Sound Wave Interference

When waves travel through the same medium, they will superpose, and the resulting interference will affect the sound we perceive. The interference will be, at the extreme, completely constructive or destructive. These two extremes yield minimum and maximum intensities of sound.

When identical waves are emitted from two sources, at certain points around the sources we see max and min constructive and destructive interference. The sound will be loudest at the max (constructive) and quietest at the minimums (destructive)

If the distance from each source is an integral number of wavelengths from each source, the waves arrive in phase, and if the distance from each source is different by an odd ½ wavelength (½, 3/2 etc.) the waves are out of phase and a min results.

If both Waves begin IN PHASE: The difference in path length (the distance the wave travels from each source) is the key factor in determining perception at a min or max. If ΔL = L2 – L1 = nλ max sound heard If ΔL = L2 – L1 = ½ λ (or 3/2, 5/2 etc) Minimum sound heard.

Another formula which will relate the lateral distance to the mth fringe (successive mins), the source separation, the wavelength of sound and the linear distance to the observer; Ym = (m-1/2)λL/d Use (m-1/2) for minimums w/ in phase sources

At a concert, a guitar player plays a single note of 494 Hz (T = 25 C) If you sit 7.00 m from one speaker and 9.10 m from another speaker, what type of sound do you hear? (Min or max). ΔL = 9.10 – 7.00 = 2.10 m V = fλ v = 331 + 0.6 (25) = 346 m/s 346m/s / 494 Hz = 0.700 m = λ ΔL = 2.10 m = 3(λ) so a max is heard.

If the waves begin OUT OF PHASE, then the opposite must be true. In order for a max to be heard, one wave would have to travel ½ wavelength (or an odd multiple of halves ) to “catch up” and get back into phase.