Consider two sources broadcasting identical waves in phase with one another. The sources create an interference pattern with bands of constructive and destructive interference.
The key to unlocking the secret of the pattern is this - what happens at any point depends on the path length difference, the distance from one source to the point minus the distance from the other source to the point. Call this path length difference ΔL.
Condition for constructive interference: ΔL = nλ, where n is any integer.
Condition for destructive interference: ΔL = (n + ½)λ, where n is any integer.
Two sources are broadcasting identical single-frequency waves, in phase. You stand 3 m from one source and 4 m from the other. What is the lowest frequency at which constructive interference occurs at your location? Take the speed of sound to be 340 m/s.
First, find the path length difference - how much further are you from one source than the other? In this case it's simply 1 m. If that's an integral number of wavelengths, constructive interference occurs.
Constructive Interference: ΔL = n λ, where n = 0, 1, 2, ...
1 = n λ
The lowest frequency corresponds to the largest wavelength, which corresponds to the smallest value of n. That is n = 1 in this case, giving a wavelength of 1 m.
With a wavelength of 1 m and a speed of 340 m/s, the frequency is:
f = v/λ = 340/1 = 340 Hz.
What are the next two smallest frequencies at which constructive interference occurs?
Just take the next two smallest n values, 2 and 3.
Substituting ΔL/n for λ gives:
fn = nv/ΔL, so:
f2 = 2v/ΔL = 680 Hz
f3 = 3v/ΔL = 1020 Hz
What are the lowest three frequencies giving destructive interference at your location?
Destructive Interference: ΔL = (n + ½)λ, where n = 0, 1, 2, ...
Solving this for wavelength gives the wavelengths at which destructive interference occurs:
λ = ΔL/(n + ½)
Substituting this into f = v/λ gives:
Frequencies at which destructive interference occurs: fn = (n + ½)v/ΔL
With v = 340 m/s and ΔL = 1 m, the lowest three frequencies at which destructive interference occurs are:
f0 = 340*1/2 = 170 Hz
f1 = 340*3/2 = 510 Hz
f2 = 340*5/2 = 850 Hz