A travelling wave is described by the equation `y = y_(0) sin ((ft - (x)/lambda))`. The maximum particle velocity is equal to four times the wave velocity if

Options

A. `lamda=(piY_0)/(4)`
B. `lamda=(piy_0)/(2)`
C. `lamda=piY_0`
D. `lamda=2piY_0`

Answer

B

Explanation

comparing the given equation with `y=asin(omegat-kx)`.
We get `a=Y_0,omega=2pif,k=(2pi)/(lamda)`. Hence maximum paricle velocity `(v_(max))_("particle")=aomega=Y_0xx2pif` and wave velocity `(v)_(wave)=(omega)/(k)=(2pif)/(2pi//lamda)=flamda`
`(v_(max)_("particl e")=4v_(wave)impliesY_0xx2pif=4flamda`
`implieslamda=(piY_0)/(2)`