Kvant Physics Problem 3
Two helical springs are made from identical pieces of steel wire.
Verified: yes
Verdicts: PASS + PASS
Solve time: 3m07s
Source on kvant.digital
Problem
Two springs were wound from two identical pieces of steel wire. The diameter of the coils of one spring is $d$, and of the other $2d$. Under the action of a load, the first spring 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 pieces of steel wire. Let the wire diameter be $a$, the shear modulus of steel be $G$, and the total length of wire in each spring be $L_w$. Since the pieces of wire are identical, both springs have the same $a$, $G$, and $L_w$.
The mean coil diameter of the first spring is $D_1=d$, and that of the second spring is $D_2=2d$.
Under the same applied load $F$, the first spring elongates by one tenth of its original length. If its length is $l_1$, then
$$\frac{\Delta l_1}{l_1}=\frac{1}{10}.$$
The required quantity is the fractional elongation of the second spring,
$$\frac{\Delta l_2}{l_2}.$$
The springs are treated as ideal close wound helical springs. Deformations are elastic and small, Hooke's law is valid, and the weight of the springs is neglected.
Physical Principles
For a close wound helical spring made from wire of diameter $a$, mean coil diameter $D$, and number of turns $N$, the spring constant is
$$k=\frac{G a^4}{8D^3N}.$$
The elongation under a load $F$ is given by Hooke's law:
$$\Delta l=\frac{F}{k}.$$
The length of a close wound spring is proportional to the number of turns:
$$l \propto N.$$
The total wire length is
$$L_w=\pi D N.$$
Since the two springs are wound from identical pieces of wire,
$$L_w=\text{const},$$
hence
$$N\propto \frac{1}{D}.$$
Derivation
Because the wire length is fixed,
$$N=\frac{L_w}{\pi D}.$$
Substituting this into the formula for the spring constant,
$$k=\frac{G a^4}{8D^3}\cdot\frac{\pi D}{L_w} =\frac{\pi G a^4}{8L_wD^2}.$$
Thus
$$k\propto \frac{1}{D^2}.$$
When the coil diameter doubles,
$$D_2=2D_1,$$
so
$$k_2=\frac{k_1}{4}.$$
For the same load $F$,
$$\Delta l=\frac{F}{k},$$
therefore
$$\Delta l_2 =\frac{F}{k_2} =\frac{F}{k_1/4} =4\frac{F}{k_1} =4\Delta l_1.$$
Next, determine how the original lengths compare. Since
$$l\propto N$$
and
$$N\propto \frac{1}{D},$$
we obtain
$$l\propto \frac{1}{D}.$$
Hence
$$l_2=\frac{l_1}{2}.$$
The fractional elongation of the second spring is
$$\frac{\Delta l_2}{l_2} = \frac{4\Delta l_1}{l_1/2} = 8,\frac{\Delta l_1}{l_1}.$$
Using the given value,
$$\frac{\Delta l_2}{l_2} = 8\cdot\frac{1}{10} = \frac{4}{5}.$$
Result
The symbolic relation is
$$\frac{\Delta l_2}{l_2} = 8,\frac{\Delta l_1}{l_1}.$$
Substituting the given fractional elongation of the first spring,
$$\frac{\Delta l_2}{l_2} = 8\left(\frac{1}{10}\right) = \frac{4}{5}.$$
The required extension is a fraction of the spring's own length, so it is dimensionless:
$$\boxed{\frac{\Delta l_2}{l_2}=\frac{4}{5}=0.80}.$$
Thus the second spring stretches by eighty percent of its original length.
Sanity Checks
The spring with larger coil diameter is softer because the stiffness scales as
$$k\propto D^{-2}.$$
Doubling the coil diameter reduces the stiffness by a factor of $4$, so a larger extension under the same load is expected.
The second spring contains only half as many turns because the same wire length is wound into larger coils. Its original length is therefore half as large. Combining a fourfold increase in elongation with a twofold decrease in original length gives an eightfold increase in fractional elongation, consistent with the result
$$\frac{4}{5}=8\cdot\frac{1}{10}.$$
The quantity $\Delta l/l$ is dimensionless, and the final answer is dimensionless as required.
The most common mistake is to use the dependence $k\propto D^{-3}$ without accounting for the fact that the number of turns changes when the wire length is fixed. Including the relation $N\propto D^{-1}$ changes the scaling to $k\propto D^{-2}$ and is essential for obtaining the correct answer.