Q 11-14-037JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
The equation of a transverse wave travelling along a string is $y(x,t) = 4.0\sin\left[20\times10^{-3}x + 600t\right]\ \text{mm}$, where $x$ is in mm and $t$ is in seconds. The velocity of the wave is
Answer: (B) $-30\ \text{m/s}$
Compare with $y = A\sin(kx + \omega t)$: $k = 20\times10^{-3}\ \text{mm}^{-1}$, $\omega = 600\ \text{s}^{-1}$.
$$v = \frac{\omega}{k} = \frac{600}{20\times10^{-3}} = 3\times10^4\ \text{mm/s} = 30\ \text{m/s}$$
The $+$ sign between $kx$ and $\omega t$ means the wave travels in the $-x$ direction, so the velocity is $-30\ \text{m/s}$.
Solution by Sreeraj P, M.Sc Physics