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Homework Help: Period of the pulley-spring system

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data
    We have the system consisting of a massless pulley and two massless springs as the picture shown below. Two springs has spring constants as [itex]/k_1[/itex] and [itex]/k_2[/itex]. The object has mass m


    Let's calculate the period of the vibration of the object if it is disturbed vertically in a small distant from the equibrilium

    2. Relevant equations

    3. The attempt at a solution
    on progress
  2. jcsd
  3. Sep 12, 2012 #2

    Simon Bridge

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    Great - get back to us when you've got stuck ;)
  4. Sep 12, 2012 #3
    My lecturer give the problem without giving a result, and so I got stuck of how to deal with it. If someone know the solution, or at least, the exact result, please let me know
  5. Sep 12, 2012 #4

    Simon Bridge

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    You are training to be able to cope with problems which have no known solution ... you cannot expect to always be told the answers in advance.

    ... you have been learning some physics in class and you have a bunch of lecture notes: use what you've learned. Give it a go and we can help where you get stuck. To do that, we need to see your reasoning and your working.
  6. Sep 16, 2012 #5
    my friend show me result http://codecogs.izyba.com/png.latex?\%20\varpi%20=%20\sqrt{%20\frac{k_1%20k_2}{m(4k_2%20+%20k_1)}%20} [Broken]
    , it seems to be true but I still do not know the solution
    Last edited by a moderator: May 6, 2017
  7. Sep 16, 2012 #6

    Simon Bridge

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    I'm sorry, nobody can help you if you don't tell us your reasoning and what you have tried.
  8. Sep 16, 2012 #7
    I am trying to analyze nature in this prob as follows
    Consider this system in equibrilium and then, disturb the object a small distance [itex]x[/itex] vertically
    and call corresponding distances of spring [itex]k_1[/itex] and [itex]k_2[/itex] as [itex]x_1[/itex] and [itex]x_2[/itex]
    then we obtain the relation between these variables


    \left\{ \matrix{
    k_1 x_1 + k_2 x_2 = m\ddot x \cr
    x = x_1 + x_2 \cr} \right.

  9. Sep 16, 2012 #8
    though I obtain two relations above but these are still not rich enough to resolve, help me
  10. Sep 16, 2012 #9

    Simon Bridge

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    Well no, you have not used all your knowledge of the system.
    It is difficult to tell what you need to do exactly because you have not yet shown me your working.

    I'll go quickly - you will need to check:

    we displace the mass a small distance x
    this moves the pulley, and pulls on the springs.

    for the mass: ##T-mg=m\ddot{x}## ...(1)

    by symmetry, the T on the mass is also the T on spring 2 forcing an extension x2:

    ##T=k_2x_2## ...(2)

    The tension forcing spring 1 is not going to be the same so I'll call it T1

    ##T_1=k_1x_1## ...(3)

    All the tensions meet at the pulley ... here I'm not certain because I think this should be unbalanced but I think it comes out in the wash if I write:

    ##T_1=2T## ...(4)

    Looking at the way the cable loops over the pulley - if spring 1 drops by x1, the pulley drops that far too, which makes x1 additional cord available on both sides to contribute to x - in addition to any extension on x2 ... so:

    ##x=2x_1+x_2## ...(5)

    Count them up - that's five simultaneous equations and five unknowns.
    Sub (4) -> (3) to eleiminate T1, then use (2) and (3) to get relations ofr x1 and x2 which you can put into (5). Solve (1) for T and put that into (5) also and you are done.

    I have $$Kx-m\ddot{x}=mg\; : \; \frac{1}{K}=\frac{4}{k_1}+\frac{1}{k_2}$$

    I am making no guarantees that these are the right ones - it is your work: you have to check. But they should show you where you need to look for additional relations.
  11. Dec 5, 2013 #10
    even i had a problem with this while trying it using the energy method.
    could anyone please help me?
    ill upload a pic of my attempt at a solution if required.
  12. Dec 6, 2013 #11

    Simon Bridge

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    The original problem seem to have gone awol.
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