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

Answer

C

Explanation

Given : `y=y_(0)sin2pi(upsilont-(x)/(lambda))`
Wave velocity, `v=upsilonlambda`
Particle velocity, `v_(p)=(dy)/(dt)=(2piupsilony_(0))cos2pi(upsilont-(x)/(lambda))`
`therefore` Maximum particle velocity, `(v_(p))_(max)=2piupsilony_(0)`
According to question, `(v_(p))_(max)=4v`
`therefore 2piupsilony_(0)=4(upsilonlambda)orlambda=(piy_(0))/(2)`