Spring Length

We will utilize Hooke's Law $F=kx$ for this comparison problem. Start with your $L$ spring that has a spring constant $k$ and compress it by a distance of $x$ . The force in the spring is now $kx$ newtons. So far so good.

Now take two of your $L-$ springs and place them end to end to form the $2L$ spring. Compress this spring by a distance $x$ . Since the bigger spring is made up of two identical $L-$ springs each of these smaller springs will compress by an identical amount. That means each individual $L-$ spring is compressed by $\frac{x}{2}$ , and therefore the force in each spring is $\frac{kx}{2}$ .

But the force in the bigger $2L-$ spring is the same as the force in both the smaller $L-$ springs that comprise it (i.e. the force in the big spring is $\frac{kx}{2}$ newtons. And that means the force constant of the big $2L-$ spring is two times smaller than the spring constant of the $L-$ springs.

Assuming the springs are made from the same material, the force constant is inversely proportional to the (relaxed) length of the spring.