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MHT CET · Physics · Waves and Sound

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 \((1.87 \pi)\). If the velocity of sound wave in air is \(330 \mathrm{~m} / \mathrm{s}\), the frequency of this wave is

  1. A \(660 \mathrm{~Hz}\).
  2. B \(550 \mathrm{~Hz}\).
  3. C \(110 \mathrm{~Hz}\)
  4. D 367 Hz.
Verified Solution

Answer & Solution

Correct Answer

(B) \(550 \mathrm{~Hz}\).

Step-by-step Solution

Detailed explanation

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,
\(\mathrm{f}=\frac{(1.8 \pi)(330 \mathrm{~m} / \mathrm{s})}{2 \pi(0.54 \mathrm{~m})}=550 \mathrm{~Hz}\)
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