(a) As the string has mass and it is suspended vertically, tension in it will be different points. For a point at a distance x from the free end,tension will be due to the weight of the string below it.So,if `m` is the mass of string of length`l`, the mass of length `x` of the string will be `((m)/(l))x`.
`:. T=((m)/(l))xg=muxg , ((m)/(l)=mu)`
`:. (T)/(mu)=xg`
or `v=sqrt((T)/(mu))=sqrt(xg)` ...(i)
At `x=0.5 m` , `v=sqrt(0.5xx9.8)`
`=2.22 m//s`
(b) From Eq. `(i)`, we can see that velocity of the wave is different at different points. So, if at point `x` the wave travels a distance `dx` in time `dt`, then
`dt=(dx)/(v)=(dx)/(sqrt(gx))`
`:. int_(0)^(t)dt=int_(0)^(l)(dx)/(sqrt(gx))`
or `t=2 sqrt((l)/(g))=2 sqrt((2.45)/(9.8))`
`=1.0 s`.
