Question

Two loudspeakers are placed above and below each other, as in the figure below, and driven by the same source at a frequency of 5.00 x 102 Hz. An observer is in front of the speakers (to the right) at point o, at the same distance from each speaker. What minimum vertical distance upward should the top speaker be moved to create destructive interference at point O? (Let h = 2.76 m and use v = 343 m/s.) 8.00 m

Answer 1

Hjdne Page No. Date Given 71 h h=2.76 m V=3 13m/s - 8m 5X 10² HZ - = wavelength = = V f a velocity frowy -) 1 343 12=0.688 M =) The distance between spenter at point ol ris ) 2 + (8) 2 - 12:44) + 64 (1.38) +64 = √ 65.9049 Fri = 8.118 m s from simlarning = (v1=2) = 8.118 m Now the top speaker is moved a distance of - Dy above its initial position i di is the new length from speaker de point o 1 Intro Tola (1.38 + y)² + 64 ]

Date Wavelength d= 0.686 0.681 = 0.343 . Now distance g. 118 +0.343 Iri = 8.461 m] Now ri2 - 64 1 68.461 -64 = Oy + 1:38 - J 7.588521 = by + 1.38 2.7547 = oy +1:38 = 1.3747 m 2. Dye 1.3742 m The required vertical distemu the top speaker should be moved = DY = 1.3747 m.