(a). The wave equation is
`y=Ae^(-(t/T-x/lambda)^(2))`
This may be expressed as
`y=Ae^(-1/lambda^(2)((x-lambda)/T^(1))^(2)`
This is the form `f(x-vt)`
Therefore,the velocity of wave `v=lambda//T=8.0//1.0=8 cm//s`
(b). substituting `t=0` in given Eq. (i),we get
`f(t)=Ae^(-(t//T)^(2)`
(c ) substituting `t=0` in given Eq.(i), we get
`g(x)=aE(-(X//LAMBDA)^(2))`
(d). At `t=0`, the displacement is maximum at the originwhile at `t=5 s`, the displacement is maximum at distance `x=vt=8xx5=40 cm`
The function `g(x)` at `t=0` and `5 s` are plotted in `s-5.2` (a) and (b)
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/1610548045277385861610548045.png)