1. ## 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

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?

2. ## 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, $\nu=\frac1{2\pi}\sqrt{\frac k m}$