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Understanding Sound Interference & Superposition in Waves

Explore the principles of sound wave interference and superposition, particularly focusing on harmonic waves, phase differences, and resultant amplitudes in the same medium. Learn how identical waves traveling different distances interfere and vary in amplitude. Dive into examples and calculations to deepen your understanding.

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Understanding Sound Interference & Superposition in Waves

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  1. Interference of Sound Waves

  2. Principle of Superposition 2 Waves In The Same Medium: The observed displacement y(x,t) is the algebraic sum of the individual displacements: y(x,t) =y1(x,t) + y2(x,t) (for a “linear medium”)

  3. What’s Special about Harmonic Waves? 2 waves, of the same amplitude, same angular frequency and wave number (and therefore same wavelength) traveling in the same direction in a medium but are out of phase: Trig: sin a + sin b = 2 cos [(a-b)/2] sin [(a+b)/2] Result: Resultant amplitude

  4. Assume  is positive: For what phase difference  is the total amplitude 2A? For what phase difference  is the total amplitude A? For what phase difference  is the total amplitude 0? For what phase difference  is the total amplitude A/2?

  5. wave 1 wave 2 Resultant: Sine wave, same f, different A, intermediate f

  6. Interference 2 waves, of the same frequency; out of phase. Eg. y1=A0sin (kx - wt) y2=A0sin (kx - wt +f) Then yR=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.

  7. Example: Two sources, in phase; waves arrive at P by paths of different lengths: P S1 x1 detector x2 At P: S2

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

  9. Example 0.35m 8.0m 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? 3m

  10. Solution

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

  12. 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/Area and power is proportional to Amp2 Two sources, each with amplitude Ao, intensity Io , phase difference f:

  13. Notes: • Maximum IR is 4 x IO • Maxima when f = 0, 2π, 4π, 6π , … • Minima (zero intensity) when f = π, 3π, 5π , … x = 0, ± λ, ± 2λ,…  x = ± λ/2, ± 3λ/2, ± 5λ/2,… Note: The sources are in phase !!!

  14. Example 6 m The detector x I 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 the intensity of a single speaker

  15. Solution

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