Subfield: Mechanics

Zero-length springs and slinky coils

A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$.

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Figure 1: the relation between the force $F$ and the spring length $L$

A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as "slinky" may be (but not necessarily) such a ZLS.

Part A: Statics
Context question:
A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring?
Context answer:
$\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$


Context question:
A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$.
Context answer:
\boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$}


Extra Supplementary Reading Materials:

Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches.

Question: A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$.

Solution:
At every point along the statically hanging spring the weight of the mass below is balanced by the tension from above. This implies that at the bottom of the spring there is a section of length $l_{0}$ whose turns are still touching each other, as their weight is insufficient to exceed the threshold force $k L_{0}$ to pull them apart. The length $l_{0}$ can be derived from the equation:

$\frac{l_{0}}{L_{0}} M g=k L_{0}$, hence $l_{0}=\frac{k L_{0}^{2}}{M g}=\alpha L_{0}$.

For $l>l_{0}$, a segment of the unstretched spring between $l$ and $l+\mathrm{d} l$ feels a weight of $\frac{l}{L_{0}} M g$ from beneath, which causes its length to stretch from $\mathrm{d} l$ to $d y=\frac{F}{k L_{0}} d l=\frac{l}{L_{0}} M g \frac{d l}{k L_{0}}=$ $\frac{M g}{k L_{0}^{2}} l d l=\frac{l}{l_{0}} d l$.

Integration of the last expression over the stretched region, up to the point $L_{0}$, gives its height when the spring is stretched

$$
H=l_{0}+\int_{l_{0}}^{L_{0}} \frac{l}{l_{0}} d l=l_{0}+\left.\frac{l^{2}}{2 l_{0}}\right|_{l_{0}} ^{L_{0}}=l_{0}+\frac{1}{2 l_{0}}\left(L_{0}^{2}-l_{0}^{2}\right)=\frac{L_{0}^{2}}{2 l_{0}}+\frac{l_{0}}{2}=\frac{L_{0}}{2}\left(\alpha+\frac{1}{\alpha}\right)
$$

Final answer: $H=\frac{L_{0}}{2}(\alpha+\frac{1}{\alpha})$