Question

Two wavelength of frequency 540 Hz travel at a speed of 330 m/s. If the sources are in phase, in the beginning, the phase difference of the wave at a point 4 m from one source and 4.4 m from the other is:

Answer 1

Phase difference $=\varphi =2\pi \frac{\Delta x}{\lambda }$ where $\Delta x=$ path difference

And $\lambda =\frac{v}{f}$

Hence,$\varphi =2\pi \frac{\Delta x.f}{v}$

$⟹\varphi =\frac{0.4\times 540}{330}\times {360}^o$

$⟹\varphi ={235.6}^o$