- #1

- 746

- 8

## Homework Statement

Find stationary state of a circular membrane, on which a coin is put.

## Homework Equations

## The Attempt at a Solution

I am not sure about the second boundary condition, so if any of you has the time, please check the following solution:

$$u_{tt}=c^2\nabla ^2u+\rho g$$ Where ##u## is the defelction from a horizontal ##x## axis, ##\rho ## is the density of the coin and ##g## is the gravity. In polar coordinates this means $$\frac 1 r \frac{\partial }{\partial r}(r\frac{\partial }{\partial r}u)=-\frac{\rho g}{c^2}$$ Solving this PDE should give me $$u(r)=-\frac{\rho g}{4c^2}r^2+Aln(r)+B$$ Now the first boundary condition is kind of the obvious one: $$u(r=R)=0=-\frac{\rho g}{4c^2}R^2+Aln(R)+B$$ if ##R## is the radius of the membrane.

Now the second boundary condition is a bit confusing for me... But here is my idea:

If I write Newton's law for the coin with radius ##a##, than the condition should be (I guess): $$mg=F_0\frac{du}{dr}$$ where I already did an approximation that the membrane will deflect by a really small angle ##\sin \varphi \approx \varphi =\frac{du}{dr}##.

**Where ##F_0## should be the tension on the membrane.**- and this is the most confusing part, because in order to determine constants ##A## and ##B## I defined a new constant ##F_0## which is also unknown.

.. Hmmm? Is this ok or is the second boundary condition a complete nonsense?