Two sound waves with wavelengths $4.0 \mathrm{~m}$ and $4.1 \mathrm{~m}$ respectively, propagate in

Question

Two sound waves with wavelengths $4.0 \mathrm{~m}$ and $4.1 \mathrm{~m}$ respectively, propagate in a gas with a velocity of $328 \mathrm{~m/s}$. The expected number of beats per second is

Professor PC Thomas Classes · Answer

(a) : The frequency of a wave is given by $f = \frac{v}{\lambda}$. For the first wave, $f_1 = \frac{328}{4.0} = 82 \mathrm{~Hz}$. For the second wave, $f_2 = \frac{328}{4.1} = 80 \mathrm{~Hz}$. The number of beats per second is equal to the difference in their frequencies: $f_1 - f_2 = 82 - 80 = 2$.