A transverse wave is given by the equation y=y_0 sin 2pi(ft-x/lambda) The maximum particle velocity will be 4 times the wave velocity if, A. lambda =(pi y_0)/4 B.lambda =(pi y_0)/2 C.lambda =pi y_0 D.lambda =2 pi y_0 ?

1 Answer
Mar 16, 2018

B

Explanation:

Comparing the given equation with y=a sin (omegat-kx) we get,

amplitude of particle motion is a=y_o , omega=2pif ,nu=f and wavelength is lambda

Now,maximum particle velocity i.e maximum velocity of S.H.M is v'=a omega=y_o2pif

And,wave velocity v=nulambda =flambda

Given condition is v'=4v

so,y_o2pif=4 f lambda

or,lambda =(piy_o)/2