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A rope of uniform linear density $\mu$ and total length L is suspended from one end. It hangs vertically under its own weight. It is lightly tapped at the lower end. How long does it take for the perturbation to reach the top of the rope?

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Answer

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Assumptions:

1) The top of the rope must support the weight of the rest of the rope.

2) Each section of the rope supports the weight of the rope that exists below it.

3) The rope is not accelerating up towards the ceiling much.

Calculation based on assumptions:

See assumption 2. Let us call the distance from the bottom of the rope the "y" axis. Imagine a segment of the rope partially up the rope.

p = "linear density"

T = "tension"

y = "distance from bottom of the rope"

m = "mass of rope"

p = m/L

m = L*p

T = m*g = p*y*g A)

Equation for velocity of wave on string:

v = velocity

v = (T/p)^(1/2) B)

Combining equation A) with equation B) and noting that velocity is the derivative of position with respect to time, we have a differential equation:

dy/dt = (p*y*g/p)^(1/2) = (y*g)^(1/2)

Rearranging and putting "y" variables on one end of the equation and "t" variables on other end of the equation and integrating we have:

integrate (from y = 0 —> y = L) [y^(-1/2)] dy = integrate (from t = 0 —> t = t) [g^(1/2)] dt

Then we have:

2*L^(1/2) = g^(1/2)*t

Solving for t:

t = 2*(L/g)^(1/2)

We have an answer! Is it right?

Let's check intuitively.

a) It takes longer for a wave to travel from the bottom of the string to the top of the string if you have a longer string. That makes sense.

b) If gravity is greater, then the wave travels faster. Gravity would increase the tension in the string, and tension is directly proportional to wave velocity. So that makes sense.

Problems?

Perhaps. It seems to me that wave velocity in the string should also be affected by the linear density of the string. So why is it not? It puts my solution in doubt.

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