The amplitude of wave disturbance propagating in
Question:

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 $\mathrm{y}$ are in meres. The shape of wave does not change during the propagation. The velocity of the wave will be______ $\mathrm{m} / \mathrm{s}$.

Solution:

At $t=0, y=\frac{1}{1+x^{2}}$

At time $\mathrm{t}=\mathrm{t}, \mathrm{y}=\frac{1}{1+(\mathrm{x}-\mathrm{vt})^{2}}$

At $\mathrm{t}=1, \mathrm{y}=\frac{1}{1+(\mathrm{x}-\mathrm{v})^{2}} \ldots$ (i)

At $\mathrm{t}=1, \mathrm{y}=\frac{1}{1+(\mathrm{x}-2)^{2}} \ldots$ (ii)

Comparing (i) \& (ii)

$v=2 \mathrm{~m} / \mathrm{s}$