Question

The amplitude of wave disturbance propagating in positive $x-$direction is given by $y=\frac{1}{1+x^2}$ at $t=0$ and $y=\frac{1}{1+(x-1{)}^2}$ at $t=2\text{s}$. The wave speed is

• A. $1.5\text{m/s}$
• B. $2\text{m/s}$
• C. $1\text{m/s}$
• D. $0.5\text{m/s}$

Answer 1

The correct option is D $0.5\text{m/s}$

Given, wave function
$y(x,0)=\frac{1}{1+x^2}....\left(i\right)$ at $t=0$
and at $t=2\text{s},y(x,2)=\frac{1}{1+(x-1{)}^2}....\left(ii\right)$
comparing Eq.$\left(ii\right)$ with the equation of travelling wave in format $y(x,t)=f(x-vt)$

$\Rightarrow$We get that $(x-1)=x-vt$
putting $t=2\text{s}$ and on equating,
$\Rightarrow 2v=1$
$\Rightarrow v=0.5\text{m/s}$
Hence, option $\left(d\right)$ is correct.