Question

The amplitude of a wave disturbance propagating in the positive x direction is given by $y=\frac{1}{1+x^2}$
at $t=0$ in metres and $y=\frac{1}{1+(x-1{)}^2}$ at $t=2s$ where $x,y$ are in metres.
If the shape of the wave does not changing with time, the velocity of the wave is:

• A. $2m{s}^{-1}$
• B. $0.5m{s}^{-1}$
• C. $3m{s}^{-1}$
• D. $1m{s}^{-1}$

Answer 1

Answer: (B)

$0.5m{s}^{-1}$

At $t=t_1=0,$ $1+x^2=\frac{1}{y}$ or $x^2=\frac{1-y}{y}$

or, $x=\sqrt{\frac{1-y}{y}}=x_1$

At $t=t_2=2s,$ $1+(x-1{)}^2=\frac{1}{y}$ or $(x-1{)}^2=\frac{1-y}{y}$

or $x=1+\sqrt{\frac{1-y}{y}}=x_2$

Thus, the velocity of the wave is $v=\frac{x_2-x_1}{t_2-t_1}=\frac{1}{2-0}=0.5m{s}^{-1}$