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JEE Mains · Physics · STD 11 - Waves and sound
  • JEE Mains 2023
  • Hard
  • Numerical

The distance between two consecutive points with phase difference of \(60^{\circ}\) in a wave of frequency \(500\,Hz\) is \(6.0\,m\). The velocity with which wave is traveling is \(.........km / s\)

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Answer & Solution

Correct Answer

(D) \(18\)

Step-by-step Solution

Detailed explanation

\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\)

\(\frac{\pi}{3}=\frac{2 \pi}{\lambda}(6 m )\)

\(\lambda=36\,m\)

\(V = f \lambda=(500\,Hz )(36\,m )\)

\(=18000\,m / s\)

\(=18\,km / s\)

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