A transverse wave is represented by `y=Asin(omegat-kx)`. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

Options

A. `piA//2`
B. `piA`
C. `2piA`
D. A

Answer

C

Explanation

The given wave equation is `y=Asin(omegat-kx)`
Wave velocity, `v=(omega)/(k)" "…(i)`
Particle velocity, `v_(p)=(dy)/(dt)=Aomegacos(omegat-kx)`
Maximum particle velocity, `(v_(p))_(max)=Aomega" "…(ii)`
According to the given question
`v=(v_(p))_(max)`
`(omega)/(k)=Aomega` (Using (i) and (ii))
`(1)/(k)=A or (lambda)/(2pi)=A" "(because k=(2pi)/(lambda))`
`lambda=2piA`