A sinusoidal wave moving along a string is shown twice in the figure. As crest A travels in the positive direction of an x axis by distance `d=6.0` cm in `4.0m`. The tick marks along the axis are separated by `10cm`, height `H=6.00mm`. The wave equation is

Options

A. `y=(3mm)sin[16x-2.4xx10^2t]`
B. `y=(3mm)sin[16x+2.4xx10^2t]`
C. `y=(3mm)sin[8x+2.4xx10^2t]`
D. `y=(3mm)sin[8x-2.4xx10^2t]`

Answer

A

Explanation

The amplitude A is half of the 6.00 mm vertical range shown in the figure, that is ,`A=3.0mm`
The speed of the wave is `v=d//t=15m//s`, where `d=0.060m` and `t=0.0040s`. The angular wave number is `k=(2pi)/(lamda)` where `lamda=0.40m`. Thus, `k=(2pi)//(lamda)=16(rad)//(m)`. The angular frequency is found from
`omega=kv=(16(rad)/(m))(15(m)/(s))=2.4xx10^2 rad//s`. We choose the minus sign (between kx and `omegat`) in the argument of the sine function because the wave is shown traveling to the right (in the `+x` direction). Therefore, with SI units understood, we obtain
`y=Asin[kx-omegat]=(3mm)sin[16x-2.4xx10^2t]`