Question

The amplitude of a wave disturbance propagating in positive direction of $x-axis$ is given by $y=\frac{1}{1+x^2}$ at $t=0$ and by $y=\frac{1}{1+{\left(x-1\right)}^2}$ at $t=2s$ , where $x$ and $y$ are in meters. The shape of the wave disturbance does not change during propagation. The velocity of the wave is

• A. $0.5m{s}^{-1}$
• B. $2.0m{s}^{-1}$
• C. $1.0m{s}^{-1}$
• D. $4.0m{s}^{-1}$

Answer 1

Answer: (A)

In a wave equation, $x$ and $t$ must be related in the form $(x-vt)$ . Therefore, we rewrite the given equation as
$y=\frac{1}{1+{\left(x-vt\right)}^2}$
For $t=0,$ it becomes $y=\frac{1}{1+x^2}$
And for $t=2$ , it becomes
$y=\frac{1}{\left[1+{\left(x-2v\right)}^2\right]}=\frac{1}{1+{\left(x-1\right)}^2}$
$\therefore 2v=1$ or $v=0.5m{s}^{-1}$