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 $t=2 \, s$ , 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. $0.5ms^{- 1}$
• B. $2.0ms^{- 1}$
• C. $1.0ms^{- 1}$
• D. $4.0ms^{- 1}$

Answer 1

Answer: (A)

$0.5ms^{- 1}$

Answer 2

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