Interference: 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 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

To have destructive interference, the path difference between the two waves must be half of the wavelength (λ/2). Therefore, we can use this formula:

path difference = d2 - d1 = (n + 1/2)λ

where d1 is the distance from the person to the first loudspeaker, d2 is the distance from the person to the second loudspeaker, and n is an integer (0,1,2,...).

When the person is at the point of destructive interference, the path difference is equal to λ/2. We can write this as:

5 - (12 - x) = (1/2)λ

where x is the distance from the first loudspeaker to the point of destructive interference. Solving for x, we get:

x = 9 - (1/2)λ

Now we need to find the wavelength of the sound. We can use the formula:

v = fλ

where v is the speed of sound (343 m/s) and f is the frequency of the sound. Solving for the wavelength, we get:

λ = v/f

Substituting this into the expression for x, we get:

9 - (1/2)(v/f) = x

Substituting the given values, we get:

9 - (1/2)(343/f) = 12 - 1

Solving for f, we get:

f = 422 Hz

Therefore, the answer is C.

Interference: Two In-Phase Loudspeakers That Emit Sound with the Same