WBJEE · Physics · Waves and Sound
A travelling acoustic wave frequency \(500 \mathrm{Hz}\) is moving along the positive \(x\) -direction with a velocity of \(300 \mathrm{ms}^{-1}\). The phase difference between two points \(x_{1}\) and \(x_{2}\) is \(60^{\circ} .\) Then the minimum separation between the two points is
- A \(1 \mathrm{mm}\)
- B \(1 \mathrm{cm}\)
- C \(10 \mathrm{cm}\)
- D \(10 \mathrm{mm}\)
Answer & Solution
Correct Answer
(C) \(10 \mathrm{cm}\)
Step-by-step Solution
Detailed explanation
By using the relation, \(v=v \lambda\) We have, \(\quad \lambda=\frac{V}{v}=\frac{300}{500}=\frac{3}{5} \mathrm{m}\) The phase difference, \(\phi=\frac{2 \pi}{\lambda}(\Delta x)\)…
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