Velocity of sound waves in air is $$330\,m/s.$$  For a particular sound wave in air, path difference of $$40\,cm$$  is equivalent to phase difference of $$1.6\,\pi .$$  The frequency of this wave is

Solution :
At a given time ($$t$$ = constant), the phase changes with position $$x.$$ The phase change $$\left( {\Delta \phi } \right)$$  at a given time for a wavelength $$\left( \lambda \right)$$  for a distance $$\Delta x$$  is given by
$$\eqalign{ & \Delta \phi = \frac{{2\pi }}{\lambda }\Delta x\,.......\left( {\text{i}} \right) \cr & {\text{From}}\,{\text{Eq}}{\text{.}}\,\left( {\text{i}} \right) \cr & \Delta x = \frac{\lambda }{{2\,\pi }} \cdot \Delta \phi \,\,{\text{or}}\,\,\lambda = 2\,\pi \cdot \frac{{\Delta x}}{{\Delta \phi }} \cr & {\text{Here,}}\,\,\Delta x = 0.4\,m \cr & \Delta \phi = 1.6\,\pi \cr & \therefore \lambda = 2\,\pi \cdot \frac{{0.4}}{{1.6\,\pi }} = 0.5 \cr & \therefore {\text{Frequency of wave is }}\nu = \frac{v}{\lambda } = \frac{{330}}{{0.5}} = 660\,Hz \cr & {\text{where,}}\,\,v = 330\,m/s = {\text{velocity}}\,{\text{of}}\,{\text{sound}} \cr} $$