Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.00 m.A person is standing 12.0 m away from the wall,equidistant from the loudspeakers.When the person moves 1.00 m parallel to the wall,she hears no sound (experiences destructive interference)for the first time.What is the frequency of the sound? The speed of sound in air is 343 m/s.
A)211 Hz
B)256 Hz
C)422 Hz
D)512 Hz
E)674 Hz

Question

Quizplus · Accepted Answer

The distance between the person and both the loudspeakers is constant and equal to 12 m.
Let's consider the path difference between the two waves when they reach the person after traveling from the two loudspeakers. The path difference should be equal to (2n+1)λ/2 (where n is an integer) for constructive interference and nλ (where n is an integer) for destructive interference. We have destructive interference when the person moves 1 m parallel to the wall, so we can say that the path difference is equal to one wavelength:

d = nλ   ==>   λ = d/n

where d is the distance between the person and the wall, and n is an integer.

The distance between the two loudspeakers is 5 m, so the distance between one loudspeaker and the person is:

x1 = sqrt(12^2 + (5/2)^2) = 12.525 m

Similarly, the distance between the other loudspeaker and the person is:

x2 = sqrt(12^2 + (5/2)^2) = 12.525 m

When the person is at a distance d = 12 ± 1 m from the midpoint between the two loudspeakers, the path difference is equal to λ, so we have destructive interference. We can find the frequency of the wave using the above equation:

λ = d/n  ==>  f = v/λ = n*v/d

Taking n = 1 and d = 13 m (because the person is at 12 m or 14 m), we get:

f = 1*343/13 = 26.38 Hz, which is not one of the answer choices.
Taking n = 2 and d = 11 m (because the person is at 10 m or 12 m), we get:

f = 2*343/11 ≈ 422 Hz, which is closest to answer choice C.

Two In-Phase Loudspeakers That Emit Sound with the Same Frequency