Q 11-14-127JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
A progressive wave travelling along the positive $x$-direction is represented by $y(x, t) = A\sin(kx - \omega t + \phi)$. Its snapshot at $t = 0$ is given in the figure. For this wave, the phase $\phi$ is:
Answer: (C) $\pi$
At $t = 0$: $y = A\sin(kx + \phi)$. From the graph $y(0) = 0$, so $\phi = 0$ or $\pi$.
The displacement becomes negative just to the right of the origin, i.e. $y \approx -Akx$ near $x = 0$. Since $\sin(kx + \pi) = -\sin kx$, $\phi = \pi$.
Solution by Sreeraj P, M.Sc Physics