Two loudspeakers are separated by a distance of 6.3 m. A listener sits directly

Question

Two loudspeakers are separated by a distance of 6.3 m. A listener sits directly in front of one speaker at a distance of 8.7 m so that the two speakers and the listener form a right triangle. Find the lowest frequency for which the path difference from the speakers to the listener is an odd number of hafl-wavelengths. Assume the speed of sound is 340 m/s.

Answer: ?

Find the second lowest frequeny for which the path difference from the speakers to the listener is an odd number of half-wavelengths.

Answer: ?

Explanation / Answer

Distance between the listener and the first speaker, x1 = 8.7 m

Distance between the listener and the second speaker, x2 = (8.72 + 6.32)1/2 = 10.74 m

Path difference, x = x2 - x1 = (10.74 - 8.7) m = 2.04 m

Now, x = n/2 where n = 1,3,5,7...

Since, wavelength, = c/f where c= 340 m/s and f = frequency of sound wave

we have,

x = n(c/2f)

=> f = n(c/2*x) = n[340/(2*2.04)] = 83.3n

For lowest frequency fl , n = 1

So, fl = 83.3 * 1 = 83.3 Hz

For second lowest frequency fsl , n = 3

So, fsl = 83.3 * 3 = 249.9 Hz

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