A transerse sinusodial wave of amplitude `2 mm` is setup in a long uniform string. Snapshot of string from `x =0` to `x = pi` meter is taken at `t = 0`, which is shown. Velocity of point `P` is in `-y` direrction. Magnitude of relative velocity of `P` with respect to `Q` is `2 cm//s`. Choose the correct options : wave equation is

Options

A. Displacement of particle at position `P` from its mean position as function of time is given by `Y = -(2xx10^(-3))sin5t(m)`
B. wave equation is `Y = (2xx10^(-3)) sin(5t+2x+(pi)/(6))(m)`
C. wave equation is `Y = (2xx10^(-3)) sin (5t+2x+(5pi)/(6))(m)`
D. wave equation is `Y = (2xx10^(-3))sin (5t-2x+(pi)/(6))(m)`

Answer

A, B

Explanation

A transverse ………………
Wave is heading towards `-ve` x-direction

equation of `R` with respect to time `= A sin (omega t +(pi)/(6))`
wave equation `= A sin (omega t + kx + (pi)/(6))`
given `|vecV_(p)-vecV_(Q)|=2omegaA=2 cm//s`
`omega(2xx10^(-3))=10^(-2)m//s`
`omega(2xx10^(-1))=1 implies omega = 5 rad//s`
from snapshot `lambda = pi m implies k = (2pi)/(lambda) = (2pi)/(pi) = 2m^(-1)`
wave equation `y = (2xx10^(-3))sin (5t+2x +(pi)/(6))(m)`