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

B

Explanation

`Y=Y_(0)sin2pi(ft-(x)/(lambda))`
Maximum particle velocity `(v_(max))_(p)`
`=Aomega=A2pif=2Y_(0)pif`
wave velocity `(v)=(omega)/(k)=(2pif)/(2pi//lambda)=flambda`
given, `(v_(max))_(P)=4v`
`implies2Y_(0)pif=4flambdaimplieslambda=(Y_(0)pi)/(2)`