What is the stiffness of the two springs connected in series? parallel?

Two springs are given, with rigidity k₁ and k₂. Some test weight gives elongation for the first spring x₁, for the second spring – x₂, causing elastic forces in them, respectively, with the moduli F₁ = k₁ ∙ x₁ and F₂ = k₂ ∙ x₂.

If two springs are connected in series, then according to Newton’s third law, the weight with the lower spring and the springs will interact with each other with forces of the same modulus: F = F₁ = F₂. The total elongation of the system of springs will be x = x₁ + x₂, then the total stiffness k will be obtained from the equality: F / k = F / k₁ + F / k₂, since x = F / k. We get:

1 / k = 1 / k₁ + 1 / k₂ or k = k₁ ∙ k₂ / (k₁ + k₂).

If two springs are connected in parallel, then the force acting from the side of the weight will be counteracted by the elastic force of the system of springs, equal to it in modulus according to Newton’s third law: F = F₁ + F₂. The total elongation of the system of springs will be x = x₁ = x₂, then the total stiffness k will be obtained from the equality:

k = F / x = (F₁ + F₂) / x = F₁ / x + F₂ / x = k₁ + k₂.

Answer: the stiffness of two springs connected in series k₁ ∙ k₂ / (k₁ + k₂), connected in parallel k₁ + k₂.