Two speakers placed 2.3 m apart produce pure tones in sync with each other at a

Question

Two speakers placed 2.3 m apart produce pure tones in sync with each other at a frequency of 520 Hz. A microphone can be moved along a line parallel to the line joining the speakers and 8.2m from it. An intensity maximum is measured a point O0 where the microphone is equidistant from the two speakers. As we move the microphone away from P0 to one side, we find intensity minima and maxima alternately. Take the speed of sound in air to be 344 m/s

a) what is the distance, in meters, between P0 and the first intensity minimum?

b) what is the distance, in meters, between P0 and the first intensity maximum?

c) what is the distance, in meters, between P0 and the second intensity minimum?

d) what is the distance, in meters, between P0 and the second intensity maximum?

e) what is the distance, in meters, between P0 and the third intensity maximum?

Explanation / Answer

lambda = v / f = (344 m/s) / 520 Hz

= 0.6615 m

(A) for intensity to minimum,

path diff = (2n + 1) lambda / 2

for first, n = 0

path diff = lambda / 2 = 0.33 m

sqrt[(8.2^2) + (2.3/2 + x)^2] - sqrt[8.2^2 + (2.3/2 - x)^2] = 0.33

x = 1.20 m

(B) for maximum, path diff = n lambd

n = 1

path diff = 0.6615 m

sqrt[(8.2^2) + (2.3/2 + x)^2] - sqrt[8.2^2 + (2.3/2 - x)^2 ] = 0.6615

x = 2.48 m

(C) path diff = (2n + 1) lambda / 2

n = 1

path diff = 0.99225

x = 3.95 m

(D) path diff = n lambda

n = 2

path diff = 1.323

x = 5.80 m

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