# PDE - Need help getting started.

1. Sep 6, 2009

### walter9459

1. The problem statement, all variables and given/known dataDerive the differential equation governing the longitudinal vibration of a thin cone which has uniform density p, show that it is
1/x/SUP] d/dx(x du/dx) = (1/c) d u/dt

Hint: The tensile force sigma = E du/dx where E is the Young's modulus (a constant), u is the longitudinal displacement and x is the longitudinal coordinate.

2. Relevant equations

3. The attempt at a solution I wasn't sure where to start so if someone give help me understand how to start this problem I would appreciate it.

2. Sep 6, 2009

### loveequation

Write $F = ma [/tex] for a longitudinal element of the rod of length [itex]dx$. The force on the left is the tensile stress times the area, i.e.,
$$E A(x) du/dx$$
pointing to the left.
Can you do the rest?

3. Sep 7, 2009

### walter9459

Okay now I have the following:
F = ma
F = EA(x)U(x)
= d/dx[EA(x)d/dx)] = m(x) d^2u/dt^2
u(x,t) = U(x)T(t)
d/dx[EA(x) d/dx[U(x)T(t)]] = d^2/dt^2 [m(x)T(t)]
t(t) d/dx [EA(x) d/dx [U(x)T(t)] = 1/T(t) d^2/dt^2 T(t) = -w^2
d/dx[EA(x) d/dx U(x)] + w^2m(x)U(x) = 0

I need to get to

1/x^2 d/dx[x^2 U(x)] = 1/c^2 U(tt)

but not sure if I'm on the right path and how I get from my result to the required result.

Regards,

Margaret

4. Sep 7, 2009

### loveequation

OK, don't introduce the separation of variables form $u(x, t) = T(t) U(x)$. At this stage we are merely trying to derive the PDE, not solve it.

The force on the left face of the differential element is
$$F(x) = EA(x)\frac{\partial u}{\partial x}$$
pointing to the left.

The force on the right face will be (using a Taylor series)
$$F(x + dx) = F(x) + \frac{\partial F}{\partial x}dx$$
pointing to the right.

Hence the net force is: (you write it down).

Next, the mass of the element is $m = \rho A(x) dx$, where $\rho$ is the density.
The acceleration of the element is $\partial^2 u/\partial t^2$.

So $F = ma$ becomes (you write it down).

Next introduce $c^2 = E/\rho$ into the above equation. This gives (you write it down).

Next write $A(x) = \pi R^2(x)$, where for a cone $R(x) = \alpha x$, $\alpha$ being the slope of the cone and $x$ is measured from the tip of the cone. Introduce this into the equation. This gives the answer. Done?

5. Sep 7, 2009

### walter9459

Thank you for your patience and help in explaining this in a manner that I was able to understand.

6. Sep 3, 2011

### daisy123

hey how do u get the net force in the above question?
I know t is the FR-FL. But what would it be?