Finding the Mass of a Hanging Rope (Wave Problem)

In summary: So the equation isTy = mBoxg + μygand then since F = kxF = -mBoxg - μygand then that leads to d²y / dt² = -g(1 + μ)and then since ω = √(k/m)T = 2π/ω = 2π√(m/k)so m = (T/2π)²kand then since k = mgm = (T/2π)²mgSo μ = (T/2π)²gThendm = μdL = (T/2π)²gdLAnd thendv = √[(T/
  • #1
Chansudesu
3
0

Homework Statement


"...you were presented with a geologist at the bottom of a mineshaft next to a box suspended from a vertical rope. The geologist sent signals to his colleague at the top by initiating a wave pulse at the bottom of the rope that would travel to the top of the rope. The mass of the box is 20.0 kg and the length of the rope is 80.0 m. If a wave pulse initiated by the geologist takes 1.26 s to travel up the rope to his colleague at the top, find the mass of the rope."

mBox = 20.0kg

L = 80.0m

T = 1.26

Homework Equations


v = √(F/μ)

∑F = ma

v = λƒ

The Attempt at a Solution


I'm still trying to figure out how to approach the problem (set up my equations)...

I know that because the rope hangs vertical, the restoring force (weight UNDER a given point) is changing as the wave travels upwards, thus, the velocity is changing as well. So

dv = √(dFRestoring/μ)

and then we need to find the restoring force:

∑F = T - mBoxg - mRopeg = 0 (because it's in equilibrium in the y-direction)

T = mBoxg + mRopeg

At the beginning, the restoring force is only the weight of the box; and at the end, the restoring force is the weight of the box AND the weight of the rope. So

dFRestoring = dT = wBox + (dmRope)g

And then I tried turning μ = m/L into:

μ = dm/dL -> dm = μdL

but this doesn't get me anywhere.. I do not know what μ is equal to since we don't know the weight and when I put the equations together I get

dv = √[(wBox + μdL)/μ] -> dv = √[L(wBox + μdL)/ m]

I'm kinda stuck here and I don't think it's in the right direction because I haven't figured out why I was given the period...

Any help would be appreciated ^^
 
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  • #2
You have one equation that relates ##\dot y## to ##F##. You can easily write one relating ##y## to ##F##, and you have almost done so, but this equation is incorrect:
Chansudesu said:
dFRestoring = dT = wBox + (dmRope)g
Pick a point ##y## on the rope. What is the tension on the rope at that point?
When you get that figured out, you can eliminate the restoring force from the two equations and solve the resulting differential equation for the time the wave takes to travel up the rope.
 
  • #3
tnich said:
You have one equation that relates ##\dot y## to ##F##. You can easily write one relating ##y## to ##F##, and you have almost done so, but this equation is incorrect:

Pick a point ##y## on the rope. What is the tension on the rope at that point?
When you get that figured out, you can eliminate the restoring force from the two equations and solve the resulting differential equation for the time the wave takes to travel up the rope.

Hmm... I can't see the tension in a rope at point y being other than Ty = mBoxg + μyg because the the mass at a point y in the rope would be the mass linear density times y...
 
  • #4
Chansudesu said:
Hmm... I can't see the tension in a rope at point y being other than Ty = mBoxg + μyg because the the mass at a point y in the rope would be the mass linear density times y...
Sounds right to me.
 
  • #5
Ahhhh! I see now. Thank you!
 

1. How do I find the mass of a hanging rope using a wave problem?

To find the mass of a hanging rope using a wave problem, you will need to know the length, tension, and frequency of the rope. You can then use the equation m = (λ/2) * (v/ω)^2, where m is the mass, λ is the wavelength, v is the velocity, and ω is the angular frequency.

2. What is the relationship between the mass of a hanging rope and its frequency?

The mass of a hanging rope is directly proportional to its frequency. This means that as the frequency increases, the mass of the rope also increases. This is because a higher frequency indicates a higher energy and therefore a greater mass is needed to create the wave.

3. Can I find the mass of a hanging rope without knowing its tension?

No, the tension of the rope is a crucial component in finding the mass using a wave problem. Without knowing the tension, you will not be able to accurately calculate the mass of the rope.

4. How does the length of the rope affect its mass in a wave problem?

In a wave problem, the length of the rope does not directly affect its mass. However, the length of the rope does affect the wavelength and velocity, which are both used in the equation to find the mass. So, a longer rope will have a longer wavelength and a slower velocity, resulting in a different mass calculation than a shorter rope.

5. Is it possible to find the mass of a hanging rope using a wave problem if it is not in equilibrium?

Yes, it is still possible to find the mass of a hanging rope using a wave problem if it is not in equilibrium. However, the calculation may be more complex as you will need to take into account the additional forces acting on the rope. It is recommended to simplify the problem by returning the rope to equilibrium before attempting to find its mass using a wave problem.

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