How does K (spring constant ) changes?

Question

How the spring constant changes when a spring is cut into equl or rational halves? What is the relation with the original K?

Answer 1

While it seems a duplicate there is probably an easier way to explain.

When a force is applied to a spring it gets longer. The more force you apply the longer it gets and the proportionality factor is the so called "compliance". The more compliant (or softer) the spring is the more it moves for the same amount of force. The spring constant is simply the inverse of the compliance and sometimes also called stiffness. The stiffer the spring, the less it moves or, conversely, the more force is required to get the same displacement.

Now let's a consider a spring that moves by 3mm when you apply 1N of force. The compliance would be displacement divided by force = .003 m/N. Now let's cut the spring into three equal pieces and glue it together again. It's the same spring as before so we still get 3mm of total displacement. The force flows through all three sub-springs equally so each sub string sees 1 N. The total displacement is split equally between each sub string, so each sub string moves by 1mm. Hence the compliance is 1mm/1N = .001 m/N.

That's our answer then: When you cut a spring into pieces the compliance gets split over all pieces proportional to their length. The spring constant or stiffness is simply the inverse of that.