Question

The amplitude of a wave disturbance propagating in the positive $X-direction$ is given by
$y=\frac{1}{2+x^2}$ at $t=0$ and $y=\frac{1}{[2+(x-1{)}^2]}$ at $t=3s$, where $x$ and $y$ are in metre. It the shape of the wave disturbance does not change during the propagation, the velocity of the wave is

• A. $\frac{1}{2}m{s}^{-1}$
• B. $\frac{1}{3}m{s}^{-1}$
• C. $\frac{1}{4}m{s}^{-1}$
• D. $\frac{1}{5}m{s}^{-1}$

Answer 1

Answer: (B)

$\frac{1}{3}m{s}^{-1}$

At $t=0,y=\frac{1}{2+x^2}$
or $2y+y{x}^2=1$
or $y{x}^2=+1-2y$
or $x=\sqrt{\frac{1-2y}{y}}=x_1$ (say)
At $t=3s,y=\frac{1}{[2+(x-1{)}^2]}$
or $2+(x-1{)}^2=\frac{1}{y}$
or $(x-1{)}^2=\frac{1}{y}-2=\frac{1-2y}{y}$
or $x=1+\sqrt{\frac{1-2y}{y}}=x_2$ (say)
$\therefore v=\frac{x_2-x_1}{t_2-t_1}=\frac{1+\sqrt{\frac{1-2y}{y}}-\sqrt{\frac{1-2y}{y}}}{3-1}=\frac{1}{3}m{s}^{-1}$.