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Homework Help: What is the mass of this vertical rope? (Mechanical Waves)

  1. Mar 13, 2015 #1
    1. The problem statement, all variables and given/known data

    A geologist is at the bottom of a mine shaft next to a box suspended by a vertical rope. The geologist sends a signal to his colleague at the top by initiating a wave pulse at the bottom of the rope that travels 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.

    msamples = 20.0 kg
    d = 80.0 m
    t = 1.26 s
    mrope = ?

    The answer is mrope = 4.30 kg.

    2. Relevant equations

    ## v = \sqrt{{\frac{Tension}{μ}}} ##

    3. The attempt at a solution

    I know the answer but I can't come up with it myself.

    At first I thought I could use kinematics to find the initial velocity of the wave speed at the bottom of the rope, but I realized that wave speed is determined by the medium and not only by a constant acceleration. E.g.:

    ## v = \sqrt{{\frac{Tension}{μ}}} = \sqrt{{\frac{m_{samples} g}{{\frac{m_{rope}}{d}}}}} ≠ {\frac{d - {\frac{1}{2}} g t^2}{t}} ##

    So I guess what I have trouble with the most is figuring out where or how I use the t = 1.26 s information.

    How do I connect these pieces?
  2. jcsd
  3. Mar 13, 2015 #2

    Simon Bridge

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    Hint: use tension and linear mass density.
  4. Mar 13, 2015 #3


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    You have an equation that relates wave speed to tension and linear density. How can you use this to find out the wave speed as a function of the position on the rope? How can you determine the travel time if you know the wave speed at each position of the rope?

    Do note that the rope mass will also affect the tension in the rope!
  5. Mar 13, 2015 #4
    Thanks for the replies.

    Hmm, in what way? I already know that ## v = \sqrt{\frac{T}{μ}} ##.

    Well I know that the tension at the top is [itex] T = (m_{samples} + m_{rope})~g [/itex], so that [itex] v = \sqrt{\frac{(m_{samples} + m_{rope})~g}{\frac{m_{rope}}{d}}} [/itex].
    The last equation I just wrote isn't a function of x.

    I believe that the mass of the system is given by [itex] M(x) = m_{sample} + \frac{m_{rope}} {d}\ dx [/itex], which is a function of x.

    Is this what you're talking about: [itex] v(x) = \sqrt{\frac{(m_{sample} + \frac{m_{rope}} {d}\ dx)~g}{\frac{m_{rope}}{d}}} [/itex] ? Seems kinda redundant. Am I over thinking it?

    ## dt = \frac{x}{v(x)}\ ## ?
    Last edited: Mar 14, 2015
  6. Mar 13, 2015 #5

    Suraj M

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    this ##t## isn't your total time. It should be ##dt## then integrate
  7. Mar 14, 2015 #6


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    ... and this dx is not a dx but an x. The tension in the rope at one point is the force required to support the sample and the rope below.
  8. Mar 14, 2015 #7
    Changed it to y so that's it's a little more intuitive.

    Mass of the system:
    [tex] M(y) = m_{sample} + \frac{m_{rope}}{d} ~ y [/tex]

    Tension on the system:
    [tex] T(y) = M(y) ~ g = (m_{sample} + \frac{m_{rope}}{d} ~ y) ~ g [/tex]

    Wave speed:
    [tex] v(y) = \sqrt{\frac{T(y)}{μ}} = \sqrt{\frac{(m_{sample} + \frac{m_{rope}}{d} y)~g}{\frac{m_{rope}}{d}}} [/tex]


    [tex] \frac{dy}{dt}\ = v(y) [/tex]

    So that

    [tex] dt = \frac{1}{\sqrt{\frac{(m_{sample} + \frac{m_{rope}}{d} y)~g}{\frac{m_{rope}}{d}}}} ~ dy [/tex]


    [tex] t = \int_{0}^{80} \frac{1}{\sqrt{\frac{(m_{sample} + \frac{m_{rope}}{d} y)~g}{\frac{m_{rope}}{d}}}} ~ dy [/tex]

    ...Integrate, plug-in my knowns, solve for unknown? Is this what I need to do to get ## m_{rope} ##?

    /EDIT: This integral is correct. If I plug in values for every variable, I do get 1.26 s. I guess I need to solve this integral! Thanks for the help, everyone.
    Last edited: Mar 14, 2015
  9. Mar 15, 2015 #8
    I solved it. I thought I'd share my solution for future reference. Thanks again for the help!

    [tex] v(y) = \sqrt{\frac{T(y)}{μ}} [/tex]
    [tex] \frac{dy}{dt}\ = v(y) [/tex]
    [tex] \sum F_{y} = T(y) = M(y) ~ g [/tex]


    [tex] T(y) = M(y) ~ g [/tex]

    [tex] M(y) = m_{sample} + \frac{m_{rope}}{d} ~ y [/tex]​
    [tex] T(y) = \left(m_{sample} + \frac{m_{rope}}{d} ~ y\right) ~ g [/tex]
    [tex] μ = \frac{m_{rope}}{d} [/tex]


    [tex] \frac{dy}{dt}\ = v(y) = \sqrt{\frac{T(y)}{μ}} [/tex]

    [tex] ~ ~ ~ ~ ~ ~ = \sqrt{\frac{\left(m_{sample} + \frac{m_{rope}}{d} y\right) ~ g}{\frac{m_{rope}}{d}}} [/tex]

    [tex] ~ ~ ~ ~ ~ ~ = \sqrt{\left(\frac{m_{sample} ~ d}{m_{rope}} + y\right) ~ g} [/tex]

    [tex] dt = \frac{1}{\sqrt{\left(\frac{m_{sample} ~ d}{m_{rope}} + y\right) ~ g}} ~ dy [/tex]

    [tex] t = \int_{y=0}^{d} \frac{1}{\sqrt{\left(\frac{m_{sample} ~ d}{m_{rope}} + y\right) ~ g}} ~ dy [/tex]

    [tex] u = \frac{m_{sample} ~ d}{m_{rope}} + y [/tex]
    [tex]du = dy[/tex]​

    [tex] ~ ~ = \int_{u_{1}}^{u_{2}} \frac{1}{\sqrt{u g}} ~ du [/tex]

    [tex] ~ ~ = \frac{2}{\sqrt{g}} ~ \sqrt{\frac{m_{sample} ~ d}{m_{rope}} + y} ~ ~ \Bigg|_{y = 0}^{d} [/tex]

    [tex] t = \frac{2}{\sqrt{g}} ~ \left(\sqrt{\frac{m_{sample} ~ d}{m_{rope}} + d} ~ - ~ \sqrt{\frac{m_{sample} ~ d}{m_{rope}}}\right) [/tex]

    So then...

    [tex] m_{rope} = \frac{16 ~ d ~ g ~ m_{samples} ~ t^2}{(4 ~ d - g ~ t^2)^2} [/tex]

    [tex] m_{samples} = 20.0 ~ kg [/tex]
    [tex] d = 80.0 ~ m [/tex]
    [tex] t = 1.26 ~ s [/tex]
    [tex] g = 9.81 ~ m/s [/tex]​

    [tex] m_{rope} = 4.30 ~ kg [/tex]
    Last edited: Mar 15, 2015
  10. Mar 15, 2015 #9
    How did you get from the final equation for t to the mass of the rope.

    I am having trouble solving for mrope and any additional help will be appreciated.
  11. Mar 15, 2015 #10
    Yeah, there's a lot of squaring and factoring going on. But I tend to over think on these algebra problems, so I might have taken the long way around. Either way, I've attached a picture of my work to this post. I hope it helps...


    M is ## m_{sample} ## and m is ## m_{rope} ##.
  12. Mar 16, 2015 #11
    I got half way through this and thought to myself I am doing waaay more than needed. Guess not. Thanks!
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