Is the spring constant k changed when you divide a spring into parts?

I've always been taught that the spring constant $k$ is a constant — that is, for a given spring, $k$ will always be the same, regardless of what you do to the spring.

My friend's physics professor gave a practice problem in which a spring of length $L$ was cut into four parts of length $L/4$. He claimed that the spring constant in each of the new springs cut from the old spring ($k_\text{new}$) was therefore equal to $k_\text{orig}/4$.

Is this true? Every person I've asked seems to think that this is false, and that $k$ will be the same even if you cut the spring into parts. Is there a good explanation of whether $k$ will be the same after cutting the spring or not? It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why?

Answer

Well, the sentence

It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why?

clearly isn't a valid argument to calculate the $k$ of the smaller springs. They're different springs than their large parent so they may have different values of an "inherent property": if a pizza is divided to 4 smaller pieces, the inherent property "mass" of the smaller pizzas is also different than the mass of the large one. ;-)

You may have meant that it is an "intensive" property (like a density or temperature) which wouldn't change after the cutting of a big spring, but you have offered no evidence that it's "intensive" in this sense. No surprise, this statement is incorrect as I'm going to show.

One may calculate the right answer in many ways. For example, we may consider the energy of the spring. It is equal to $k_{\rm big}x_{\rm big}^2/2$ where $x_{\rm big}$ is the deviation (distance) from the equilibrium position. We may also imagine that the big spring is a collection of 4 equal smaller strings attached to each other.

In this picture, each of the 4 springs has the deviation $x_{\rm small} = x_{\rm big}/4$ and the energy of each spring is

$$E_{\rm small} = \frac{1}{2} k_{\rm small} x_{\rm small}^2 = \frac{1}{2} k_{\rm small} \frac{x_{\rm big}^2}{16}$$ Because we have 4 such small springs, the total energy is $$E_{\rm 4 \,small} = \frac{1}{2} k_{\rm small} \frac{x_{\rm big}^2}{4}$$ That must be equal to the potential energy of the single big spring because it's the same object $$= E_{\rm big} = \frac{1}{2} k_{\rm big} x_{\rm big}^2$$ which implies, after you divide the same factors on both sides, $$k_{\rm big} = \frac{k_{\rm small}}{4}$$ So the spring constant of the smaller springs is actually 4 times larger than the spring constant of the big spring.

You could get the same result via forces, too. The large spring has some forces $F=k_{\rm big}x_{\rm big}$ on both ends. When you divide it to four small springs, there are still the same forces $\pm F$ on each boundary of the smaller strings. They must be equal to $F=k_{\rm small} x_{\rm small}$ because the same formula holds for the smaller springs as well. Because $x_{\rm small} = x_{\rm big}/4$, you see that $k_{\rm small} = 4k_{\rm big}$. It's harder to change the length of the shorter spring because it's short to start with, so you need a 4 times larger force which is why the spring constant of the small spring is 4 times higher.