Two identical loudspeakers are driven in phase by a common o

Two identical loudspeakers are driven in phase by a common oscillator at 820 Hz and face each other at a distance of 1.20 m. Locate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Take the speed of sound in air to be 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker. Enter \'none\' in all unused answer boxes.)

Solution

v = f lambda = f * 2 d

343 = 820 * 2 d

d =0.209 m

If the speakers vibrate in phase, the point halfway between them is anantinode of pressure at a distance from either speaker of = 1.20 / 2 = 0.60 m

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0.60 - (1/2 * 0.209) = 0.496 m

0.496 - 0.209 = 0.287 m

0.287 - 0.209 = 0.0775 m

0.496 + 0.209 = 0.705 m

0.705 + 0.209 = 0.914 m

0.914 + 0.209 =1.12 m