A spring of force constant $k$ is cut into two equal halves. The force constant of each half is:
- $k/2$
- $k$
- $2k$ (stiffer when shorter)
- $4k$
Show answer and solution
Correct answer: C. $2k$ (stiffer when shorter)
Spring constant is inversely proportional to length: $k \propto 1/L$.
Halving the length doubles the spring constant.
**Useful corollary:** cutting a spring into n equal parts gives each part a constant of $nk$.