Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of 1536 Hz. You move between the speakers, along the line joining them, at a constant speed of 2.8 m/s. What beat frequency do you observe? The speed of sound in the room is 330 m/s.
A) 13 Hz
B) 431 Hz
C) 26 Hz
D) 4.7 Hz
E) 118 Hz

Question

Quizplus · Accepted Answer

The beat frequency is given by the absolute difference between the frequencies emitted by the two loudspeakers, which in this case is 0 Hz because they emit the same frequency, minus the Doppler shifted frequency due to the motion of the observer.

The Doppler shift in frequency is given by:

f_obs = f*(v_sound + v_obs)/(v_sound + v_src)

where:
- f is the emitted frequency (1536 Hz in this case)
- v_sound is the speed of sound (330 m/s in this case)
- v_obs is the speed of the observer (2.8 m/s in this case)
- v_src is the speed of the source (0 m/s in this case, since the speakers are stationary)

Plugging in the numbers, we get:

f_obs = 1536*(330 + 2.8)/(330 + 0) ≈ 1548 Hz

Finally, the beat frequency is:

f_beat = |f - f_obs| = |1536 - 1548| ≈ 12 Hz

However, since the two loudspeakers are in phase, we actually hear a sinusoidal tone at the frequency of 1536 Hz, not a beat tone. Thus, there is no correct answer among the choices given.

Two In-Phase Loudspeakers Are Some Distance Apart