Question

Two sound waves, originating from the same source, travel along different paths in air and then meet at a point. If the source vibrates at a frequency of $1\text{kHz}$ and one path is $83\text{cm}$ shorter than the other, then the phase difference between the waves arriving at the point of observation is ____$\pi$. The speed of sound in air is $332{\text{ms}}^{-1}.$

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Answer 1

Given,
$f=1\text{kHz}=1000\text{Hz}$
$v=332{\text{ms}}^{-1}$

Path difference, $\Delta x=83\text{cm}=0.83\text{m}$

Now, wavelength of sound wave,

$\lambda =\frac{v}{f}=\frac{332}{1000}=0.332\text{m}$

Phase difference, $\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$

$\Rightarrow \Delta \varphi =\frac{2\pi }{0.332}\times 0.83$

$\therefore \Delta \varphi =5\pi$

Hence, $5$ is the correct answer.

Why this question ? Concept : Phase difference is equal to product of angular wave number and path difference.