Question

The equation of a wave disturbance propagating in the positive Y-direction is given by $y=\frac{1}{1+x^2}$ at $t=0$ and $y=\frac{1}{[1+(x-1{)}^2]}$ at $t=2s$, where $x$ and $y$ are in $\text{metres}$. If the shape of the wave disturbance does not change during the propagation, what is the velocity of the wave?

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

Answer 1

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

Given:
At $t=0,y=\frac{1}{1+x^2}$
or $x=\sqrt{\frac{1-y}{y}}=x_1$
At $t=2\text{s},y=\frac{1}{[1+(x-1{)}^2]}$
or $x=1+\sqrt{\frac{1-y}{y}}=x_2$

Since the shape of the wave doesn't change during propagation, we can find velocity of the wave by calculating the distance travelled by a point having certain displacement $y$ in given time interval.
$\therefore v=\frac{x_2-x_1}{t_2-t_1}=\frac{1+\sqrt{\frac{1-y}{y}}-\sqrt{\frac{1-y}{y}}}{2-0}=\frac{1}{2}=0.5\text{m/s}$
Hence, the correct answer is option (c).