1. Sep 10, 2006

### spock0149

I've searched through about 5 math books but don't know how to start this one:

I have a drumskin of radius a, and small transverse oscillations of amplitude:

$$\nabla^2 z = \frac{1}{c^2}\frac{\partial^2 z }{dt^2}$$

Ok, so I can write the normal mode as
$$z=Z(\rho)cos(\omega t)$$

Questions:

1) If I want to find the differential equation for $$z=Z(\rho)$$, do I just plug the second equation into the first, but use the polar coordinate version of nabla?

2) If I want to obtain an estimate for the lowest normal mode frequency using a trial function of form $$a^{\nu}-\rho^{\nu}$$ with $$\nu$$ an adjustable parameter...where do I start???

Thanks!

Last edited: Sep 10, 2006
2. Sep 11, 2006

### Astronuc

Staff Emeritus
Answer to #1 is yes, one needs the polar version of nabla.

It's been decades since I solved this problem, so I have to refresh my memory on it's solution, but one should get a Bessel's function, J(r), IIRC, and one looks for the first zero for the fundamental mode.

3. Sep 11, 2006

### spock0149

Thanks astronuc! The polar version works out nicely.

I'm still not sure how to use the trial function. Is the suggested trial function something I should use as z and plug into the equation?

Thanks!

:)

4. Sep 17, 2006

### spock0149

Ok, so writing out nabla in polar I get:

$$\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial z }{\partial \rho})+\frac{1}{\rho^2} \frac{\partial^2 z}{\partial \phi^2}+\frac{\partial^2 z}{\partial t^2} = \frac{1}{c^2}\frac{\partial^2 z}{\partial t^2}$$

I am trying to find a normal mode independant of azimuthal angle which is why I let the phi derivative go to zero.

after some re-arranging I get:

$$\rho^2 Z''+\rho Z'+\rho^2 \omega^2(\frac{1}{c^2}-1)Z=0$$

now, this kind of looks like Bessels equation, except the last term should look like $$(\rho^2-\nu^2)Z$$

with $$\nu$$ some real number

I can't quite make the jump from my equation to Bessels...can someone give me any ideas?

Thanks!

5. Sep 19, 2006

### Astronuc

Staff Emeritus
Yes, this is fine. The assumes no azimuthal dependency.

The temporal term on the left side is not correct. Nabla is spatial only. So, the correct equation should be,

$$\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial z }{\partial \rho})\, =\, \frac{1}{c^2}\frac{\partial^2 z}{\partial t^2}$$

Now we can try a trial function, $$Z(\rho)\,T(t)$$

$$\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial (Z(\rho)\,T(t)) }{\partial \rho})\, =\, \frac{1}{c^2}\frac{\partial^2 (Z(\rho)\,T(t))}{\partial t^2}$$

which then yields

$$\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial Z(\rho) }{\partial \rho})T(t)\,=\,\frac{1}{c^2}\frac{\partial^2 T(t)}{\partial t^2}Z(\rho)$$.

Dividing through by $$Z(\rho)\,T(t)$$

$$\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial Z(\rho) }{\partial \rho})\,\frac{1}{Z(\rho)}\,=\,\frac{1}{c^2}\frac{\partial^2 T(t)}{\partial t^2}\,\frac{1}{T(t)} = -\lambda^2$$

which then give two equations.

$$\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,\frac{1}{Z(\rho)}\,=\, -\lambda^2$$

and

$$\frac{1}{c^2}\frac{d^2 T(t)}{dt^2}\,\frac{1}{T(t)} = -\lambda^2$$

The second equation can be written

$$\frac{d^2 T(t)}{dt^2}\,+\,\omega^2\,T(t)\,=\,0$$

and the other equation can be written as

$$\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,\frac{1}{Z(\rho)}\,=\, -\lambda^2$$

which becomes

$$\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,+\,{\rho}\lambda^2\,Z(\rho)\,=\,0$$

The last equation should simplify to

$$\rho^2 Z''+\rho Z'+\rho^2 \lambda^2Z\,=\,0$$

Then letting $$x\,=\lambda\rho$$

One can write

$$x^2 Z''(x)\,+\,x Z'(x)\,+\,x^2Z\,=\,0$$

which is Bessel's equation of order zero.

Last edited: Sep 19, 2006
6. Sep 19, 2006

### spock0149

mmmmm....lovely. :)

Thanks astronuc. Thanks for pointing out that the nabla term was spatial only. I 'knew' that...but obviously it didn't surface from the deep cobwebs of my brain! I probably would have been stumped on this for ages without that!

The solutions makes a lot of sense. I saw that you used separation of variables and Bessels equation came out nicely!

Thanks again.

PS - cool beard!

7. Oct 4, 2006

### spock0149

Thanks again for the help.

Now, I have to find an estimate for the lowest normal more using the adjustable parameter:

$$a^\nu-\rho^\nu$$

Here is my effort, please tell me what you think:

$$\rho^2 Z'' + \rho Z'+\rho\frac{\omega^2}{c^2}Z=0$$

Let: $$Z=a^\nu-\rho^\nu$$
So,
$$Z'=-\nu\rho^{\nu-1}$$
and
$$Z''=-\nu^2\rho^{\nu-1}$$

Plugging into our original equation we get:

$$-\nu^2\rho^2\rho^{\nu-2}-\rho\nu\rho^{\nu-1}+\rho^2\frac{\omega^2}{c^2}(a^\nu - \rho^\nu)=0$$

...some lines of algebra...

$$\omega=\sqrt\frac{\rho^{\nu-2}c^2\nu(\nu+1)}{a^\nu-\rho^\nu}$$

Is this the correct way to get the normal mode? It makes sense to me but seems like its a little to simple.

Last edited: Oct 4, 2006
8. Oct 14, 2006

### spock0149

anyone care to take a stab?