Spring Constant After Cutting

The result

A spring of natural length $L$ and spring constant $k$ is cut to a shorter length $l$. The spring constant of the cut piece:

$k' = \frac{L}{l} \cdot k$

Shorter spring → larger spring constant. Cutting a spring makes it stiffer.

Why cutting increases stiffness

Think of a spring as $N$ identical tiny springs connected in series, each of length $\delta = L/N$ and spring constant $k_0$.

For the full spring (series combination of $N$ tiny springs):

$\frac{1}{k} = \frac{N}{k_0} \implies k_0 = Nk$

For a cut piece of length $l$ (contains $n = l/\delta = lN/L$ tiny springs):

$\frac{1}{k'} = \frac{n}{k_0} = \frac{lN/L}{Nk} = \frac{l}{kL}$

$k' = \frac{kL}{l}$

Cutting into two pieces

A spring of length $L$ and constant $k$ cut into two pieces of lengths $l_1$ and $l_2$ ($l_1 + l_2 = L$):

$k_1 = \frac{kL}{l_1}, \quad k_2 = \frac{kL}{l_2}$

Verification — reconnect in series:

$\frac{1}{k_{series}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{l_1}{kL} + \frac{l_2}{kL} = \frac{l_1+l_2}{kL} = \frac{L}{kL} = \frac{1}{k}$

$k_{series} = k$ — reconnecting the pieces gives back the original spring. ✓

Cutting in ratio $m:n$

If the spring is cut in ratio $m:n$:

$l_1 = \frac{m}{m+n}L, \quad l_2 = \frac{n}{m+n}L$

$k_1 = \frac{(m+n)k}{m}, \quad k_2 = \frac{(m+n)k}{n}$

Example: Spring of constant $k$ cut in ratio $1:2$:

$k_1 = 3k \quad (shorter\ piece), \quad k_2 = \frac{3k}{2} \quad (longer\ piece)$

The shorter piece is stiffer — consistent with the general result.