ExamBro
ExamBro
AP EAMCET · PHYSICS · Waves and Sound

A progressive wave of frequency \(500 \mathrm{~Hz}\) is travelling with a velocity of \(360 \mathrm{~ms}^{-1}\). The distance between the two points, having a phase difference of \(60^{\circ}\) is .............

  1. A 1.2 m
  2. B 12 m
  3. C 0.12 m
  4. D 0.012 m
Verified Solution

Answer & Solution

Correct Answer

(C) 0.12 m

Step-by-step Solution

Detailed explanation

\[ \begin{gathered} f=500 \mathrm{~Hz}, v=360 \mathrm{~ms}^{-1} \\ \lambda=\frac{v}{f}=\frac{360}{500} \end{gathered} \] Now, a phase difference of \(60^{\circ}\) corresponds to a path difference of \(\frac{60}{360} \times \lambda\). So, distance between 2 particles is…
Same subject
Explore more questions on app