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Do the spring constant depend upon the length of the spring? No. of coils? Like what happens to the spring constant if you cut it in the half?

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    $\begingroup$ Instead of cutting the spring in half, imagine what happens if you paint a red dot on the middle. That thought experiment should tell you what happens to the halves above and below the dot. $\endgroup$ – MSalters Apr 8 '15 at 12:17
  • $\begingroup$ Half the length equals double the stiffness generally. As commented above the displacement of the middle is half of the displacement of the end. $\endgroup$ – ja72 Apr 8 '15 at 14:53
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In the standard approximation it does not depend on the length but on the number of active windings. As you can find e.g. here, the spring constant $k$ is

$$k=\frac{G d^4}{8 n D^3}$$

whre $d$ is the wire diameter, $D$ the coil diameter, $n$ the number of active coils, and $G$ the shear modulus. So if you cut it in half $n \rightarrow n/2$ the spring constant doubles $k\rightarrow 2k$. The maximum length decreases of course and you get into the non-linear regime with less displacement.

To calculate spring constants also check out one of many online spring calculators.

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