Two sound waves of wavelengths $1.00 \mathrm{~m}$ and $1.02 \mathrm{~m}$ produce 6 beats per second

Question

Two sound waves of wavelengths $1.00 \mathrm{~m}$ and $1.02 \mathrm{~m}$ produce 6 beats per second in a certain gas. The velocity of sound in the gas is

Professor PC Thomas Classes · Answer

(a) : Let the velocity of sound in the gas be $v$. The frequencies of the two waves are $f_1 = \frac{v}{\lambda_1} = \frac{v}{1.00}$ and $f_2 = \frac{v}{\lambda_2} = \frac{v}{1.02}$. The beat frequency is $f_1 - f_2 = 6$. Substituting the frequencies, we get $v\left(\frac{1}{1.00} - \frac{1}{1.02}ight) = 6$. This simplifies to $v\left(\frac{1.02 - 1.00}{1.02}ight) = 6 \Rightarrow v\left(\frac{0.02}{1.02}ight) = 6$. Solving for $v$, we get $v = \frac{6 	imes 1.02}{0.02} = 306 \mathrm{~m/s}$.