Two sound waves with wavelengths $$5\,m$$  and $$5.5\,m$$  respectively, each propagate in a gas with velocity $$330\,m/s.$$  We expect the following number of beat per second

Solution :
Let $${\lambda _1} = 5.0\,m,v = 330\,m/s\,{\text{and}}\,{\lambda _2} = 5.5\,m$$
The relation between frequency $$\left( n \right),$$  wavelength $$\left( \lambda \right)$$ and velocity $$\left( v \right)$$ is given by
$$\eqalign{ & v = n\,\lambda \cr & \Rightarrow n = \frac{v}{\lambda }\,......\left( {\text{i}} \right) \cr} $$
The frequency corresponding to wavelength $${\lambda _1},$$
$${n_1} = \frac{v}{{{\lambda _1}}} = \frac{{330}}{{5.0}} = 66\,\,Hz$$
The frequency corresponding to wavelength $${\lambda _2},$$
$${n_2} = \frac{v}{{{\lambda _2}}} = \frac{{330}}{{5.5}} = 60\,\,Hz$$
Hence, number of beats per second
$$ = {n_1} - {n_2} = 66 - 60 = 6$$