Question

The amplitude of a wave disturbance propagating in the positive $x$-directions is given by
$y=\frac{1}{1+x^2}$ at time $t=0$ and by
$y=\frac{1}{[1+(x-1{)}^2]}$ at $t=2$ seconds
where $x$ and $y$ are in metres. The shape of the wave disturbance does not change curing the propagation, then the velocity of the wave is

• A. $0.5m{s}^{-1}$
• B. $1.0m{s}^{-1}$
• C. $2.0m{s}^{-1}$
• D. none of the above

Answer 1

The correct option is A $0.5m{s}^{-1}$

All the wave equations are of the form $f(x\pm vt)$ .
where v is the velocity.
here the equation is $\frac{1}{1+{(x-vt)}^2}$

$(x-1)=(x-vt)$
since at t =2,

$x-2v=x-1$;

$v=0.5m/s$