Two sound waves with wavelengths $5.0 ext{ m}$ and $5.5 ext{ m}$, respectively, propagate in a gas with a

Question

Two sound waves with wavelengths $5.0 	ext{ m}$ and $5.5 	ext{ m}$, respectively, propagate in a gas with a velocity of $330 	ext{ m/s}$. How many beats per second can we expect?

Accepted Answer

Given:
Velocity of sound, $v = 330 	ext{ m/s}$
Wavelength of the first wave, $\lambda_1 = 5.0 	ext{ m}$
Wavelength of the second wave, $\lambda_2 = 5.5 	ext{ m}$

The frequencies of the two sound waves can be calculated using the relation $f = \frac{v}{\lambda}$:
$$f_1 = \frac{v}{\lambda_1} = \frac{330}{5.0} = 66 	ext{ Hz}$$
$$f_2 = \frac{v}{\lambda_2} = \frac{330}{5.5} = 60 	ext{ Hz}$$

The number of beats produced per second is the difference between their frequencies:
$$	ext{Number of beats} = f_1 - f_2 = 66 - 60 = 6$$

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