The amplitude of wave disturbance propagating in the positive $x$-direction is given by $y=\frac{1}{(1+x)^{2}}$ at time $t=0$ and $y=\frac{1}{1+(x-2)^{2}}$ 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 $...\,{m} / {s}.$
JEE MAIN 2021, Diffcult
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At ${t}=0, {Y}=\frac{1}{1+{x}^{2}}$
$\text { At time } t=t, y=\frac{1}{1+(x-v t)^{2}}$
$\text { At } t=1, y=\frac{1}{1+(x-v)^{2}}....(i)$
$\text { At } t=1, y=\frac{1}{1+(x-2)^{2}}....(ii)$
comparing $(i) \& (ii)$
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