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

Answer

D

Explanation

`y=y_(0) sin 2pi [ft -x/lambda]`
`:. (dy)/(dt)=v_("Particle")=y_(0) 2pi f cos 2 pi [ft -x/lambda]`
for `v_("Particle"|"max")=y_(0) 2 pi t` ...(i)
Wave velocity `v=omega/k=(2pi f)/(2pi //lambda) implies v=f lambda` ...(ii)
Acording to the question,
`v_("Particle|max")=4v`
`implies y_(0) 2 pi f=4 f lambda` from (i) and (ii)
`:. lambda =(y_(0) pi)/2`