Kvant Physics Problem 3

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Problem

From two identical pieces of steel wire, two springs were made. The diameter of the coils of one of them is equal to $d$, and of the other $2d$. The first spring, under the action of a load, stretched by one tenth of its length. By what fraction of its length will the second spring stretch under the action of the same load?

Setup and Assumptions

Two helical springs are made from identical steel wire segments of equal total wire length $L_w$ and identical wire diameter. The springs differ only in coil diameter: the first has coil diameter $D_1 = d$, the second has $D_2 = 2d$.

Both springs are assumed ideal: the wire obeys Hooke’s law, deformations are purely elastic, the coils are closely wound with uniform pitch, and end effects are neglected. The shear modulus of steel is $G$, and it is the same for both springs. The applied load is a force $F$ acting along the spring axis.

The first spring elongates by a fraction $\varepsilon_1 = \frac{\Delta L_1}{L_1} = \frac{1}{10}$. The goal is to determine $\varepsilon_2 = \frac{\Delta L_2}{L_2}$ for the second spring under the same force $F$.

Physical Principles

For a close-coiled helical spring made of wire, the axial stiffness is determined by torsional deformation of the wire. The spring constant is

$k = \frac{G r^4}{8 n D^3},$

where $r$ is the wire radius, $n$ is the number of turns, and $D$ is the coil diameter.

The axial extension satisfies Hooke’s law

$\Delta L = \frac{F}{k}.$

The total wire length is related to geometry by

$L_w = n \pi D,$

so the number of turns scales as $n = \frac{L_w}{\pi D}$.

The macroscopic spring length scales with the number of turns when pitch is fixed, so

$L \propto n.$

Derivation

Substituting $n = \frac{L_w}{\pi D}$ into the stiffness formula gives

$k = \frac{G r^4}{8 D^3} \cdot \frac{\pi D}{L_w} = \frac{G r^4 \pi}{8 L_w} \cdot \frac{1}{D^2}.$

All factors except $D$ are identical for both springs, so

$k \propto \frac{1}{D^2}.$

For the second spring with $D_2 = 2d$,

$\frac{k_2}{k_1} = \frac{D_1^2}{D_2^2} = \frac{d^2}{(2d)^2} = \frac{1}{4},$

hence

$k_2 = \frac{k_1}{4}.$

Under the same force $F$, the extensions satisfy

$\Delta L_1 = \frac{F}{k_1}, \quad \Delta L_2 = \frac{F}{k_2} = \frac{F}{k_1/4} = 4 \Delta L_1.$

The first spring’s relative extension is

$\frac{\Delta L_1}{L_1} = \frac{1}{10},$

so

$\Delta L_1 = \frac{L_1}{10}.$

Thus,

$\Delta L_2 = 4 \cdot \frac{L_1}{10} = \frac{2L_1}{5}.$

The second spring’s length scales with the number of turns. Since $n \propto \frac{1}{D}$, doubling $D$ halves $n$, hence the second spring length is

$L_2 = \frac{L_1}{2}.$

Therefore the relative extension of the second spring is

$\frac{\Delta L_2}{L_2} = \frac{\frac{2L_1}{5}}{\frac{L_1}{2}} = \frac{2}{5} \cdot 2 = \frac{4}{5}.$

Result

$\varepsilon_2 = \frac{\Delta L_2}{L_2} = \frac{4}{5} = 0.80.$

The second spring stretches by $0.80$, i.e. $80%$ of its length under the same load.

Sanity Checks

The dependence $k \propto D^{-2}$ is consistent with torsional energy storage in a helical spring, since increasing coil diameter increases lever arm for torque and reduces stiffness strongly. Doubling the diameter reduces stiffness by a factor of four, matching the derived scaling.

The extension increasing by a factor of four is consistent with reduced stiffness under constant force. The additional factor of two in relative elongation arises from the halved number of turns and thus halved spring length.

Dimensional consistency holds throughout since $k$ has units of $\mathrm{N/m}$ and all geometric scaling factors are dimensionless. The final result is a pure fraction, as required for relative deformation.