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Math Help - Multiple Spring system?

  1. #1
    Super Member fardeen_gen's Avatar
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    Multiple Spring system?

    A light rod is supported by two springs fixed on ceiling. A mass m hangs from centre of the rod with a third spring as shown in the diagram. Find the natural frequency of the system:

    (i) When spring 1 and 2 share the weight mg equally

    (ii) When the bar is assumed to stay horizontal all the time

    Answer:
    Spoiler:
    (i) \frac{1}{2\pi}\sqrt{\frac{4k_1k_2k_3}{m(4k_1k_2 + k_1k_3 + k_2k_3)}}
    Spoiler:

    (ii) \frac{1}{2\pi}\sqrt{\frac{k_1k_3 + k_2k_3}{m(k_1 + k_2 + k_3)}}


    I am not getting the force equations of the springs. Any ideas?
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
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    408

    Basic Assumptions

    First, two previously known lemmas about multiple spring systems that you can assume to be true or derive for yourself:

    (1) In parallel, F_{eq}=F_1+F_2 so k_{eq}=k_1+k_2

    (2) In series, F_1=F_2 so \frac1{k_{eq}}=\frac1{k_1}+\frac1{k_2}

    Start with F_1=k_1x_1, F_2=k_2x_2, F_3=k_3x_3

    F_{12}=F_1+F_2=F_3 because these are in series, so k_{12}x_{12}=k_3x_3. Since the load is split evenly between springs #1 and #2, we can assume x_{12}=\frac12(x_1+x_2).

    (i) Assume F_1=F_2 so WLOG F_{12}=2F_1=F_3. So 2k_1x_1=k_3x_3=k_{12}\frac12(x_1+x_2) or 4k_1=k_{12}(1+\frac{x_2}{x_1}). Since k_1x_1=k_2x_2, 4k_1=k_{12}(1+\frac{k_1}{k_2}) . Solving for k_{12} gives k_{12}=\frac{4k_1k_2}{k_1+k_2}. By lemma 2, (k_{123})^{-1}=\frac1k_3+\frac{k_1+k_2}{4k_1k_2}=\frac{4k_1k_2  +k_1k_3+k_2k_3}{4k_1k_2k_3}

    (ii) By lemma 1, k_{12}=k_1+k_2 and by lemma 2, (k_{123})^{-1}=\frac1k_3+\frac1{k_1+k_2}=\frac{k_1+k_2+k_3}{k_  1k_3+k_2k_3}

    In each instance of course, \nu=\frac1{2\pi}\sqrt{\frac k m}
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