Discussion
If a helical spring is halved in length,its spring stiffness is:
Answer: C
The spring stiffness doubles.
The stiffness of a helical spring is \(k = \frac{Gd^{4}}{8D^{3}n}\), where n is the number of active coils. Halving the length halves n, so
\[ k' = \frac{Gd^{4}}{8D^{3}(n/2)} = 2k \]
Put another way, each half carries the full load but stretches only half as much. The stiffness does not stay the same, halve or triple.
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