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Equation of motion from a Lagrangian

  1. Apr 19, 2016 #1
    1. The problem statement, all variables and given/known data

    A string of length ##a##, mass per unit length ##\sigma## and under tension ##T## is fixed at each end. The Lagrangian governing the time evolution of the transverse displacement ##y(x, t)## is

    ##L = \int_{0}^{a} dx \bigg[ \frac{\sigma}{2} \Big( \frac{\partial y}{\partial t}\Big)^{2} - \frac{T}{2} \Big( \frac{\partial y}{\partial x}\Big)^{2} \bigg]##

    where ##x## identifies position along the string from one end point. By expressing the displacement as a sine series Fourier expansion in the form

    ##y(x,t) = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \text{sin} \big(\frac{n\pi x}{a}\big) q_{n}(t)##

    show that the Lagrangian becomes

    ##L = \sum_{n=1}^{\infty} \bigg[ \frac{\sigma}{2} \dot{q}_{n}^{2} - \frac{T}{2} \big( \frac{n \pi}{a}\big)^{2} q_{n}^{2} \bigg]##.

    Derive the equations of motion. Hence show that the string is equivalent to an infinite set of decoupled harmonic oscillators with frequencies

    ##\omega_{n} = \sqrt{\frac{T}{\sigma}} \big( \frac{n \pi}{a} \big)##.

    2. Relevant equations

    3. The attempt at a solution

    Before I solve the problem, I want to convince myself that the form of the Lagrangian is correct. I can see that the kinetic term is correct because it is simply half times the mass density times speed squared for each infinitesimal segment integrated over the entire length.

    But, I am having difficulty understanding the form of the potential term. Why, for each infinitesimal segment, is the half of the tension multiplied by the square of the gradient to obtain the potential energy?
     
  2. jcsd
  3. Apr 19, 2016 #2

    TSny

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    It has to do with the amount of stretching of an infinitesimal segment as the segment moves away from its equilibrium position. The tension T does work on the segment as it stretches. The potential energy term represents this work.

    Take a small segment of the string between ##x## and ##x+dx##. Compare the length of this segment when the string is in the equilibrium position (zero slope) with the length of the segment when it has a slope ##\frac{\partial y}{\partial x}##.
     
  4. Apr 20, 2016 #3
    The potential energy of an infinitesimal segment of the string arises from the work done on the infinitesimal segment of the string by the tension.

    The work done on the infinitesimal segment of the string = (tension)(extension of the infinitesimal segment)

    Now, extension ##= ds - dx##

    ##= \sqrt{(dx)^{2}+(dy)^{2}}-dx##
    ##= dx \bigg(\sqrt{1+ \Big(\frac{\partial y}{\partial x}\Big)^{2}}-1\bigg)##
    ##= dx \bigg( 1+ \frac{1}{2} \Big(\frac{\partial y}{\partial x}\Big)^{2} + \mathcal{O}\Big( \Big(\frac{\partial y}{\partial x}\Big)^{4} \Big) -1 \bigg)##
    ##= dx \bigg( \frac{1}{2} \Big(\frac{\partial y}{\partial x}\Big)^{2} + \mathcal{O}\Big( \Big(\frac{\partial y}{\partial x}\Big)^{4} \Big) \bigg)##

    Under the assumption that the displacement of the string from equilibrium is rather small, so that ##\frac{\partial y}{\partial x} \ll 1##,

    potential energy of the infinitesimal segment ##= \frac{1}{2} T \ dx \Big(\frac{\partial y}{\partial x}\Big)^{2}##.

    So, total potential energy of the string ##= \int_{0}^{a}\ dx \frac{1}{2} T \Big(\frac{\partial y}{\partial x}\Big)^{2}##.

    Am I correct?
     
  5. Apr 20, 2016 #4

    TSny

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    Yes, that looks very good.
     
  6. Apr 24, 2016 #5
    So, the Lagrangian governing the time evolution of the transverse displacement ##y(x,t)## is ##L = \int_{0}^{a} dx \bigg[ \frac{\sigma}{2} \Big( \frac{\partial y}{\partial t}\Big)^{2} - \frac{T}{2} \Big( \frac{\partial y}{\partial x}\Big)^{2} \bigg]##.

    Now, the displacement ##y(x,t)## can be expressed as a sine series Fourier expansion of the form ##y(x,t) = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \text{sin} \big(\frac{n\pi x}{a}\big) q_{n}(t)##, because the boundary condition ##y(0,t)=0## ensures that all the amplitudes for the cosine terms of the Fourier expansion are zero. The factor of ##\sqrt{\frac{2}{a}}## can be absorbed into the amplitudes ##q_{n}(t)##, but the factor of ##\sqrt{\frac{2}{a}}## has been pulled out from the amplitudes ##q_{n}(t)## for reasons that will become clear later. The reason for rewriting the displacement ##y(x,t)## and hence the Lagrangian ##L## in terms of the Fourier amplitudes ##q_{n}(t)## will also become clear later.

    Now,

    ##y(x,t) = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \text{sin} \big(\frac{n\pi x}{a}\big) q_{n}(t)##
    ## \implies \frac{\partial y(x,t)}{\partial t} = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \text{sin} \big(\frac{n\pi x}{a}\big) \dot{q}_{n}(t)##
    ##\implies \Big(\frac{\partial y(x,t)}{\partial t}\Big)^{2}= \frac{2}{a} \sum_{m,n=1}^{\infty} \text{sin} \big(\frac{m\pi x}{a}\big) \text{sin} \big(\frac{n\pi x}{a}\big) \dot{q}_{m}(t) \dot{q}_{n}(t)##

    and

    ##y(x,t) = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \text{sin} \big(\frac{n\pi x}{a}\big) q_{n}(t)##
    ## \implies \frac{\partial y(x,t)}{\partial x} = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \Big(\frac{n\pi}{a}\Big) \text{cos} \big(\frac{n\pi x}{a}\big) q_{n}(t)##
    ## \implies \Big(\frac{\partial y(x,t)}{\partial x}\Big)^{2}= \frac{2}{a} \sum_{m,n=1}^{\infty} \Big(\frac{m\pi}{a}\Big)\Big(\frac{n\pi}{a}\Big) \text{cos} \big(\frac{m\pi x}{a}\big) \text{cos} \big(\frac{n\pi x}{a}\big) q_{m}(t) q_{n}(t)##

    so that

    ##L = \int_{0}^{a} dx \bigg[ \frac{\sigma}{2} \Big( \frac{\partial y}{\partial t}\Big)^{2} - \frac{T}{2} \Big( \frac{\partial y}{\partial x}\Big)^{2} \bigg]##
    ## = \frac{\sigma}{a} \sum_{m,n=1}^{\infty}\dot{q}_{m}(t)\dot{q}_{n}(t)\int_{0}^{a}dx\ \text{sin}\big(\frac{n\pi x}{a}\big)\text{sin}\big(\frac{m\pi x}{a}\big) - \frac{T}{a} \sum_{m,n=1}^{\infty} \Big(\frac{m\pi}{a}\Big)q_{m}(t)\Big(\frac{n\pi}{a}\Big)q_{n}(t)\int_{0}^{a}dx\ \text{cos}\big(\frac{n\pi x}{a}\big)\text{cos}\big(\frac{m\pi x}{a}\big)##.

    Now,

    ##\int_{0}^{a} dx\ \text{sin}\big(\frac{n\pi x}{a}\big)\text{sin}\big(\frac{m\pi x}{a}\big)##
    ##=\frac{1}{2}\int_{0}^{a}dx\ \text{cos}\big[\frac{(n-m)\pi x}{a}\big]-\text{cos}\big[\frac{(n+m)\pi x}{a}\big]##
    ##=\frac{1}{2}\Big[\frac{a}{(n-m)\pi}\text{sin}\big[\frac{(n-m)\pi x}{a}\big]-\frac{a}{(n+m)\pi}\text{sin}\big[\frac{(n+m)\pi x}{a}\big]\Big]_{0}^{a}##
    ##=\frac{1}{2}\Big[\frac{a\ \text{sin}(n-m)\pi}{(n-m)\pi}-\frac{a\ \text{sin}(n+m)\pi}{(n+m)\pi}\Big]##
    ##=\frac{a}{2}\delta_{mn}##

    and

    ##\int_{0}^{a} dx\ \text{cos}\big(\frac{n\pi x}{a}\big)\text{cos}\big(\frac{m\pi x}{a}\big)##
    ##=\frac{1}{2}\int_{0}^{a}dx\ \text{cos}\big[\frac{(n-m)\pi x}{a}\big]+\text{cos}\big[\frac{(n+m)\pi x}{a}\big]##
    ##=\frac{1}{2}\Big[\frac{a}{(n-m)\pi}\text{sin}\big[\frac{(n-m)\pi x}{a}\big]+\frac{a}{(n+m)\pi}\text{sin}\big[\frac{(n+m)\pi x}{a}\big]\Big]_{0}^{a}##
    ##=\frac{1}{2}\Big[\frac{a\ \text{sin}(n-m)\pi}{(n-m)\pi}+\frac{a\ \text{sin}(n+m)\pi}{(n+m)\pi}\Big]##
    ##=\frac{a}{2}\delta_{mn}##

    so that

    ##L = \frac{\sigma}{a} \sum_{m,n=1}^{\infty}\dot{q}_{m}(t)\dot{q}_{n}(t)\frac{a}{2}\delta_{mn} - \frac{T}{a} \sum_{m,n=1}^{\infty} \Big(\frac{m\pi}{a}\Big)q_{m}(t)\Big(\frac{n\pi}{a}\Big)q_{n}(t)\frac{a}{2}\delta_{mn}##
    ##= \frac{\sigma}{2}\sum_{n=1}^{\infty}(\dot{q}_{n}(t))^{2}-\frac{T}{2}\sum_{n=1}^{\infty}\big(\frac{n\pi}{a}\big)^{2}(q_{n}(t))^{2}##
    ##=\sum_{n=1}^{\infty}\frac{\sigma}{2}(\dot{q}_{n}(t))^{2}-\frac{T}{2}\big(\frac{n\pi}{a}q_{n}(t)\big)^{2}##

    Therefore, the Lagrangian ##L## in terms of the Fourier amplitudes ##q_{n}(t)## consists of ##n## terms, one for each amplitude ##q_{n}(t)##. Therefore, the motion of the system in Fourier-transformed space is decoupled. This can be demonstrated more explicitly by calculating the Euler-Lagrange equations of motion for the system:

    ##\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{q}_{m}}\Big)-\frac{\partial L}{\partial q_{m}}=0##
    ##\implies \frac{d}{dt}(\sigma \dot{q}_{m})+T \big(\frac{m\pi}{a}\big)^{2} q_{m}=0##
    ##\implies \sigma \ddot{q}_{m}+T\big(\frac{m\pi}{a}\big)^{2}q_{m}=0##
    ##\implies \ddot{q}_{m}=-\frac{T}{\sigma}\big(\frac{m\pi}{a}\big)^{2}q_{m}##

    Therefore, the dynamics of the string, i.e., the motion of the transverse displacement ##y(x,t)## of the string has been Fourier-transformed into the motion of an infinite set of decoupled harmonic oscillators with frequencies ##\omega_{n}=\sqrt{\frac{T}{\sigma}}\big(\frac{n\pi}{a}\big)##.

    The harmonic oscillators are decoupled from each other because each of the equations of motion contains only a single index and therefore describes the dynamics of only one of the infinitely many harmonic oscillators.

    Would you please comment on my solution?
     
  7. Apr 24, 2016 #6

    TSny

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    Looks very, very good. Extremely well written out!
     
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