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. `lambda = pi Y_(0)//4`
B. `lambda = pi Y_(0)//2`
C. `lambda = pi Y_(0)`
D. `lambda = 2pi Y_(0)`

Answer

B

Explanation

`v = (dY)/(dt) = Y_(0)cos[2pi(ft-(x)/(lambda))]xx2pif`
or `v = 2pi fV_(0) cos[2pi(ft-(x)/(lambda))]`
The particle velocity is maximum, when
`cos[2pi(ft - x/(lambda))] = 1`
`:. v_(max) = 2pifY_(0)"….."(i)`
We know that, `Y = a sin (omegat - kx)`
The wave velocity V is given by
`V = (omega)/(k) = (2pif)/(2pi//lambda) = flambda"......."(2)`
Given that `v_(max) = 4V`
`:. 2 pif Y_(0) = 4flambda`
or `lambda = (piY_(0))/(2)`