One end of a string of length `L` is tied to the ceiling of a lift acceleration upwards with an acceleration `2g`. The other end of the string is free. The linear mass density of the string varies linearly from `0 to lambda` from bottom to top. Choose the correct option(s)

Options

A. The velocity of the wave in the string will be 0.
B. The acceleration (relative to string) of the wave on the string will be `3g//4` every where.
C. The time taken by a pulse to reach from bottom to will be `sqrt(8L//3g)`.
D. The time taken by a pulse to reach from bottom to top will be `sqrt(4L//3g)`.

Answer

B, C

Explanation

One end of …………..

`mu = (lambda)/(L)xxy`
`dm = mu dy`
`m_(y)= int_(0)^(y)(lambda)/(L)ydy = (lambda y^(2))/(2L)`
`T =m_(y)g+m_(y)2g = m_(y)3g`
`v_(y) = sqrt((T)/(mu)) = sqrt((lambda y^(2))/(2L).3g.(L)/(lambda y))`
`V_(y)^(2) = (3)/(2)gy = 0^(2)+2.a.y implies a = (3)/(4)g`
`v_(y) = sqrt((3gy)/(2))`
`int_(0)^(l) (dy)/(sqrt(y)) = int_(0)^(t) sqrt((3g)/(2))dt implies t = sqrt((8l)/(3g))`