Correct Answer - A::B::C
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 = (pi)/(lambda) = 78.5 m^(-1)`
and `omega =2pif = 23562 rad//s`
At `t = 0`,`x = 0 , (dely)/(delx) = "positive"`
and the given curve is a sine 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(x, t) = (0.06 m) sin [(78.5 m^(-1) x = (23562 s^(-1)) t] m`
