Two identical loudspeakers are driven in phase by the same amplifier. The speake

Question

Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2m apart. A person stands 4.1m away from one speaker and 4.8m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be v = 340m/s.
1) PLEASE EXPLAIN THE EQUATIONS YOU USE AND ESPECIALLY HOW YOU PICK YOUR N VALUES
2) The answer is 729 Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2m apart. A person stands 4.1m away from one speaker and 4.8m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be v = 340m/s.
1) PLEASE EXPLAIN THE EQUATIONS YOU USE AND ESPECIALLY HOW YOU PICK YOUR N VALUES
2) The answer is 729
1) PLEASE EXPLAIN THE EQUATIONS YOU USE AND ESPECIALLY HOW YOU PICK YOUR N VALUES
2) The answer is 729

Explanation / Answer

The path difference is:

d = r2 - r1 = 4.8 - 4.1 = 0.7

For the destructive interference,

d = r2 - r1 = n x lambda/2

where, n = 1,2,3... ; lambda = wavelength

we know that, v = f lambda => lambda = v/f

d = n x v/2f

f = n x v/2d

for the second lowest frequency, n = 3

f = 3 x 340/2 x 0.7 = 728.6 Hz = 729 Hz(aprox)

Hence, f = 729 Hz.

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