Q 11-14-086JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
The amplitude of wave disturbance propagating in the positive $x$-direction is given by $y = \dfrac{1}{(1+x)^2}$ at time $t = 0$ and $y = \dfrac{1}{1 + (x - 2)^2}$ at $t = 1$ s, where $x$ and $y$ are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ______ $\text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 2
The pulse peak moves from $x = 0$ at $t = 0$ (taking the intended form $y = \dfrac{1}{1 + x^2}$) to $x = 2$ m at $t = 1$ s, with unchanged shape.
$v = \dfrac{2\ \text{m}}{1\ \text{s}} = 2\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics