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Unformatted text preview: homework 38 – BAUTISTA, ALDO – Due: May 3 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. The acoustical system shown in the figure is driven by speaker emitting a 369 Hz note. The speed of sound is 345 m / s. S Speaker Receiver r 1 r 2 If destructive interference occurs at a par ticular instant, by what minimum amount must the difference of two path lengths in the Ushaped tube be increased in order to hear constructive interference? Correct answer: . 46748 m (tolerance ± 1 %). Explanation: Given : f = 369 Hz and v = 345 m / s . The velocity of the wave is v = fλ . To produce constructive interference the dif ference must be increased by Δ l = λ 2 = v 2 f = 345 m / s 2 (369 Hz) = . 46748 m . Question 2 Part 2 of 2. 10 points. By what minimum amount must the differ ence of the path lengths be increased in order to hear destructive interference again? Correct answer: 0 . 934959 m (tolerance ± 1 %). Explanation: With destructive interference currently tak ing place, one can increase the path difference by one full wavelength to produce destructive interference again, so Δ l = λ = v f = 345 m / s 369 Hz = . 934959 m . Question 3 Part 1 of 1. 10 points. Two loudspeakers are placed a distance of 8 m above and below one another, as shown, and driven by the same source. A detector is positioned a perpendicular distance of 17 m from the two speakers and a vertical height 6 . 3 m from the lower speaker. This results in constructive interference at the first order maximum. The velocity of sound is 345 m / s . 8 m 6 . 3 m 17 m center line intensity What minimum distance (in any direction) should the top speaker be moved in order to create destructive interference between the two speakers? Correct answer: 0 . 522511 m (tolerance ± 1 %). Explanation: Let : h = 6 . 3 m , d = 8 m , L = 17 m , y = h d 2 = (6 . 3 m) (8 m) 2 = 2 . 3 m , y d 2 = (2 . 3 m) (8 m) 2 homework 38 – BAUTISTA, ALDO – Due: May 3 2006, 4:00 am 2 = 1 . 7 m , and y + d 2 = (2 . 3 m) + (8 m) 2 = 6 . 3 m v s = 345 m / s . Constructive and destructive interference. Phase difference: Maxima: φ = 2 m π . Minima: φ = (2 m + 1) π . φ = k δ , and k = 2 π λ . δ is difference in path length. Double Slit interference. The rules for determining interference max imum or minimum are the same for sound waves and light waves. Thus, the path length difference is δ = d 2 d 1 = n λ , (1) where n = 1 for the first maximum. y d 2 y y + d 2 L d 1 d 2 d ⊗ ⊗ We cannot use the formula sin θ = δ d since the detector is not extremely far from the speakers compared to the distance between the speakers! This formula is only valid when the detector is far enough from the sources that the signals are traveling nearly parallel to each other. We must go back to the defini tions and basic concepts of constructive and destructive interference, since we know how to relate the difference in path lengths to the...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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