Helical spring stiffness after cutting — closely coiled spring: A closely-coiled helical spring has stiffness k. It is cut into n equal parts (each part having 1/n of the original coils). What is the stiffness of each part?

Introduction:Spring stiffness depends strongly on the number of active coils. Cutting a spring changes its coil count and thus its force–deflection characteristics. This is frequently used to tune stiffness in mechanical assemblies.Given Data / Assumptions:

• Closely coiled helical spring, small pitch angle.
• Original stiffness: k; total active coils: N.
• Spring is cut into n equal parts, each with N/n coils.

Concept / Approach:For closely coiled springs, stiffness k is inversely proportional to the number of active coils: k ∝ 1/N (with k = Gd^4 / (8D^3N) for round wire, where d is wire diameter, D is mean coil diameter, and G is shear modulus). Reducing N increases k proportionally.Step-by-Step Solution:

Verification / Alternative check:If n = 2, halving coil count doubles stiffness; if n = 4, quarter of coils gives 4k. This matches practical experience and the k ∝ 1/N law.Why Other Options Are Wrong:

• k/n and k/n^2: Predict decreased stiffness after cutting, contrary to 1/N dependence.
• n^2k: Overstates the increase; proportionality is linear in n, not quadratic.
• k: Ignores coil-count effect.

Common Pitfalls:Confusing series vs parallel combination ideas with coil-count changes; mixing up wire diameter or coil diameter effects, which remain constant when simply cutting length.Final Answer: