You are given four springs, one each of 3.5, 6, 8.5, and 11 newtons per meter [N/m]. How do I find the smallest equivalent stiffness that can be made using only three of these springs?
You have two possible ways to connect the springs: in parallel or in series.
The equivalent stiffness of three springs in parallel is given by: k_eq = k₁ + k₂ + k₃ In order to keep this number the smallest possible, you need to take the three springs with smaller k: k_eq_min = 3.5 + 6 + 8.5 = 18 N/m
The equivalent stiffness of three springs in series is given by: 1 / k_eq = 1 / k₁ + 1 / k₂ + 1 / k₃ In order to get the smallest k_eq possible, 1 / k_eq must be the biggest possible, therefore you need to take again the three springs with smaller k:
