A transverse wave is propagating along `+x` direction. At `t=2` sec the particle at `x=4`m is at `y=2` mm. With the passage of time its `y` coordinate increases and reaches to a maximum of 4 mm. The wave equation is (using `omega` and `k` with their usual meanings)

Options

A. `y = 4sin[omega(t +2) +k(x -2) + (pi)/(6)]`
B. `y = 4sin[omega(t +2) +k(x ) + (pi)/(6)]`
C. `y = 4sin[omega(t - 2) - k(x ) + (pi)/(6)]`
D. `y = 4sin[omega(t - 2) - k(x -4 ) + (pi)/(6)]`

Answer

D

Explanation

SHM equation of the particle at x = 4 is
`y = 4 sin(omega(t-2) + (pi)/(6))`
Wave equation replacing t by `[t - ((x-4)/(v))]` , is
`y = 4 sin(omega(t - ((x-4))/(v) -2) + (pi)/(6))`
`4 sin(omega(t -2) - k (x -4) + (pi)/(6))`