For the wave shown in figure, write the equation of this wave if its position is shown at `t= 0`. Speed of wave is `v = 300m//s`.

Options

A. `y = (0.06 m) sin [(78.5 m^(-1))x + (23562 s^(-1))t] m`
B. `y = (0.06 m) sin [(78.5 m^(-1))x - (23562 s^(-1))t] m`
C. `y = (0.06 m) sin [(78.5 m^(-1))x + (23562 s^(-1))t] m`
D. `y = (0.86 m) sin [(70.5 m^(-1))x - (28562 s^(-1))t] m`

Answer

B

Explanation

The amplitude, `A = 0.06 m`
`5/(2) lambda = 0.2 m`
`:. lambda = 0.08 m`
`:. F = (v)/(lambda) = (300)/(0.08) = 3750 Hz`
`k = (2pi)/(lambda) = 78.5 m^(-1)` and `omega = 2 pi f = 23562 "rad"//s`
At `t = 0 , x = 0, (dy)/(dx) = "positive"`
and the given curve is a since curve.
Hence, equation of wave travelling in positive x-direction should have the form,
`y (x,t) = A sin (kx - omegat)`
Substituting the values, we have
`y = (0.06m) sin [(78.5 m^(-1)) xx - (23562 s^(-1))t]m`