Two transverse waves A and B superimposed to produce a node at x = 0. If the equation of wave A si `y= a cos (kx-omegat)`, then the equation of wave B is

Options

A. `y_(2)=a sin(omega t+kx+(pi)/(3))`
B. `y_(2)=a cos(omega t+kx+(pi)/(3))`
C. `y_(2)=a cos(omega t+kx+(3pi)/(3))`
D. `y_(2)=a cos (omega t+kx+(4pi)/(3))`

Answer

D

Explanation

A standing …………
At `x = 0` the phase difference should be `pi` .
`:.` the correct option is D.
Altermate solution
`y_(2)=a cos (omega t+ kx + phi 0)`
`:. y = y_(1)+y_(2) = a cos (omega t - kx + (pi)/(3))+a cos (omega t+kx+ phi_(0))`
`= 2a cos [omega t+((pi)/(3)+varphi_(0))/(2)]xxcos [kx +(varphi_(0)-(pi)/(3))/(2)]`
`:. y = 0 at x = 0` for ant `t`
`implies kx +(varphi_(0)-(pi)/(3))/(2) = (pi)/(2) at x = 0`
`:. phi_(0) = (4pi)/(3)`. Hence `y_(2) = a cos (omega t + kx +(4pi)/(3))`