Sample Problem
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 |
= |
| n v
| |
| 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 |
= |
| 1
| |
| 2
|
|
* 340 = 170 Hz |
| f1 |
= |
| 3
| |
| 2
|
|
* 340 = 510 Hz |
| f2 |
= |
| 5
| |
| 2
|
|
* 340 = 850 Hz |