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# ?

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Answer 1 · 2018-03-16T14:58:10

#B#

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#