Correct Answer - C
Here, amplitude `r=2cm =0.02m, upsilon=128m//s`
`lambda=(4)/(5)=0.8m, v=(upsilon)/(lambda)=(128)/(0.08)=160Hz,`
`omega=2piv=2xx(22)/(7)xx160=1005`
and `k=(2pi)/(lambda)=(2xx(22//7))/(0.8)=7.85`
Now `y=rsin (kx-omegat)`
`=0.02 sin (7.85x-1005t)`