Two loudspeakers S₁ and S₂, 3.0 m apart, emit the same single-frequency tone in phase at the speakers. A listener L directly in front of speaker S₁ notices that the intensity is a minimum when she is 4.0 m from that speaker see figure). What is the lowest frequency of the emitted tone? The speed of sound in air is 340 m/s. A) 85 Hz B) 0.17 kHz C) 0.26 kHz D) 0.34 kHz E) 0.51 kHz

Question

Quizplus · Accepted Answer

When the two loudspeakers emit a single-frequency tone in phase, we get an interference pattern of alternating constructive and destructive interference. The condition for destructive interference is given by:
Δx = nλ/2  (1)
where Δx is the path difference between the two speakers, n is an integer, and λ is the wavelength of the tone.
In this problem, the two speakers are 3.0 m apart, and the listener is at a distance of 4.0 m from one of the speakers. So, the path difference is:
Δx = 4.0 m - 3.0 m = 1.0 m
Substituting this and the speed of sound in air (340 m/s) into equation (1), we get:
1.0 m = nλ/2
λ = 2(1.0 m)/n
The lowest frequency corresponds to n = 1, which gives:
λ = 2.0 m
f = v/λ = 340 m/s / 2.0 m = 170 Hz = 0.17 kHz
Therefore, the answer is B.

Two Loudspeakers S1 and S2, 3

Two Loudspeakers S₁ and S₂, 3