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 .............
- A 1.2 m
- B 12 m
- C 0.12 m
- D 0.012 m
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…
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