Two identical loudspeakers are driven in phase by the same amplifier.The speakers are positioned a distance of 3.2 m apart.A person stands 4.1 m away from one speaker and 4.8 m 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 340 m/s.
A)245 Hz
B)735 Hz
C)1225 Hz
D)490 Hz
E)1470 Hz

Question

Quizplus · Accepted Answer

The path difference between the two speakers at the point where the person is standing is given by:
Δx = x2 - x1 = 4.8 - 4.1 = 0.7 m

For destructive interference, the path difference must be an odd multiple of half-wavelength, i.e., Δx = (2n + 1)λ/2, where n is an integer.

Rearranging the above equation, we get:
λ = 2(Δx)/(2n + 1)

Substituting the given values, we get:
λ = 2(0.7)/(2n + 1) = 1.4/(2n + 1)

For the second lowest frequency, n = 1, since we want the second odd multiple of half-wavelength. Hence,
f = v/λ = v(2n + 1)/2Δx = 340(2(1) + 1)/2(0.7) = 735 Hz

Therefore, the answer is B.

Two Identical Loudspeakers Are Driven in Phase by the Same