For a particular sound wave propagating in air, a path difference between two points is $0.54 \mathrm{~m}$…
For a particular sound wave propagating in air, a path difference between two points is $0.54 \mathrm{~m}$ which is equivalent to phase difference of $7$. If the velocity of sound wave in air is $330 \mathrm{~m} / \mathrm{s}$, the frequency of this wave is
$660 \mathrm{~Hz}$.
$550 \mathrm{~Hz}$.
$110 \mathrm{~Hz}$
367 Hz.
Solution
The correct option is (B).
Concept: Path difference and phase difference are related by,
$\frac{2 \pi \mathrm{X}}{\lambda}=\phi$
where $\mathrm{x}$ is the path difference, $\phi$ the phase difference and $\lambda$ the wavelength.
The velocity of a wave is given by: $v=\lambda f$, where $f$ is the frequency.
Combining the above two equations:
$f=\frac{\phi v}{2 \pi x}$
Given, $x=0.54 \mathrm{~m}, \phi=(1.8 \pi)$ and $\mathrm{v}=330 \mathrm{~m} / \mathrm{s}$, therefore,
$\mathrm{f}=\frac{(1.8 \pi)(330 \mathrm{~m} / \mathrm{s})}{2 \pi(0.54 \mathrm{~m})}=550 \mathrm{~Hz}$