# Thread: Wave equation - vibrations of a beam

1. ## Wave equation - vibrations of a beam

Hi, I have this question:

The longitudinal displacement u(x,t) for longitudinal vibrations of a beam of uniform cross-section satisfies the wave equation:

v$\displaystyle u_{xx}$ - p$\displaystyle u_{tt}$ = 0

where v is the modulus of elasticity of the beam and p is the mass density. For a beam of length L with the end x = 0 fixed and the end x = L free, the boundary conditions are:

u(0,t) = 0 , $\displaystyle u_x$(L,t) = 0.

Determine the natural angular frequencies of vibration of such a beam.

Using separation of variables u(x,t) = F(x)G(t) I get two decoupled first order equations:

F"(x) - (c/v)F(x) = 0 and G"(t) - (c/p)G(t) = 0.

I'm not sure where to go from here though. In all my examples from lectures, both ends have been fixed so the form of F(x) is known to be a function of sine and cosine. But I have no idea how to deal with the free end.

Katy

2. Hi, I'm unsure where the c has come from in your decoupled equations so starting with

$\displaystyle v u_{xx} - p u_{tt} = 0$

and following with $\displaystyle u = F(x)G(t)$ then substituting into the first equation

$\displaystyle \frac{F''}{F} = \frac{p}{v}\frac{G''}{G}$

and as both sides are independant of each other can be written equal to a constant, $\displaystyle -k^2$ say. This could be any constant, but the square allows for neatness in a minute, the negative allows for a solution in the cosine / sine form rather than hyperbolic solutions.

You would then get a solution of the form

$\displaystyle F(x) = A \cos(kx) + B \sin(kx)$

As I said I don't quite follow where the c comes from but assuming it's important I'm guessingyour lecture notes lead to something similar to

$\displaystyle F(x) = A \cos(\sqrt{c/v}x) + B \sin(\sqrt{c/v}x)$?

Regardless the G function can be solved similarly but it's t dependant and t is unimportant so leave it as G.

$\displaystyle u(x,t) = \left(A \cos(kx) + B \sin(kx)\right)G(t)$

Boundary condition at x=0 leads to B=0

$\displaystyle u(x,t) = A \cos(kx) G(t)$

Then differentiate u wrt x for the second BC

$\displaystyle u_x = -Ak \sin(kx) G(t)$

This equals zero for x=L, either G=0 which would be trivial so

$\displaystyle -Ak\sin(kL)=0$

k can't be equal to zero so it is set so the sine function is zero for all L,
i.e.
$\displaystyle kL = n\pi$
$\displaystyle k = n\pi / L$

Then
$\displaystyle u = A \cos(\frac{n\pi}{L}x)G(t)$

and (I think) you have the answer assuming you have $\displaystyle \sqrt{c/v}$ instead of my k?