2025/8/5,
Hiizu Nakanishi
Note on the Falling Motion of a Slinky Coil 2
Case of the coil stretched by a constant tension $T$
In the case where the lower end of a slinky coil is pulled by a
constant tension instead of gravity, we determine (i) the speed at
which the coil contracts after the top end is released and (ii) the
wave speed that propagates through the coil.
Equilibrium state of the coil
Let the total mass and spring constant of the coil be $M$ and $K$,
respectively, and specify each part of the coil by the continuous
variable $\ell\in[0,1]$. The mass $m$ and spring constant $k$ of an
infinitesimal segment $d\ell$ are
\[
m = M\,d\ell, \qquad k = {K\over d\ell}.
\]
If the entire coil is under a uniform tension $T$, the elongation $dz$
of the segment $d\ell$ is
\[
dz = {T\over k} = {T\over K}\, d\ell,
\]
so the equilibrium configuration of the coil is
\[
z^{0}(\ell)
= \int_0^{z^{0}}dz'
= \int_0^{\ell} {T\over K}\, d\ell'
= {T\over K}\,\ell.
\tag{1}
\]
Contraction speed of the slinky coil
After the top end is released,
the part $0<\ell<\ell_{\rm top}$ collapses and moves together,
while $\ell_{\rm top}<\ell<1$ remains at rest.
The center‑of‑mass position $z_G(\ell_{\rm top})$ of the whole coil is then
\[
z_G(\ell_{\rm top})
= {1\over M}\left(
M\,\ell_{\rm top}\, z^{0}(\ell_{\rm top})
+ \int_{\ell_{\rm top}}^{1} M\, z^{0}(\ell')\, d\ell'
\right).
\]
At time $t$, the total momentum of the coil is $Tt$, thus using the
expression above one finds
\[
Tt
= M\,{d z_G(\ell_{\rm top})\over dt}
= M\,\ell_{\rm top}\,
{d z^{0}(\ell_{\rm top})\over dt},
\]
which yields
\[
\ell_{\rm top}(t) = \sqrt{K\over M}\; t.
\]
The velocity of the top end of the coil is therefore
\[
v_{\rm top}
= {d z^{0}(\ell_{\rm top})\over dt}
= {T\over M}\,{t\over \ell_{\rm top}}
= {T\over \sqrt{M K}},
\tag{2}
\]
which is equal to the propagation speed of waves in a coil under tension $T$.
Wave‑propagation speed in a slinky coil
Because the elongation is $dz$, the tension $\sigma$ is
\[
\sigma = k\,dz = K\,{\partial z\over \partial\ell}.
\]
A wave travelling through the coil satisfies
\[
M\,d\ell\, \ddot z
= K\left(
{\partial z\over \partial\ell}\Bigg|_{\ell+d\ell/2}
- {\partial z\over \partial\ell}\Bigg|_{\ell-d\ell/2}
\right)
\;\Longrightarrow\;
{\partial^{2} z\over \partial t^{2}}
= {K\over M}\,{\partial^{2} z\over \partial \ell^{2}}.
\]
Using (1), the sound speed $c$ in the coil is
\[
c = \sqrt{K\over M}\,{d z^{0}\over d\ell}
= {T\over \sqrt{M K}},
\tag{3}
\]
which is the same as the contraction speed (2).
Thus, under a uniform tension with no external field such as gravity,
The contraction speed of a slinky coil equals the information‑propagation speed within the coil.
In contrast, when the coil is hung vertically, the elongation becomes smaller
toward the bottom, and accordingly the sound speed in the coil decreases.
However, the deceleration of the contracted coil falling under gravity is
smaller than that, so the contracted part—already behaving as a single body—
falls ahead of the pulse propagating through the coil.