# How does cutting a spring increase spring constant? [duplicate]

I know that on cutting a spring into n equal pieces, spring constant becomes n times.

But I have no idea why this happens.

Please clarify the reasons

• Commented May 1, 2020 at 8:42
• There is no conservation of (the absolute value of) mechanical force... Commented May 2, 2020 at 12:50

## 6 Answers

Equations are nice, but if you're looking for a conceptual answer :

• Great answer, easy to understand. This should be the accepted one. Commented Jul 26, 2020 at 6:14

In general, the spring constant or stiffness of a coiled spring is given as $$k=\frac{\pi Gd^4}{64R^3n}$$

Where $$G$$ is the modulus of rigidity of spring material

$$d$$ is the diameter of the spring wire

$$R$$ is the mean radius of the coil

$$n$$ is the effective number of coils in spring which is directly proportional to the length of coiled spring i.e. $$n\propto L$$

The above formula simply shows that if the other parameters are kept constant then $$k$$ is inversely proportional to $$n$$ number of coils. Since the number of coils $$n$$ is directly proportional to the length $$L$$ of the coiled spring, its spring constant $$k$$ is inversely proportional to its length. Thus

$$k\propto \frac 1n\iff k\propto \frac1L$$ Therefore if a spring is broken into a certain no. of pieces, all parameters $$G$$, $$R$$ & $$d$$ remain constant for all the pieces except the number of coils or turns $$n$$ decreases hence stiffness $$k$$ increases.

In general, if a coiled spring of length $$L$$ & stiffness $$k$$ is broken into $$m$$ no. of pieces of lengths $$L_1, L_2, L_3, \ldots L_m$$ then their respective spring constant or stiffness is given as follows $$k_1=k\left(\frac{L}{L_1}\right), \ \ \ k_2=k\left(\frac{L}{L_2}\right),\dots k_i=k\left(\frac{L}{L_i}\right), \ldots , k_m=k\left(\frac{L}{L_m}\right)$$
Relation among spring constants of broken pieces & original spring: $$\implies L1=\frac{kL}{k_1}, \ L2=\frac{kL}{k_2}, \ldots , Li=\frac{kL}{k_i},\ldots, Lm=\frac{kL}{k_m}$$ If we add all $$m$$ number of lengths of pieces we get the original length $$L$$ of spring i.e. $$L_1+L_2+\ldots +L_i+\ldots +L_m=L$$ $$\frac{kL}{k_1}+\frac{kL}{k_2}+\ldots+\frac{kL}{k_i}+\ldots+\frac{kL}{k_m}=L$$ $$KL\left(\frac{1}{k_1}+\frac{1}{k_2}+\ldots+\frac{1}{k_i}+\ldots+\frac{1}{k_m}\right)=L$$ $$\color{blue}{\frac{1}{k_1}+\frac{1}{k_2}+\ldots+\frac{1}{k_i}+\ldots+\frac{1}{k_m}=\frac1k}$$

The above relation of spring constants is analogous to the parallel connection of $$m$$ number of electric resistors.

• Always a sincere and affectionate thank you for your cooperation and presence also in my questions on Math.SE and Physics.SE. Commented Feb 7, 2021 at 18:50

Let us consider that you join these n pieces of springs in series. Now you know that you have got the original spring whose spring constant is $$k$$ (say). Now joining springs in series is like joining resistors in parallel (identical formulae) which can easily proven by balancing forces. Hence,

$$\frac{1}{k} = \frac{1}{k'} + \frac{1}{k'} + \frac{1}{k'}+\dots n ~\rm times$$ where $$k'$$ is the spring constant of individual cut springs.

On solving the above equation you will get that the spring constant becomes $$n$$ times.

• Do you want the proof to the formula which I used? Commented May 1, 2020 at 8:09
• No I know this one..... Just couldn't relate that to the question. Thank you Commented May 1, 2020 at 8:09

This happens because spring constant is not really a constant. If you consider any normal elastic material, when a force F is applied, the strech is given by hooke's law :

$$\frac{{F}/{A}}{{\Delta L}/{L}}=Y$$

where $$Y$$ is the young's modulus of material, which is upto a limit, constant and depend only on the material.

This means the strech

$$\Delta L = \frac{FL}{AY}$$

or

$$F = \frac{AY}{L}\Delta L$$

we see that the proportionality constant for spring is thus $$\frac{AY}{L}$$. Thus, for a spring of $$\frac{1}{2}$$ the length, spring constant would be double.

For a given deformation, the distance change between two adjacent particles (molecules/atoms) is more if you decrease the length of the spring. Thus, if you keep the displacement small enough so as the intermolecular force is linearly proportional to the intermolecular distance, the force required to produce the same deformation in a short spring is more compared to a longer spring. If you cut a spring into $$n$$ pieces, the distance change between two particles would have to be $$n$$ times more to keep the total deformation the same, and linearity tells us that the force required will be $$n$$ times greater.

Note that the above model may fail as in a real metal spring, there are grain boundaries, dislocations, etc. But it is presents a good intutive picture.

• Can the downvoter please explain what is wrong with the answer so that it can be corrected? Commented May 9, 2020 at 8:46

spring constant is inversely proportional to its length hence when a spring of constant $$k$$ is cut into $$n$$ number of pieces, the length becomes $$\frac1n$$ times initial length so spring constant becomes $$k/(1/n)=nk$$. therefore $$k$$ becomes $$n$$ times on cutting a spring.