A rope of total mass m and length L is suspended vertically. Show that a transverse pulse travels the length of the rope in a time interval `Deltat=2sqrt(L//g`. Sugestion: first find an expression for the wave speed at any point a distance x from the lower end by considering the rope's tension as rejulting from the weight of the segment below that point.

Explanation

The tention will increase as elevation `x` increase, because the upper part of the hanging rope must support the weight of the lower part. Then the wave speed increase with height.
we will need to do an integral based on `v=dx//dt` to find the travel time of the variable-speed pulse.
we define `x=0` at the bottom of the rope and `x=L` at the top of the rope. the tension in the rope at any point is the weight of the rope.below the point.we can thus write the tension unit length of the rope, which we assume uniform. the speed of the wave pulse at each point along the rope's length is, therefore,
but at each point, the wave propagates at the rate of
`v=(dx)/(dt)`
so we substitute for `v` and generate the differential equation equation:
`(dx)/(dt)=sqrt(gx) or dt=(dx)/sqrt(gx)`
`Integrating both sides,
`Deltat=(1)/sqrt(g)int_(0)^(L)(dx)/sqrt(x)=[2sqrtx]_(0)^(L)/sqrt(g)=2sqrtL/g`
what you have learned in mathematic class is another operation that can be done to both sides of an equation: integrating from one physical point to another. here the integral extends in time and space from the situation with the opulse reaching the top.