MHT CET · Physics · Waves and Sound
A progressive wave of frequency 400 Hz is travelling with a velocity \(336 \mathrm{~m} / \mathrm{s}\). How far apart are the two points which are \(60^{\circ}\) out of phase?
- A 0.14 m
- B 0.21 m
- C 0.24 m
- D 0.28 m
Answer & Solution
Correct Answer
(A) 0.14 m
Step-by-step Solution
Detailed explanation
Phase difference \((\phi)=60^{\circ}\) or \(\frac{\pi}{3} \mathrm{rad}\)
\(\therefore \quad\) Path difference \(=\frac{\lambda}{2 \pi} \times \phi=\frac{\lambda}{2 \pi} \times \frac{\pi}{3}=\frac{\lambda}{6}\)
\(\therefore \quad\) Path difference \(=\frac{\lambda}{6}=\frac{v}{n} \times \frac{1}{6} \quad \ldots(\because v=\mathrm{n} \lambda)\)
Substituting given values,
Path difference \(=\frac{336}{400} \times \frac{1}{6}=0.14 \mathrm{~m}\)
\(\therefore \quad\) Path difference \(=\frac{\lambda}{2 \pi} \times \phi=\frac{\lambda}{2 \pi} \times \frac{\pi}{3}=\frac{\lambda}{6}\)
\(\therefore \quad\) Path difference \(=\frac{\lambda}{6}=\frac{v}{n} \times \frac{1}{6} \quad \ldots(\because v=\mathrm{n} \lambda)\)
Substituting given values,
Path difference \(=\frac{336}{400} \times \frac{1}{6}=0.14 \mathrm{~m}\)
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