Rubber Physics (Part 1)

There is a simple intuitive solution which is hinted at by the information that the red band are much stiffer than the blue ones.

Imagine that the links with only blue bands stretch really easily, and any link with a red band (even if it also has a blue band in parallel) is so stiff it hardly stretches at all.

Then it is easy to see that chains B and C will stretch a lot when a force is applied because they have weak blue links. Chain A, which has a red band in every link will hardly stretch at all.

If two rubber bands with force constants $k_1$ and $k_2$ are linked in parallel, the force $f = f_1 + f_2 = - k_1 x_1 - k_2 x_2$ is divided between the two bands, while the displacement $x = x_1 = x_2$ is the same for both rubbers. Therefore, also the force constants add up: $k_\text{parallel} = -\frac{f}{x} = k_1 + k_2$ However, if the two rubber bands are chained in series, on both act the same force $f = f_1 = f_2$ , but both bands have different displacements $x_1$ and $x_2$ , which add up to a net deflection $x = x_1 + x_2$ . The ratio of the displacements is equal the (inverse) ratio of the force constants: $\begin{aligned} & & f = - k_1 x_1 &= - k_2 x_2 \\ \Rightarrow & & \frac{x_2}{x_1} &= \frac{k_1}{k_2} \\ \Rightarrow & & x &= x_1 + x_2 = \left(1 + \frac{k_1}{k_2} \right) x_1 \\ \Rightarrow & & f &= - k_1 x_1 = - \frac{k_1}{1 + \frac{k_1}{k_2}} x = -\frac{k_1 k_2}{k_1 + k_2} x \\ \Rightarrow & & k_\text{series} &= \frac{k_1 k_2}{k_1 + k_2} \end{aligned}$

We can represent the different rubber chains A, B and C by combining parallel and series connection:

In applying the rules derived above, we get the following force constants for the sketched chain elements: $\begin{aligned} & & k_A &= \frac{(k_1 + k_2)(k_1 + k_2)}{(k_1 + k_2)+(k_1 + k_2)} = \frac{k_1 + k_2}{2} \\ & & k_B &= \frac{(k_1 + k_1)(k_2 + k_2)}{(k_1 + k_1)+(k_2 + k_2)} = 2 \frac{k_1 k_2}{k_1 + k_2} \\ & & k_C &= \left(\frac{k_1 k_2}{k_1 + k_2}\right) + \left(\frac{k_1 k_2}{k_1 + k_2}\right) = 2 \frac{k_1 k_2}{k_1 + k_2}\\ \Rightarrow & & \frac{k_A}{k_B} = \frac{k_A}{k_C} &= \frac{(k_1 + k_2)^2}{4 k_1 k_2} > 1 \quad \text{for} \quad k_1 \not= k_2 \end{aligned}$ This inequality in the last calculation step follows from $\begin{aligned} & & (k_1 - k_2)^2 = k_1^2 + k_2^2 -2 k_1 k_2 &> 0 \\ \Rightarrow & & (k_1 + k_2)^2 = k_1^2 + k_2^2 + 2 k_1 k_2 &> 4 k_1 k_2 \\ \Rightarrow & & \frac{(k_1 + k_2)^2}{4 k_1 k_2} &> 1 \end{aligned}$

Chains B and C are not essentially different. This alone suggests that A is the answer.

In the limit where a red link is infinitely stiff, chain A cannot stretch at all, while in chain B and C the blue sections will stretch.

Either way, we see that the answer is $\boxed{\mathbf A}$ .

Look at the edge case of red not stretching at all and blue stretching with no effort. The blue links can essentially be removed. The only one that would still be held together is A, and it would be held together with no stretching at all.

Intuitively, a chain can only be as strong as its weakest link must extend to “a chain can only be a flexible as its most flexible link”. So chain A will be the least flexible.

Chain B and chain C are effectively identical: therefore chain A is the "odd one out".

Without all the numbers and math it appears that chain "A" is the only one that pairs the rubber bands such that each red band cancels out the flexibility of the blue ones. On a practical note how does Amy get the rubber bands linked? They're just regular circle type bands, right ?