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 $t=1\text{s}$, where $x$ and $y$ are in meter. The shape of wave does not change during the propagation. The velocity of the wave in $\left(\text{m/s}\right)$ will be _______.

• A. 2.00
• B. 2.0
• C. 2

Answer 1

Answer: (A), (B), (C)

The wave equation, in general, is given by,

$y=f(kx-vt)$

$\therefore$ the equation of the given wave is, $y=\frac{1}{1+(x-vt{)}^2}...\left(1\right)$

Now,

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

At $t=1\text{s},y=\frac{1}{1+(x-2{)}^2}...\left(2\right)$

Comparing equations $\left(1\right)$ and $\left(2\right)$, we get,

$vt=2$

$\Rightarrow v\times 1=2$

$\Rightarrow v=2\text{m/s}$