Question

The displacement of a wave disturbance propagating in the positive x-direction is given by $y=\frac{1}{1+x^2}$ at t = 0 and $y=\frac{1}{1+(x-1{)}^2}$ at t = 2s. Where, x and y are in metre. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is $\frac{x}{2}m/s$. Find the value of x.

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

The correct option is A 1

$y(x,0)=\frac{1}{1+x^2}$
$y(x,2)=\frac{1}{1+{(x-1)}^2}$
If the speed of the pulse is $v$,
$y(x,t)=y(x-vt,0)$
$y(x,2)=y(x-2v,0)$
Therefore,
$\frac{1}{1+{(x-1)}^2}=\frac{1}{1+{(x-2v)}^2}$
Comparing LHS and RHS,
$v=\frac{1}{2}m/s^2$
Hence, $x=1$