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I was watching a documentary about reptiles, at some point, it showed how chameleons catch their prey.

Approximately, their tongue has an acceleration of $2500 \frac{m}{s^2}$, length of $0.7m$. They contract their accelerator muscle to shoot out their tongue, so they use elasticity to produce this incredibly fast motion.

It kept me thinking if one can approximate their tongue as a mass attached to a spring with a specific yet unknown Young's modulus $E$ and known period $T$ (imagine it launches its tongue off and retracts it back analogous to harmonic oscillator) since $k=\frac{AE}{L}$.

Having the information about the length, the area, mass, period and equation of motion of simple harmonic oscillator, can we determine the Young's modulus of their tongue? How precise would it be?

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Young's modulus $E$ is defined as:

$$E=\frac{\sigma}{\varepsilon}$$

where $\sigma$ is the stress ($\mathrm{Nm^{-2}}$) and $\varepsilon$ the strain. To be sure:

$$\sigma=\frac{F}{A}$$ and: $$\varepsilon=\frac{L_0-L}{L_0}$$

From videos on the Chameleon's tongue that I've watched that system can be approximated by a short piece of string with a longer piece of elastic band attached to it. At the end of the elastic band is a bob with mass $m$:

Chameleon tongue model

At some point the bob is launched and when both string and elastic are tout, the bob has a velocity $v_0$. Call this $t=0$.

The elastic now starts exerting a restoring force, so that the Newtonian equation of motion (NoM) becomes:

$$ma=-kx\tag{1}$$

(we ignore the mass of the string and elastic)

Here $k$ is the spring constant of the elastic band and $x=L-L_0$ ($L_0$ is the unstretched length of the elastic and $L$ its stretched length.

$(1)$ is indeed the NoM of an SHO:

$$m\ddot{x}+kx=0$$

Or:

$$\ddot{x}+\omega^2 x=0$$

where $\omega^2=\frac{k}{m}$

With initial conditions $\dot{x}(0)=v_0$ and $x(0)=0$, it has the solution:

$$x(t)=A\sin(\omega t+\phi)$$

$x(0)=0$ tells us that $\phi=0$

And $\dot{x}(0)=v_0$:

$$A\omega\cos(0)=v_0\Rightarrow A=\frac{v_0}{\omega}$$

So that:

$$x(t)=\frac{v_0}{\omega}\sin \omega t$$

So if the period $T$ could be measured, then because:

$$\omega=\frac{2\pi}{T}=\sqrt{\frac{k}{m}}$$

$k$ and thus $E$ could be estimated.

When the mass for the first time comes to a halt, an amount of time $T/4$ has elapsed.


And here's an important complication, arising from the high strains the tongue is subjected to.

We wrote initially that:

$$\sigma=\frac{F}{A}$$

Assuming the material is incompressible:

$$A_0L_0=A(L)L\Rightarrow A(L)=A_0\frac{L_0}{L}$$

where $A$ is the cross-section.

$$E=\frac{\sigma}{\varepsilon}=F\frac{L}{A_0L_0}\times \frac{L_0}{L_0-L}$$

$$E(L)=\frac{F(L)L}{A_0(L_0-L)}$$

This means that:

$$\frac{\rm{d}E}{\rm{d}F}\neq\text{constant}$$

because $E$ is not invariant to $\varepsilon$. In plain English, the elastic band doesn't behave like a Hookean spring with constant $k$: instead $k$ depends on $x$.

Stress-strain curve

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  • $\begingroup$ Thanks for the detailed answer. Assuming that spring and elastic has no mass would cause some error (but otherwise its harder to solve) but I wonder how accurate this approximation would be for $E$ if we had a slow motion video of him catching his prey. $\endgroup$
    – Monopole
    Commented Feb 2, 2021 at 8:15
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    $\begingroup$ Well, there's another source of error, not mentioned hitherto. Because the strain of the elastic section is large, its cross-section isn't constant. I'll add a point about that in the answer. $\endgroup$
    – Gert
    Commented Feb 2, 2021 at 14:30

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