Question

Two guitarists attempt to play the same note of wavelength 64.8 cm at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 65.2 cm instead. What is the frequency of the beats these musicians hear when they play together?

Answer 1

When two guitarists play the same note with wavelengths of 64.8 cm and 65.2 cm, respectively, the beat frequency heard is the absolute difference between the frequencies of these two notes, which is calculated to be 3.25 Hz.

Step-by-step explanation:

The question refers to the phenomenon of beat frequencies, which occur when two waves of slightly different frequencies interfere with each other. The situation involves two guitarists playing the same note with wavelengths of 64.8 cm and 65.2 cm, respectively. The beat frequency is the absolute difference between the frequencies of these two notes.

To calculate the frequencies of the notes played by the guitarists, we use the formula for the speed of a wave: v = f \u22c5 \u03bb, where v is the speed of the sound, f is the frequency, and \u03bb is the wavelength. The speed of sound at room temperature (≈ 20°C) is approximately 343 meters per second (m/s). Converting the wavelength from centimeters to meters, we have:

\u03bb1 = 64.8 cm = 0.648 m
\u03bb2 = 65.2 cm = 0.652 m

Now we can calculate the frequencies:

f1 = v / \u03bb1 = 343 m/s / 0.648 m ≈ 529.32 Hz
f2 = v / \u03bb2 = 343 m/s / 0.652 m ≈ 526.07 Hz

The beat frequency is the difference between these two frequencies:

Beat frequency = |f1 - f2| = |529.32 Hz - 526.07 Hz| = 3.25 Hz

Therefore, the musicians will hear a beat frequency of 3.25 Hz when they play together.