Wave equation and falling bar

classical-mechanicsordinary differential equationspartial differential equationswave equation

I want to understand the following: consider an (elastic) bar of one dimension (length only) with length $L=1$m floating at some height $h=2m$:

  1. Vertically
  2. Horizontally

I know that the equation describing the deformation $u(x,t)$ of its points is
$$
u_{tt} = u_{xx} + \gamma u_{txx}.
$$
For parameter $\gamma =0$ this is the 1-dimensional wave equation. The initial conditions for each point $x \in L = [0,1]$ must be

  • zero initial deformation $u(x,0)=x$
  • zero initial velocity $u_t(x,0)= 0$

Aren't these i.c.s sensible?

Then I let the bar fall either vertically or horizontally. I am curious, shouldn't the impact and thus the resulting solution for $u$ depend on gravity constant $g$ somehow?

Since the bar is 1d if it fall horizontally I guess there is no sulution to $u$ since the impact will be only a function of $y$ (height). But what would the deformation be if the bar fall vertically or at some angle?

Note: I normally would expect some solution proportioanl to $e^{-(x-t)+e^{-(x+t)}}$ but with the initial conditions as above I get a linear solution on Mathematica.

Best Answer

Yes. This problem isn't really 1D. The bar is described by the position vector $$ {\bf y}({\bf x}, t) = {\bf x} + {\bf u}({\bf x}, t) $$ where $\bf u$ denotes the displacement field. Of course the displacement of the bar's particles is oriented along its length (a unit vector $\bf n$ orientating the bar might be introduced here). The material is viscoelastic of Kelvin-Voigt type, and the equation of motion reads $$ \rho {\bf y}_{tt} = \operatorname{div}{\boldsymbol\sigma} + \rho {\bf g} $$ where ${\bf g}$ is the gravity force, and the stress tensor reads $$ {\boldsymbol\sigma} = (E \partial_{\bf n} {\bf y} + \eta \partial_{\bf n} {\bf y}_{t})\cdot {\bf n} \, ({\bf n}\otimes {\bf n}) \, . $$ The constants $\rho$, $E$, $\eta$ are respectively the mass density, the elastic modulus, and the viscosity. In the vertical case where ${\bf y} = y {\bf e}_3$, ${\bf n} = {\bf e}_3$ and ${\bf g} = -g {\bf e}_3$, we have $$ \rho y_{tt} = E y_{zz} + \eta y_{zzt} - \rho g $$ along the vertical direction ${\bf e}_3$ with coordinate $z$.

Note: the corresponding Lagrangian density is $$ L = \tfrac12\rho (y_t)^2 - \tfrac12 E (y_z)^2 +\rho g z $$ if $\eta=0$. The dissipative case is less straightforward as non-conservative forces arise.

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