At time `t=0`,`y(x)` equation of a wave pulse is
`y=(10)/(2+(x-4)^(2))`
and at `t=2s`, `y(x)` equation of the same wave pulse is
`y=(10)/(2+(x+4)^(2))`
Here, `y` is in mm and `x` in metres. Find the wave velocity.

Answer

B, C

Explanation

From the given `y(x)` equations at two different times we can see that value of `y` is maximum `(=(10)/(2)or 5mm)` at `x=2m`at time `t=0`and at `x=-4` at time `t=2s`.
So,peak of the wave pulse has travelled a distance of `6m (from `x=2m` to `x = -4m`) in `2s` along negative x-direction.
Hence, the wave velocity is
`v=(-6)/(2)=-3m//s`