Question

Two loudspeakers are mounted on a rack, one h = 3.42 m above the other. Exactly 8.00 meters to the right of the midpoint, a listener rests at point o. Point O is equally distant from each loudspeaker. 8.00 m The loudspeakers are driven by the same tone generator and vibrate in phase at 510 Hz. It is possible to create a condition of destructive interference at Point O by changing one or both of the path lengths (r, and rz) between speaker and listener. Suppose that this is done by raising the upper speaker while leaving the lower speaker in place. What is the smallest vertical distance (inm) that you would need to raise the upper speaker by, in order to create destructive interference at Point 07 (The speed of sound waves in air is 343 m/s.)

Answer 1

condition for destructive interference is, beta difference (2 + 1) x 7 for smallest path difference, 120. pate difference & I/ :. 348 x 2 a bath difference. $10 12 Now, the upper, mil should be rased & distanee (say) ( 17x) + 8 343 [1.71) which is equal to, = 0.336 s should be 0.336 greater than ry Now, so ſ 87 + (3:42) 2 8.18 m - 8.516 m neight Raised willke, AX2 2.92 -171 = 1' 21 m.