A transverse wave is given by the equation y=y_0 sin 2pi(ft-x/lambda)y=y0sin2π(ftxλ) The maximum particle velocity will be 44 times the wave velocity if, A. lambda =(pi y_0)/4λ=πy04 B.lambda =(pi y_0)/2λ=πy02 C.lambda =pi y_0λ=πy0 D.lambda =2 pi y_0λ=2πy0 ?

1 Answer
Mar 16, 2018

BB

Explanation:

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

amplitude of particle motion is a=y_oa=yo , omega=2pifω=2πf ,nu=fν=f and wavelength is lambdaλ

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