Two closely coiled helical springs A and B are identical except for number of turns. Spring A has twice the number of turns as spring B. How does the stiffness of A compare with B?

Introduction / Context:
Helical spring stiffness determines load–deflection behavior in suspensions, valves, and precision mechanisms. For closely coiled springs, stiffness depends sensitively on geometry, especially wire diameter, mean coil diameter, and number of active turns.

Given Data / Assumptions:

• Springs A and B have the same wire diameter d, same mean coil diameter D, same material (same shear modulus G).
• Only difference: number of active turns n_A = 2 n_B.
• Close-coiled helical spring, small pitch angle.

Concept / Approach:
Spring stiffness k for a close-coiled helical spring is k = G d^4 / (64 D^3 n). Thus, stiffness is inversely proportional to the number of active turns n when other parameters are fixed.

Step-by-Step Solution:

Verification / Alternative check:
Physical reasoning: more coils increase flexibility and reduce stiffness; doubling coils halves stiffness if all else is unchanged.

Why Other Options Are Wrong:
Fractions like one-eighth or one-sixteenth suggest changes in d or D (since k ∝ d^4 and ∝ 1/D^3), which are not given here.

Common Pitfalls:
Forgetting which variables most strongly affect k; confusing active turns with total turns; ignoring squared or fourth-power dependencies.

Final Answer:

one-half of B