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\begin{document}

Treat a metal rod with sounds waves going down it as a serious of
coupled masses and springs. \( V=\sum _{n}\frac{k_{s}}{2}(n_{n+1}+n)^{2} \) and \( T=\sum _{n}\frac{m}{2}n_{n}^{'2} \). The lagrangian gives a solution
of the form \( n_{n}=Ae^{-i\omega t}e^{i\beta n} \) . This gives a dispersion relation of the form: \( w^{2}=\frac{2k_{s}}{m}(1-cos(\beta )) \) for
\( \beta =ka<<1 \), you get \( w^{2}=\frac{k_{s}}{m}k^{2}a^{2} \). Then \( \frac{w}{k}=\sqrt{\frac{Y}{\mu }} \) where \( \mu =\frac{m}{a} \)= linear mass density, and \( Y=k_{s}a \)= Young's
modulus.

Instead of having a discrete \( n_{i} \), have a continuous \( n(x) \). Then \( V=\frac{k_{s}}{2}\sum _{n}(\frac{\partial n}{\partial x}|_{x=na}a)^{2}=\frac{Y}{2}\int dx(\frac{\partial n}{\partial x})^{2} \). The kinetic
energy becomes: \( T=\frac{1}{2}\mu \int dx(\frac{\partial n}{\partial t})^{2} \). So \( \delta S=\int dt\int dx[\mu \frac{\partial n}{\partial t}\frac{\partial \delta n}{\partial t}-Y\frac{\partial n}{\partial x}\frac{\partial \delta n}{\partial x}] \) \( =\int dt\int dx[\mu \frac{\partial ^{2}n}{\partial t^{2}}-Y\frac{\partial ^{2}n}{\partial x^{2}}]\delta n(x,t) \). This gives the wave equation directly.m

\end{document}
