A transverse wave is derscried by the equation `y=y_(0) sin 2 pi (ft - (x)/(lamda))`. The maximum particle velocity is equal to four times the wave velocity if :-

Options

A. `lamda = (pi y_(0))/(2)`
B. `pi y_(0)`
C. `lamda = (4)/(pi y_(0))`
D. `lamda = (4pi)/(y_(0))`

Answer

A

Explanation

`y = y_(0) sin 2pi (n t - (x)/(lamda))`
`:. V = (dy)/(dt) = y_(0) cos 2pi (nt - (x)/(lamda)) xx 2pi n`
This is maximum if `cos 2pi (nt - (x)/(lamda)) = 1`
`:. V_("max") = y_(0) 2pi n` and wave velocity `= n lamda`
If `n lamda = (1)/(4) (y_(0) 2pi n), and lamda = (y_(0) .2pi n)/(4n) = (pi y_(0))/(2)`