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Wave in a Dangling Rope

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data

    A uniform rope of length L and negligible stiffness hangs from a solid fixture in the ceiling

    The free lower end of the rope is struck sharply at time t=0. What is the time t it takes the resulting wave on the rope to travel to the ceiling, be reflected, and return to the lower end of the rope?
    Express your answer in terms of L and constants such as g (the magnitude of the acceleration due to gravity), π, etc.

    2. Relevant equations

    v = sqrt (t/[itex] \mu [/itex] ) Where [itex] \mu [/itex] is mass per unit length.

    3. The attempt at a solution
    Starting off with the given equation:

    v = [itex] \frac {T}{\mu} [/itex]

    Using T = ma, I can replace T with [itex] \mu [/itex] z * g, where z is a length of string and g is the gravitational acceleration.

    Now, equating a few equations:

    v = [itex] \frac {dx} {dt} = \sqrt {gz} [/itex]

    Solving for dt:

    dt = [itex] \frac {dz} {\sqrt {gz} } [/itex]
    From here, I integrate dt, so:
    [itex] \int {dt} = \int { \frac {dz} {\sqrt {gz} } } [/itex]

    Finally, I end with:
    [itex] t = 2 \sqrt {gz} [/itex]

    I just wanted someone to check my work to make sure I didn't make any silly mistakes in this. This was a long process and it was pretty hard, however, I am confused about one thing. Should "z" become L to represent the length of the entire string? Thank you for reading, and any help you can provide.
     
  2. jcsd
  3. Oct 12, 2014 #2

    TSny

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    That looks good. Think about the limits of integration for finding the time for the wave pulse to go from the bottom to the top. The lower limit on the time integral will be 0 and the upper limit will be the time, ##t_1##, when the pulse reaches the top. The two limits on the ##\small z##-integral should be values of ##\small z## that correspond to the time limits. Eventually, you want to find the time for the pulse to travel up and back down.

    This is not the correct result for the integration. One way to see that it cannot be right is to note that the units for the expression on the right do not reduce to a unit of time.
     
  4. Oct 12, 2014 #3
    Yup, I noticed that but I couldn't find the edit button anymore. Currently I have:

    [itex] t = 2 \frac {\sqrt {z} } {\sqrt {g} }[/itex]

    I'm guessing that that's only the amount of time it takes to go up, so I need to multiply that by 2?
     
  5. Oct 12, 2014 #4

    TSny

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    No guessing allowed :)

    Did you think about the values of the lower and upper limits on the integrals? Note that you need to express your answer in terms of the length of the rope, ##\small L##.
     
  6. Oct 12, 2014 #5
    Hmm, yes, I did think about that, and if the limits of the time integral are 0 and [itex] t _1 [/itex] , then the limits of the z integral should be 0 (because when time is 0 there is no length of rope, and by the time it reaches [itex] t_1 [/itex], then the upper limit should be L.
     
  7. Oct 12, 2014 #6

    TSny

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    Sounds good.
    What did you get for the integrals after plugging in the limits? What do you get for the answer to the problem?
     
  8. Oct 12, 2014 #7
    I got [itex] t_1 = 2 \frac {\sqrt {L} } {\sqrt {g} } [/itex] , and then I multiply it by 2 to get:

    [itex] t_1 = 4 \frac {\sqrt {L} } {\sqrt {g} } [/itex]

    I think that's right... right? :P
     
  9. Oct 12, 2014 #8

    TSny

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    Right. Good work!
     
  10. Oct 12, 2014 #9
    Thanks for all your help! =D
     
  11. May 16, 2015 #10
    I have a question related to this topic, I see there are many threads about it and figured it was wiser to continue a similar one.

    I went about trying to solve this problem in a different manner as follows (EDIT: my question only asked for the time the wave takes to go the length of the rope, no reflection):

    Δt = Δd/Vavg

    Δd = L

    Vavg = 1/L 0L √gy dy = (2/3)√(gL)

    ∴ Δt = L/(2/3)√(gL) = (3/2)√(L/g)

    I understand the method posted above, which parallels my textbook's SSM, but I can't see what's wrong with my method. I've been staring at the two different methods for 30 minutes and my brain is completely blanking.
     
    Last edited: May 16, 2015
  12. May 16, 2015 #11

    TSny

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    Welcome to PF!

    In kinematics, vavg is defined as Δd/Δt.

    This is equivalent to a time average of the instantaneous velocity v(t). That is, vavg = (1/T) ∫v(t)dt.

    However, this is not the same as taking a spatial average of the velocity. That is, if v(y) is the instantaneous velocity at position y, then the spatial average is (1/L) ∫v(y)dy.

    But this spatial average is not generally equal to the time average.
     
  13. May 16, 2015 #12
    Thank you! Okay, so I'm now trying to figure out what use the spatial average could have been just for some random musing...

    v(y) = √gy
    V(y) = 2/3(gy)3/2 + C
    v'(y) = (1/2)√g/L

    wait what? what did I do here? Integrating the velocity function gives me m2/s, and differentiating it gives me s-1.

    ok... so I see now that v'(y) = 1/t, which is interesting because it produces the right answer without any integration. Is that a coincidence or is that sound math?

    t = y/v ⇒ t = dy/dv ⇒ 1/t = dv/dy = (1/2)√g/L ⇒ t = 2√L/g

    Anyway, I don't know where I'm headed, but I want to end up at something like Δx/Δt = Vavg but for spatial average instead.

    Alright so by graphing the function, and the average value integral formula, and my above calculus, (and.. common sense) it is apparent I am looking for something like Δm2/s / Δm = spatial average. m2/s is not a unit I am familiar with.



     
    Last edited: May 16, 2015
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