= Rod wavelength in a Kundt tube
{c}
{title2=$c_{\rm rod}/c_{\rm gas}=\lambda_{\rm rod}/\lambda_{\rm gas}$}
A freely moving rod end has zero axial <stress>, not zero axial <displacement>. For a uniform slender rod of length $\ell$, its low-frequency extensional modes have shapes $\xi_r(x)\propto\cos(n\pi x/\ell)$ and <wavelengths> $\lambda_r=2\ell/n$. A centre clamp is compatible with the $n=1$ displacement node and $\lambda_r=2\ell$; clamps at $\ell/4$ and $3\ell/4$ are compatible with $n=2$ and $\lambda_r=\ell$. The support positions therefore matter before a gas <wavelength> can determine a rod <speed of sound>.
For a chosen model with $E=200\,\mathrm{GPa}$, $\rho_r=8000\,\mathrm{kg/m^3}$ and $\ell=1.60\,\mathrm m$, the slender-rod extensional relation $c_r=\sqrt{E/\rho_r}$ gives $5000\,\mathrm{m/s}$. The two-quarter-support mode then has $f=3125\,\mathrm{Hz}$. If the gas has stipulated $c_g=343\,\mathrm{m/s}$, its half-<wavelength> is $d_g=54.88\,\mathrm{mm}$, and
$$
\frac{c_r}{c_g}=\frac{\lambda_r}{\lambda_g}=\frac{1.60\,\mathrm m}{2(0.05488\,\mathrm m)}\simeq14.5773.
$$
The centre-supported fundamental would instead have $f=1562.5\,\mathrm{Hz}$ and twice that gas spacing. The same rod material has not changed its wave speed. In this slender-rod model, $E$ is <Young's modulus>; a bulk compressional wave involves a different elastic-modulus combination, so the two wave speeds are not interchangeable.
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