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Published byColin Charles Modified about 1 year ago

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Interference of Sound Waves

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Interference 2 waves, of the same frequency; out of phase. Eg. y 1 =A 0 sin (kx - t) y 2 =A 0 sin (kx - t + ) Then y R =A R sin(kx- t+ /2), and the resultant amplitude is A R =2A 0 cos(½ . Identical waves which travel different distances will arrive out of phase and will interfere, so that the resultant amplitude varies with location.

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Quiz y 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’ (difference of π). What will you hear? A) no change – same loud sound B) no sound C) something between ‘no sound’ and ‘loud sound’

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Example: Two sources, in phase; waves arrive at P by paths of different lengths: At P: detector S1S1 S2S2 P x1x1 x2x2

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Phase difference : Then (using trig), at detector: Define x to be the path difference = kx terms don’t cancel!

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Example A pair of speakers is separated by 3.0m and driven by the same oscillator. The listener walks perpendicular from a point on the centerline 8m away to a distance of 0.35m before reaching the first minimum in sound intenstiy. What is the frequency of the oscillator? 8.0m 0.35m 3m

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Solution

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Intensity I I = Power per unit area Units: W / m 2 (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 A 2 Two sources, each with amplitude A o, intensity I o, phase difference

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Notes: 1)Maximum I R is 4 x I O 2)Maxima when = 0, 2π, 4π, 6π, … 3)Minima (zero intensity) when = π, 3π, 5π, … x = ± λ/2, ± 3λ/2, ± 5λ/2,… x = 0, ± λ, ± 2λ,… Note: The sources are in phase !!!

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2 speakers, same intensity, in phase; f = 170 Hz (so = 2.0 m when speed of sound is 340 m/s) At the position x=9 m, find the intensity in terms of the intensity of a single speaker 6 m x The detector Example I think I hear something!

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Solution

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