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Springs energy

  1. Sep 20, 2014 #1
    our teacher gave us this exercise but no one could solve it.
    he said :
    I hang the weight of 100 g in a spring. it elongates 1 cm.
    then i hang a 200 g in the spring. it elongates 2cm .
    what is the original length of the spring?
    i tried solving it but it just didn't go right.. how do i solve it??
     
  2. jcsd
  3. Sep 20, 2014 #2

    ShayanJ

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    Show us what you did!
     
  4. Sep 20, 2014 #3
    well I didn't know what formula to use ... I tried
    F= k*x so we have 2F= k2x but its just stupid attempts theyre all false idk even what I was trying
     
  5. Sep 20, 2014 #4

    ShayanJ

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    Let [itex] l_0 [/itex] denote the original length and [itex] l_1 [/itex] and [itex] l_2 [/itex] the two secondary lengths. Then the problem as given us [itex] l_1-l_0 [/itex] and [itex] l_2-l_0 [/itex] and [itex] m_1 [/itex] and [itex]m_2 [/itex]. So we have four unknowns ([itex] k, l_0, l_1, l_2 [/itex]) and need four equations.
    Two of them are [itex] k(l_i-l_0)=m_i g [/itex].
    For the other two, we can use conservation of energy. But if we assume that the spring is at rest in after the elongation i.e. in its second equilibrium length, we will have [itex] \frac 1 2 k (l_i-l_0)^2=m_i g (l_i-l_0) [/itex], which are in contradiction with our first two equations. So that assumption is wrong and its easy to understand. When you attach a weight to the spring, its equilibrium length changes and so the fact that its now shorter, before actually reaching its new equilibrium length, means that its stretched and so it starts to vibrate. So when it reaches its new equilibrium length, it will have a kinetic energy which should be considered in the conservation of energy equations.
    And for calculating those speeds, you should solve the equations of motion [itex] m\ddot{x}_i=-kx_i-m_ig \ \ \ (x_i=l_i-l_0) [/itex] and then find at what time, the spring has length [itex] l_0+\frac{m_i g}{k}[/itex] and put that time into [itex] \dot{x}_i(t) [/itex] to find the speeds.
     
  6. Sep 20, 2014 #5
    that makes sense... but I couldn't use these laws cz I didn't take them.. XD thanks a lot :) ill try today and tell u what happened :D
     
  7. Sep 21, 2014 #6

    Doc Al

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    As written, it seems like there is insufficient information given to solve the problem. Are you stating the problem word-for-word, exactly as given?
     
  8. Sep 21, 2014 #7

    ShayanJ

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    I actually meant "compressed". Sorry for that.
    The point is, after you attach the mass to the spring, the spring+mass has a equilibrium length [itex] l_0+\frac{m_i g }{k} [/itex](assuming a point mass). So the fact that at time t=0, [itex] l=l_0 [/itex], means the spring is compressed by an amount equal to [itex] \frac{m_i g}{k} [/itex] and is at rest there and so it should vibrate with the given initial conditions given in the last sentence.

    The fact that you don't know those laws, means your instructor...or teacher (or whatever!!!), should have given you the data obtained by those calculations, i.e. the speeds. So s/he didn't give you enough data for solving the problem.
     
    Last edited: Sep 21, 2014
  9. Sep 21, 2014 #8

    Borek

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    As worded it is pure static, why do you introduce speeds into that? It is only confusing and doesn't add anything to the problem that seems to be incomplete.
     
  10. Sep 21, 2014 #9
    yea. he just gave us this. but ive solved it ill post the answer so u can check it:)
     
  11. Sep 21, 2014 #10

    ShayanJ

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    Well, I think there is something vague here.
    I interpreted the problem such that numbers given are the displacements of the equilibrium position from its original value. In this case, the problem isn't static.
    Another interpretation is that those numbers are giving the amplitudes of the oscillations around the new equilibrium position plus the amount of the shift of the equilibrium position. In this case, the problem is pure static.
    So it depends on the meaning of "elongates"!

    EDIT:
    But if, after hanging the mass, we don't let the spring oscillate and take the mass and slowly bring it to the new equilibrium position and don't give it a speed, then it won't oscillate and there is no speed. It seems this is the case and I was overcomplicating things! But these things should be explicitly stated.
     
    Last edited: Sep 21, 2014
  12. Sep 21, 2014 #11
    yea ur right. I solved it at school today, cz the teacher changed the numbers. and showed him the answer (after u gave me pointers) I didn't do exactly what u said but idk look at it :P

    the distance between the original state and the new state is x1 which equals L1 - L0.
    the distance between the original state and the new state is x2 which equals L2 - L0
    we have F1=Kx1, after substitution we have : F1=k(L1 - L0)
    and the other one is : F2=Kx2 , after substitution : F2=K(L2 - L0)
    so ,
    F2/F1 = K(L2 - L0) / K(L1 - L0)
    then we remove k with k and we have
    F2/F1 = (L2 - L0) / (L1 - L0)

    then I put the numbers and its solved ! thanks a lot ^-^
     
  13. Sep 21, 2014 #12

    Borek

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    To what?
     
  14. Sep 21, 2014 #13
    yea ur right. I solved it at school today, cz the teacher changed the numbers. and showed him the answer (after u gave me pointers) I didn't do exactly what u said but idk look at it :P

    the distance between the original state and the new state is x1 which equals L1 - L0.
    the distance between the original state and the new state is x2 which equals L2 - L0
    we have F1=Kx1, after substitution we have : F1=k(L1 - L0)
    and the other one is : F2=Kx2 , after substitution : F2=K(L2 - L0)
    so ,
    F2/F1 = K(L2 - L0) / K(L1 - L0)
    then we remove k with k and we have
    F2/F1 = (L2 - L0) / (L1 - L0)

    then I put the numbers and its solved ! thanks a lot ^-^
     
  15. Sep 21, 2014 #14
    f1 = 2n f2 = 6n
    l1 = 22cm l2= 26cm
     
  16. Sep 22, 2014 #15

    Simon Bridge

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    energy stored in the spring is \frac{1}{2}kx^2 < mgx
    isn't there a particular ratio about how much of the mgx gets to be kinetic?
    I seem to remember a thread on this somewhere...
    ...but I have a funny feeling that the teacher is (erroneously) thinking that all the gpe gets stored in the spring.
    Either that or the additional information is supposed to be inferred from the classwork/notes so far.

    What has the class learned about springs besides Hook's Law?
     
  17. Sep 22, 2014 #16

    Borek

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    Please read later posts, data given were changed.
     
  18. Sep 22, 2014 #17
    we're just starting the school year .. well we learned that k=fx and that's hook's law :D:p we didn't learn much we've just started the lesson.. but thank you :):)
     
  19. Sep 22, 2014 #18
    It's not that he changed the numbers only.
    He changed the problem by giving the lengths and not the elongations, as originally stated.
    As formulated originally, this method will be useless as you don't know l1 and l2 but (l1-lo) and (l2-lo).
     
  20. Sep 22, 2014 #19

    Simon Bridge

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    ... post #14, I saw ... actually didn't read it properly: it looked like just the numbers were changed but it could be, as nasu observes, that what the numbers were is different. That would certaily qualify as additional information not included before.

    Um no - that is not Hook's Law. At least, not how it is normally written down. Maybe k=F/x?

    In one formulation:
    Hook's Law is: "the restoring force on a spring is directly proportional to it's extension (provided the extension is within the elastic limit)."
    The constant of proportionality is called "the spring constant" or "the coefficient of extension" or whatever your teacher said to call it.

    If the un-stretched length of the spring is ##l_0## and the stretched length is ##l## then ##\vec x = (l - l_0)\hat x## is called the extension, and Hook's law reads: ##\vec F = -k\vec x## ... the minus sign just means the restoring force is in the opposite direction to the extension.

    To keep the spring stretched, therefore, a force equal and opposite the restoring force is needed.
    If that is provided by a mass ##m## hanging off it's end, then we can say that ##mg-kx=0## and so on.

    Notice that your teacher has given you two values for stretched lengths and not two values for the extension.
    That akes the problem much simpler - you can write out Hook's law for each case, giving you two equations with two unknowns.
     
    Last edited: Sep 27, 2014
  21. Sep 26, 2014 #20
    [QUOTE

    Um no - that is not Hook's Law. At least, not how it is normally written down. Maybe k=F/x?
    .[/QUOTE]
    yes right. ... wait I didn't put the (/) :eek:
     
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