Geometric elasticity · A course with experiments

L3 · Chapter 4

How fast a snap happens

Put inertia back into the beam equation and time comes in units of $t^{*}=L^{2}(12\rho_{s}/Eh^{2})^{1/2}$ , the time a bending wave needs to feel both ends of the strip. At the moment of release the strip sits on an unstable shape, so a small disturbance grows as $e^{\sigma t}$ . For a deep arch $\sigma\approx24.113$ whatever the depth; near the buckling threshold $\sigma$ falls towards zero. After the jump the inverted arch rings at about 44 in the same units. A 240 mm plastic strip snaps in about ten milliseconds and rings near 25 Hz.

Sandbox Physics · 2026-10-30

Putting inertia back

A strip of density $\rho_{s}$ and thickness $h$ carries mass $\rho_{s}h$ per unit area. Adding its acceleration to the beam equation of chapter 2 and measuring time in units of $t^{*}$ leaves an equation with no material constants at all. The Lab adds a damping term $\gamma\,\partial w/\partial t$ so that the strip settles after each snap.

Pandey et al. (2014) write the time unit with the sound speed $c=(E/\rho_{s})^{1/2}$ and read it as the time for the strip to feel its ends. Gomez, Moulton and Vella (2017) reach the same $t^{*}$ by balancing inertia against bending. It grows with the square of the length and falls with thickness, so a long thin strip is slow and a short thick one is fast.

\frac{\partial^{2}w}{\partial t^{2}}+\frac{\partial^{4}w}{\partial x^{4}}+\tau^{2}\frac{\partial^{2}w}{\partial x^{2}}=0,\qquad t^{*}=L^{2}\Bigl(\frac{12\rho_{s}}{Eh^{2}}\Bigr)^{1/2}=\frac{2\sqrt{3}\,L^{2}}{c\,h}

The growth rate at the moment of release

When the blade loses contact, the strip is at the flat compressed shape (shallow arch) or the pure twisted shape (deep arch). Write the shape as that state plus a small piece $\epsilon\,w_{p}(x)\,e^{\sigma t}$ , perturb the load as well, and keep terms of first order in $\epsilon$ . The result is an eigenvalue problem for $\sigma$ with the clamped conditions on $w_{p}$ , Eqs. 4 and 5 of Pandey et al.

From the twisted shape there is a single growing mode, with $\sigma\approx24.113$ for every $d$ and $S$ . From the flat shape, $\sigma$ depends on how far $d/S$ is above $4\pi^{2}$ . Just above that threshold Pandey et al. find Eq. 6, so the growth rate rises from zero as the strip is confined more, then levels off at 24.113 once $d/S$ passes 80.76. Eq. 6 is a limit formula: solving the full eigenvalue problem here gives 8.56 at $d/S=45$ , where Eq. 6 gives 8.38, and 16.67 at 60, where it gives 15.55.

\sigma\approx\frac{4\pi^{3/2}}{\sqrt{3}}\Bigl[\Bigl(\frac{d}{S}\Bigr)^{1/2}-2\pi\Bigr]^{1/2}\quad\bigl(d/S\ \text{just above}\ 4\pi^{2}\bigr),\qquad\sigma\approx24.113\quad\bigl(d/S>80.76\bigr)

Damping slows the start

With the damping term, a perturbation $e^{\lambda t}$ must satisfy $\lambda^{2}+\gamma\lambda=\sigma^{2}$ . The positive root is the growth rate the Lab measures from the midpoint as it leaves the clamp line. Damping lowers it but cannot stop it: for any $\gamma$ the root stays positive as long as $\sigma>0$ .

For the deep arch preset, $\gamma=6$ and $\sigma=24.113$ give 21.30. The Lab’s 48-segment strip has $\sigma=24.04$ , 0.3 % low, which gives 21.23; the midpoint’s measured rate is 21.14, 0.4 % below that.

\lambda=-\frac{\gamma}{2}+\sqrt{\frac{\gamma^{2}}{4}+\sigma^{2}}

Ringing after the snap

The inverted arch is stable. Linearizing about it gives $\sigma^{2}<0$ , so the strip vibrates at $\omega=(-\sigma^{2})^{1/2}$ . Just above the threshold this is Eq. 7 of Pandey et al. In their Kirchhoff calculation, which allows large slopes, $\omega$ keeps rising until $d/S=122.4$ and then stays at 42.38, and beyond that point the lowest mode also moves sideways. Their supplement gives 44.36 for the same plateau in the shallow-angle model used here, where the plateau mode is the antisymmetric one.

The ringing period is $t_{r}=(2\pi/\omega)\,t^{*}$ , which is $0.148\,t^{*}$ with 42.38. For a snapping spherical cap the same paper predicts $0.133\,t^{*}$ , with the base diameter as the length, and its Fig. 5 shows measured arches and shells, toy poppers among them, each following its own line. In the Lab, the deep arch rings at a measured 43.0, the lightly damped preset at 46.1: the first swings are large, and large swings ring at a slightly different frequency from small ones.

\omega\approx\frac{2^{3/2}\pi}{3^{1/2}}\Bigl(\frac{d}{S}-4\pi^{2}\Bigr)^{1/2}\quad\bigl(d/S\ \text{just above}\ 4\pi^{2}\bigr),\qquad t_{r}=\frac{2\pi}{\omega}\,t^{*}

A snap that slows down near its threshold

Gomez, Moulton and Vella (2017) clamp one end of a strip at an angle $\alpha$ and measure confinement by $\mu=\alpha\,(\Delta L/L)^{-1/2}$ . The inverted shape exists only while $\mu<1.782$ . Just past that fold the strip starts from rest near a shape that is no longer an equilibrium and has no growing mode. Its midpoint moves away quadratically in time, not exponentially.

The time to get clear grows as the fold is approached, $T_{b}\approx0.179\,\Delta\mu^{-1/4}$ in units of $t^{*}$ , with $\Delta\mu=\mu-1.782$ . This is critical slowing down without any dissipation. Pandey et al.’s strip is released from a genuinely unstable shape, so its start is exponential and its time, $1/\sigma$ , stays finite for every deep arch.

T_{b}\approx0.179\,\Delta\mu^{-1/4},\qquad\Delta\mu=\mu-1.782

Worked example

A 240 mm plastic strip

Gomez et al. used PET film with $E=5.707\ \mathrm{GPa}$ , $\rho_{s}=1337\ \mathrm{kg\,m^{-3}}$ and $h=0.35\ \mathrm{mm}$ , in lengths including 240 mm. The sound speed is $c=(5.707\times10^{9}/1337)^{1/2}=2066\ \mathrm{m\,s^{-1}}$ and $t^{*}=2\sqrt{3}\times0.240^{2}/(2066\times3.5\times10^{-4})=0.276\ \mathrm{s}$ .

Its stretchability is $S=(h/L)^{2}/12=1.77\times10^{-7}$ , so any arch deeper than $80.76\,S$ , an end shortening of 0.0014 %, counts as a deep arch. The undamped snap therefore grows by a factor $e$ every $t^{*}/24.113=11.4\ \mathrm{ms}$ . The inverted arch rings with period $(2\pi/44.36)\,t^{*}=39.1\ \mathrm{ms}$ , about 25.6 Hz; with Kirchhoff’s 42.38 it is 40.9 ms. Clamped at an angle and set at $\Delta\mu=0.1$ , the same strip would need $0.179\times0.1^{-1/4}\,t^{*}=0.318\,t^{*}=87.8\ \mathrm{ms}$ .

t^{*}=0.276\ \mathrm{s},\qquad \frac{t^{*}}{24.113}=11.4\ \mathrm{ms},\qquad \frac{2\pi}{44.36}\,t^{*}=39.1\ \mathrm{ms}

An experiment to think through

Predict, change, explain

Rank the presets “Deep arch twists, then snaps”, “Barely buckled: a slow snap” and “Little damping: it rings” by their measured growth rate $\lambda$ . Which one has the largest?
In the arch Lab play the three presets in turn and read the growth rate and the ringing frequency after each snap. Then set the damping to 0 on the deep arch and compare again.
Use $\lambda=-\gamma/2+(\gamma^{2}/4+\sigma^{2})^{1/2}$ with $\sigma=24.113$ and Eq. 6 to account for each measured rate, and say why the barely buckled strip leaves its flat shape so slowly.

Press a clamped arch until it snaps

The arch Lab solves the clamped beam equation with midline stretching on 48 segments, pushed at the middle by a blade that can only press.

Try explaining it yourself

1. Cut the PET strip to 120 mm. What are $t^{*}$ , the e-folding time and the ringing frequency now?

Read an explained answer

$t^{*}$ scales with $L^{2}$ , so it falls by four to 0.069 s. The e-folding time becomes 2.86 ms and the ringing frequency four times higher, about 102 Hz.

2. Gomez et al.’s steel shim has $E=203.8\ \mathrm{GPa}$ , $\rho_{s}=7881\ \mathrm{kg\,m^{-3}}$ and $h=0.1\ \mathrm{mm}$ . For a 140 mm strip, find $t^{*}$ and the e-folding time of a deep snap.

Read an explained answer

$c=(203.8\times10^{9}/7881)^{1/2}=5085\ \mathrm{m\,s^{-1}}$ , $t^{*}=2\sqrt{3}\times0.14^{2}/(5085\times10^{-4})=0.134\ \mathrm{s}$ , so $t^{*}/24.113=5.5\ \mathrm{ms}$ . Steel’s faster sound speed is partly offset by the thinner sheet.

3. The barely buckled preset has $d/S=45$ and $\gamma=3$ . What growth rate does Eq. 6 predict with damping, and why does the Lab measure a little more?

Read an explained answer

Eq. 6 gives $\sigma=8.38$ , so $\lambda=-1.5+(2.25+8.38^{2})^{1/2}=7.02$ . The Lab measures 7.18 because Eq. 6 is a limit formula valid only as $d/S$ approaches $4\pi^{2}$ ; at 45 the full eigenvalue is 2.1 % higher, and the Lab’s own linear analysis on its grid predicts 7.22.

4. In Gomez et al.’s setup, how much longer does the snap take at $\Delta\mu=0.001$ than at $\Delta\mu=0.1$ ?

Read an explained answer

The ratio is $(0.1/0.001)^{1/4}=100^{1/4}=3.16$ : from $0.318\,t^{*}$ to $1.01\,t^{*}$ . The quarter power makes the slowing gentle; even a hundredfold closer approach only triples the time.

Sources to read alongside

Pandey, Moulton, Vella & Holmes, Dynamics of snapping beams and jumping poppers, EPL 105, 24001 (2014), arXiv:1310.3703v2
Read in full, with the supplementary information: the dynamic beam equation and the integrated Hooke law (Eqs. 1 and 2), the three ranges of d/S and the force on the twisted branch (Eq. 3), the linear stability problem (Eqs. 4 and 5), the growth rate 24.113 and the near-threshold formulas (Eqs. 6 and 7), the ringing frequencies 42.38 and 44.36, the popper equation (Eq. 9) and Figs. 1 to 5.
Gomez, Moulton & Vella, Critical slowing down in purely elastic “snap-through” instabilities, Nature Physics (2017), arXiv:1609.03809v2
Main text and Methods read: the elastic time scale (Eq. 2), the fold where the inverted arch stops existing, the amplitude equation (Eq. 7) whose solution starts by growing as the square of time, the snap time that grows as the distance to the fold to the power minus one quarter (Eq. 9, Fig. 4) and the PET and steel strip properties. The appended supplementary derivation was not read in detail.

Putting inertia back

The growth rate at the moment of release

Damping slows the start

Ringing after the snap

A snap that slows down near its threshold

A 240 mm plastic strip

Predict, change, explain

Try explaining it yourself

1. Cut the PET strip to 120 mm. What are t∗t^{*}t∗, the e-folding time and the ringing frequency now?

2. Gomez et al.’s steel shim has E=203.8 GPaE=203.8\ \mathrm{GPa}E=203.8 GPa, ρs=7881 kg m−3\rho_{s}=7881\ \mathrm{kg\,m^{-3}}ρs​=7881 kgm−3 and h=0.1 mmh=0.1\ \mathrm{mm}h=0.1 mm. For a 140 mm strip, find t∗t^{*}t∗ and the e-folding time of a deep snap.

3. The barely buckled preset has d/S=45d/S=45d/S=45 and γ=3\gamma=3γ=3. What growth rate does Eq. 6 predict with damping, and why does the Lab measure a little more?

4. In Gomez et al.’s setup, how much longer does the snap take at Δμ=0.001\Delta\mu=0.001Δμ=0.001 than at Δμ=0.1\Delta\mu=0.1Δμ=0.1?

Sources to read alongside

1. Cut the PET strip to 120 mm. What are $t^{*}$ , the e-folding time and the ringing frequency now?

2. Gomez et al.’s steel shim has $E=203.8\ \mathrm{GPa}$ , $\rho_{s}=7881\ \mathrm{kg\,m^{-3}}$ and $h=0.1\ \mathrm{mm}$ . For a 140 mm strip, find $t^{*}$ and the e-folding time of a deep snap.

3. The barely buckled preset has $d/S=45$ and $\gamma=3$ . What growth rate does Eq. 6 predict with damping, and why does the Lab measure a little more?

4. In Gomez et al.’s setup, how much longer does the snap take at $\Delta\mu=0.001$ than at $\Delta\mu=0.1$ ?