Why does a short pulse sent on a rope speed up as it climbs height?

Question

Following from the hanging rope model here, there was a question I was doing in which the rope is jerked from the bottom , and from the relation between velocity, tension and mass density, we get velocity as:

$$ v(x) = \sqrt{ \frac{T(x)}{\mu}}$$

And using relation, $ T = \frac{Mgx}{L}$,

$$ v \propto \sqrt{x}$$

So, it turns out that as you go up the rope, the velocity of the pulse speeds up.. but why? How can we understand this result intuitively? By the way I know that equation in premise is an approximation (I'm asking how do we understand why the equation is true in the approximated form)

Answer 1

The tension is due to the weight of portion of the rope below the point $x$ so as you go up there’s a greater portion of the rope and thus a greater portion of the mass of the rope to pull the little portion of the rope at $x$. Note that since the tension is “local” (it changes at every $x$ rather than being constant throughout), you also need the “local” mass parameter $\mu$ (the linear mass density) rather than the total mass of the rope.

The rest is thus a question of understanding how the tension (here, the local tension) enters in the velocity: the tension is basically the restoring force so greater tension means greater restoring force.