How Does Wave Speed Vary Along a Hanging Rope?

[phil ess]

Homework Statement

A uniform rope of mass m and length L hangs vertically from the ceiling. The distance along the rope, as measured from the bottom of the rope is y (i.e., the bottom of the rope is y = 0 and the top is y = L).

Homework Equations

v = sqrt(T/u) ?

The Attempt at a Solution

ok so to find the speed of the rope i used the above equation, with:

u = m/L (linear density of string)
T = mg (tension in string)

So substituting gives: v₁ = sqrt (gL)

now since gravity is accelerating downward, and we need v as a function of y:

v₂² = v₁² + 2ad

then substituting:

v₂² = gL - 2gy

v(y) = sqrt (gL -2gy)

Does this look correct? The concern i have is that this equation says the wave can't go past half the length of the rope, which seems kinda wonky, though it may be the case. Can anyone clear this up please!

Thanks!

EDIT: ok the next question says how long does it take to get to the top of the string so I know this can't be right, since y =/= L in my equation.

 

[Carid]

As the wave goes up the rope it encounters greater tension. The velocity changes.

 

[nlightner]

Your solution looks correct for the speed of the wave along the rope. However, there are a few things to consider:

1. The equation v = sqrt(T/u) assumes that the rope is under tension and that the wave travels along the rope without any external forces acting on it. In this case, the only external force acting on the rope is gravity, so the equation may not be entirely accurate.

2. The wave can travel along the entire length of the rope, not just half of it. The equation v(y) = sqrt(gL - 2gy) gives the speed of the wave at any point along the rope, including the top (y = L).

3. To find the time it takes for the wave to reach the top of the rope, you can use the equation d = vt, where d is the distance traveled (L) and v is the speed of the wave (which is a function of y). You can then solve for t to find the time it takes to reach the top of the rope.

Overall, your solution looks correct with the caveat that the equation may not be entirely accurate due to the external force of gravity.