A heavy but uniform rope of lenth L is suspended from a ceiling. (a) Write the velocity of a transverse wave travelling on the string as a function of the distance from the lower end. (b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach teh celling ? (c ) A particle is dropped from the ceiling at the instant the bottom end is given the jerk where will the particle meet the pulse ?

Explanation

Let m be the mass ofthe hanging rope, then its linear mass density will be
`mu = (M)/(L)`
(a) At a distance x above the lower end if we consider a cross section A then tension at point A will be due to the weight of the lower part and it is given as
Tension at A is, `T = (M)/(L) xg`
Now velocity oftransverse waves at point A is given as
`v = sqrt((T)/(mu)) = sqrt((M/L)/(M//L)) = sqrt(xg)`

(b) If ajerk is given at the lower end ofrope, it propagation in upward direction and its velocity at a distance x from lower end is given byequation-(6.56). Wecan find the time taken by pulse ofjerk to reach the top by integration expression in equation- (6.56) as
`(dx)/(dt) = sqrt(xg)`
`(dx)/(sqrt(x)) = sqrt(g) dt`
Integrating this expression in proper limits we get
`underset(0)overset(L)(int)(dx)/(sqrt(x)) = underset(0)overset(t)(int) sqrt(g)dt`
`[2sqrt(x)]_(0)^(L) = sqrt(g) t`
`t = 2sqrt((L)/(g))`
(c) When a particle is dropped from the top it falls by a distance (L- x) in time t. When it will meet the pulse and ifpulse has travelled a distance x. Thus time taken by pulse to travel a distance X from bottom is
`t = 2sqrt((x)/(g))`
In this time the distance fallen by particle in its free fall motion is `(L - x) = (1)/(2)"gt"^(2)`
`L - x = (1)/(2)g (2sqrt((x)/(g)))^(2)`
`L - x = 2x`
`x = (L)/(3)`
Thus particle and the pulse meet at a distance `(L)/(3)` from the bottom.