2Interference 2 waves, of the same frequency; out of phase. Eg. y1=A0sin (kx - wt) y2=A0sin (kx - wt +f)ThenyR=ARsin(kx-wt+f /2),and the resultant amplitude is AR=2A0cos(½f ).Identical waves which travel different distances will arrive out of phase and will interfere, so that the resultant amplitude varies with location.
3Quiz You are located at position y, where you can hear a loud sound - the first maximum in intensity from two speakers. The speakers are then connected ‘out of phase’ (differenceof π). What will you hear? A) no change – same loud sound B) no sound C) something between ‘no sound’ and ‘loud sound’y
4Example:Two sources, in phase; waves arrive at P by paths of different lengths:S1Px1detectorx2At P:S2
5Phase difference :=Define x to be the path differenceThen (using trig), at detector:kx terms don’t cancel!
6Example A pair of speakers is separated by 3.0m and driven by the same oscillator. The listener walks perpendicular from apoint on the centerline 8m away to a distance of 0.35mbefore reaching the first minimum in sound intenstiy.What is the frequency of the oscillator?8.0m0.35m3m
8Intensity I I = Power per unit area Units: W / m2 (the area is measured perpendicular to the wave velocity)Intensity is proportional to (resultant amplitude)2 , since I=P/A and power is proportional to A2Two sources, each with amplitude Ao, intensity Io , phase difference f:
9Notes:Maximum IR is 4 x IOMaxima when f = 0, 2π, 4π, 6π , …Minima (zero intensity) whenf = π, 3π, 5π , …x = 0, ± λ, ± 2λ,… x = ± λ/2, ± 3λ/2, ± 5λ/2,…Note: The sources are in phase !!!
10I think I hear something! Example6 mxThe detectorI think I hear something!2 speakers, same intensity, in phase; f = 170 Hz(so l = 2.0 m when speed of sound is 340 m/s)At the position x=9 m, find the intensity in terms of theintensity of a single speaker