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16.2: The Continuous Uniform Linear Chain

  • Page ID
    9657
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    The Lagrangian for the discrete lattice chain, for longitudinal modes, is given by equation \((14.10.3)\) to be

    \[L = \frac{1}{2} \sum^{n+1}_{j=1} \left( m\dot{q}^2_j - \kappa (q_{j-1} – q_j)^2\right)\label{16.1}\]

    where the \(n\) masses are attached in series to \(n+1 \) identical springs of length \(d\) and spring constant \(\kappa \). Assume that the spring has a uniform cross-section area \(A\) and length \(d\). Then each spring volume element \(\Delta \tau = Ad\) has a mass \(m\), that is, the volume mass density \(\rho = \frac{m }{\Delta \tau}\) or \(m = \rho \Delta \tau \). Chapter \(16.5\) will show that the spring constant \(\kappa = \frac{EA}{ d}\) where \(E\) is Young’s modulus, \(A\) is the cross sectional area of the chain element, and \(d\) is the length of the element. Then the spring constant can be written as \(\kappa = \frac{E\Delta \tau }{ d^2} \). Therefore Equation \ref{16.1} can be expressed as a sum over volume elements \(\Delta \tau = Ad\)

    \[L = \frac{1}{2} \sum^{n+1}_{j=1} \left( \rho \dot{q}^2_j − E \left(\frac{q_{j-1} -q_j}{d} \right)^2 \right) \Delta \tau \label{16.2}\]

    In the limit that \(n \rightarrow \infty\) and the spacing \(d = dx \rightarrow 0\), then the summation in Equation \ref{16.2} can be written as a volume integral where \(x = jd\) is the distance along the linear chain and the volume element \(\boldsymbol{\Delta} \boldsymbol{\tau} \rightarrow \mathcal{0}\). Then the Lagrangian can be written as the integral over the volume element \(d\tau\) rather than a summation over \(\Delta \tau \). That is,

    \[L = \frac{1}{ 2} \int \left( \rho \dot{q}^2 − E \left(\frac{dq(x, t)}{ dx }\right)^2 \right) d\tau\label{16.3}\]

    The discrete-chain coordinate \(q(t)\) is assumed to be a continuous function \(q(x, t)\) for the uniform chain. Thus the integral form of the Lagrangian can be expressed as

    \[L = \frac{1}{2} \int \left( \rho \dot{q}^2 − E \left( \frac{dq(x, t) }{dx } \right)^2 \right) d\tau = \int \mathfrak{L}d\tau \label{16.4}\]

    where the function \(\mathfrak{L}\) is called the Lagrangian density defined by

    \[\mathfrak{L} \equiv \frac{1}{2} \left( \rho \dot{q}^2 − E \left(\frac{dq(x, t) }{dx }\right)^2 \right) \label{16.5}\]

    The variable \(x\) in the Lagrangian density is not a generalized coordinate; it only serves the role of a continuous index played previously by the index \(j\). For the discrete case, each value of \(j\) defined a different generalized coordinate \(q_i\). Now for each value of \(x\) there is a continuous function \(q(x, t)\) which is a function of both position and time.

    Lagrange’s equations of motion applied to the continuous Lagrangian in Equation \ref{16.4} gives

    \[\rho \frac{ d^2q}{ dt^2} − E \frac{d^2q}{ dx^2} = 0 \label{16.6}\]

    This is the familiar wave equation in one dimension for a longitudinal wave on the continuous chain with a phase velocity

    \[v_{phase} = \sqrt{\frac{E}{ \rho}} \label{16.7}\]

    The continuous linear chain also can exhibit transverse modes which have a Lagrangian density were the Young’s modulus \(E\) is replaced by the tension \(\tau\) in the chain, and \(\rho\) is replaced by the linear mass density \(\mu\) of the chain, leading to a phase velocity for a transverse wave \(v_{phase} = \sqrt{\frac{\tau}{\mu}}\).


    This page titled 16.2: The Continuous Uniform Linear Chain is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.