A transverse wave is described by the equation `y=y_(0)sin2pi(ft-(x)/(lambda))`. The maximum particle velocity is equal to four times the wave velocity, if

Options

A. `lambda=(piy_(0))/(4)`
B. `lambda=(piy_(0))/(2)`
C. `lambda=piy_(0)`
D. `lambda=2piy_(0)`

Answer

B

Explanation

`y=y_(0)sin2pi(ft-(x)/(lambda))...(i)`
For particle velocity,
`(dy)/(dt)=2pify_(0)cos2pi(ft-(x)/(lamda))`
Maximum particle velocity,
`((dy)/(dt))_(max)=2pify_(0)`
Wave velocity = `flambda`
Accordingly,
`2pify_(0)=4(flambda)`
or `lambda=(piy_(0))/(2)`