2 Interference 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.
3 Quiz 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
4 Example:Two sources, in phase; waves arrive at P by paths of different lengths:S1Px1detectorx2At P:S2
5 Phase difference :=Define x to be the path differenceThen (using trig), at detector:kx terms don’t cancel!
6 Example 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
8 Intensity 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:
9 Notes: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 !!!
10 I 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