The amplitude of wave disturbance propagating in the positive $x$-direction is given by $\mathrm{y}=\frac{1}{1+(\mathrm{x})^{2}}$ at time $\mathrm{t}=0$ and $\mathrm{y}=\frac{1}{1+(\mathrm{x}-2)^{2}}$ at $\mathrm{t}=1 \mathrm{~s}$, where x and y are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be \_\_\_\_ $\mathrm{m} / \mathrm{s}$.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Waves & Sound

The amplitude of wave disturbance propagating in the positive

x

-direction is given by

y = \frac{1}{1 + ( x ) ^{2}}

at time

t = 0

and

y = \frac{1}{1 + ( x - 2 ) ^{2}}

t = 1 s

, where x and y are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ____

m / s

Answer: 2

The disturbance peaks where the denominator is smallest. At

t = 0

that happens at

x = 0

, and at

t = 1

s it happens where

x - 2 = 0

, that is at

x = 2

m. The shape is unchanged, so the whole profile has merely been translated

2

m along the positive

x

-direction in

1

s. The wave speed is

\frac{2}{1} = 2

m/s.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer