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Homework Help: Spring's' constant

  1. Dec 17, 2004 #1


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    I haven't done anything with the following problem because I really don't know where to start from. My book doesn't give any guideline. The only thing I am aware of is F=-Kx and E=Kx^2/2 but I cannot use them to proceed in some way.

    '' A spring of length l it is consisted of 2 springs. The first has constant k and length l/2 and the second has constant 2k and length l/2.
    Prove that the constant of the hole spring is 2k/3.''

    I am sorry for the bad english. My mother tongue is greek.

    Thank you.
  2. jcsd
  3. Dec 17, 2004 #2
    Welcome to PF joy!

    1. Given two springs (and say a mass). How many ways can you connect the springs to the mass? (Hint: don't think about a mass sandwiched between two springs.)

    2. Can you set up the equation for an equivalent spring and for the two "sub-springs"?

    Hope that helps...(second hint: E = (1/2)kx^2 needn't be used)
  4. Dec 19, 2004 #3


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    Thanks for your help!

    I have tried to use energy but it is still difficult. I realised though, that it is maybe a problem that needs oscillation...

  5. Dec 19, 2004 #4


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    Well, since the combined spring has length L and the two springs are length L/2, they must be attached end-to-end, or in series. So what does that mean. Consider Spring 1(Sp1) of spring constant k1, and Spring 2 (Sp2) with constant k2.

    Sp 1 . . Sp2
    |/\/\/|/\/\/\/| <-- F

    Now force F pushes (or pulls) on the combined spring. What can we say about the force in each spring, in relationship to F. Let force in Spring 1 = F1 and Spring = F2. The forces are directly transmitted from F - to Sp 2 and from Sp 2 to Sp 1, so F = F1 = F2.

    Now consider displacements x1 in Sp 1 and x2 in Sp 2.

    F1 = k1x1 and F2 = k2x2, and the total displacement of the combined spring x = x1 + x2.

    In the combined spring F = kx (and you wish to find k in terms of k1 and k2).

    From the equilibrium conditions, remember x = x1 + x2, or

    F/k = F1/k1+ F2/k2.

    Now should be able to take it from here.
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