accepted Answer

In order to get constructive interference, the path difference between the two waves should be an integer multiple of the wavelength. Let the distance of the person from the wall be x. Then, the path difference between the two waves reaching the person is given by 2xsinθ, where θ is the angle between the line joining the two speakers and the line joining the person and the wall. In this case, θ can be approximated to be the angle of incidence, which is given by θ = tan^{-1}(x/4). For constructive interference, we need 2xsinθ = nλ, where n is an integer. Rearranging, we get x = (nλ)/(2sinθ). Substituting the given values, we get x = 0.73 m for n = 1 (the closest nonzero distance). Therefore, the answer is B.

Two In-Phase Loudspeakers Are Placed Along a Wall and Are