Superposition and interference?

Question

Two identical loudspeakers are driven by the same oscillator of frequency 200Hz. The speakers are located on a vertical pole a distance of 4.00m from each other. A man walks straight toward the lower speaker in a direction perpendicular to the pole.

a) How many times will he hear a minimum in sound intensity?

b) How far is he from the pole at these moments? Take the speed of the sound to be 330m/s and ignore any sound reflection from the ground.

Answer 1

Wavelength λ = c/F = 330/200 = 1.65 m
He hears a minimum every time the two paths differ in length by an odd multiple of λ/2. We assume the speakers operate in-phase.
If the man walks straight toward the lower speaker (speaker 1) we must assume his ears are at the same height as speaker 1. Define x1 = distance to speaker 1, x2 = distance to speaker 2, and the difference Δx = x2-x1. x2 is the hypotenuse of the right triangle with adjacent sides x1 and 4.
x2 = sqrt(x1^2 + 4^2)
Δx = x2 - x1
sqrt(x1^2 + 16) = x1 + Δx
x1^2 + 16 = (x1 + Δx)^2 = x1^2 + 2x1Δx + Δx^2
16 = 2x1Δx + Δx^2
x1 = (Δx^2 - 16)/(-2Δx)
Solve this for Δx = (1, 3, 5 etc) * λ/2. Eventually x1 will be negative. In fact there are only two positive solutions, one near 9 m (Δx = 1λ/2) and one near 2 m (Δx = 3λ/2).