A travelling wave in a stretched string is described by the equation `y = A sin (kx - omegat)` the maximum particle velocity is

Options

A. `Aomega`
B. `omega//k`
C. `domega//dk`
D. `x//l`

Answer

A

Explanation

Given that, the displacement of a particle is
`y=Asin(omegat-kx)" "......(i)`
The particle velocity
`v_(p)=(dy)/(dt)" "......(ii)`
Now, on differentiating Eq. (i) w.r.t.t,
`(dy)/(dt)=Acos(omegat-kx).omega`
`implies(dy)/(dx)=Aomegacos(omegat-kx)`
From Eq. (ii)
`impliesv_(p)=Aomegacos(omegat-kx)`
For maximum particle velocity,
`cos(omegat-dx)=1`
So, `v_(p)=Aomegaxx1`
`impliesv_(p)=Aomega`