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