Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of 536 Hz. A microphone is moved between the speakers along the line joining the two speakers with a
Constant speed of 1.60 m/s. What beat frequency is observed? The speed of sound in air is 343 m/s.
A) 5.00 Hz
B) 9.00 Hz
C) 7.00 Hz
D) 8.00 Hz
E) 6.00 Hz

Question

Quizplus · Accepted Answer

The beat frequency is given by:$$f_b = \frac{v}{d} \sin 	heta$$where:* $v$ is the speed of sound* $d$ is the distance between the speakers* $	heta$ is the angle between the line joining the speakers and the microphoneIn this case, the speakers are in-phase, so $	heta = 0$. Therefore, the beat frequency is:$$f_b = \frac{v}{d}$$Substituting the given values, we get:$$f_b = \frac{343 	ext{ m/s}}{0.500 	ext{ m}} = 686 	ext{ Hz}$$The microphone is moving at a constant speed of 1.60 m/s, so the beat frequency will vary as the microphone moves between the speakers. The maximum beat frequency will occur when the microphone is halfway between the speakers, and the minimum beat frequency will occur when the microphone is at one of the speakers. The difference between the maximum and minimum beat frequencies is:$$\Delta f_b = 2f_b = 2(686 	ext{ Hz}) = 1372 	ext{ Hz}$$The beat frequency observed by the microphone will be between $f_b - \Delta f_b/2$ and $f_b + \Delta f_b/2$, which is between 5.00 Hz and 9.00 Hz. Therefore, the correct answer is A.

Two In-Phase Loudspeakers Are Some Distance Apart