Q 11-14-054JEE MainJEE Main 2023 (25 Jan, Shift 1)Easy
The distance between two consecutive points with phase difference of $60^\circ$ in a wave of frequency $500\ \text{Hz}$ is $6.0\ \text{m}$. The velocity with which wave is travelling is ______ $\text{km s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 18
A phase difference of $60^\circ=\dfrac{\pi}{3}$ corresponds to $\dfrac\lambda6$, so $\lambda=36\ \text{m}$.
$v=f\lambda=500\times36=18000\ \text{m s}^{-1}=18\ \text{km s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics