Solving Drumhead PDE: Normal Modes and Estimation Techniques

[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:

[tex] \nabla^2 z = \frac{1}{c^2}\frac{\partial^2 z }{dt^2} [/tex]

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

Questions:

1) If I want to find the differential equation for [tex]z=Z(\rho) [/tex], 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 [tex]a^{\nu}-\rho^{\nu}[/tex] with [tex]\nu[/tex] an adjustable parameter...where do I start?

Thanks!

 

[Astronuc]

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.

 

[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!

:)

 

[spock0149]

Ok, so writing out nabla in polar I get:

[tex]\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}[/tex]

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:

[tex]\rho^2 Z''+\rho Z'+\rho^2 \omega^2(\frac{1}{c^2}-1)Z=0[/tex]

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

with [tex]\nu[/tex] some real number

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

Thanks!

 

[Astronuc]

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

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

Ok, so writing out nabla in polar I get:

[tex]\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}[/tex]

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

[tex]\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial z }{\partial \rho})\, =\, \frac{1}{c^2}\frac{\partial^2 z}{\partial t^2}[/tex]

Now we can try a trial function, [tex]Z(\rho)\,T(t)[/tex]

[tex]\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}[/tex]

which then yields

[tex]\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)[/tex].


Dividing through by [tex]Z(\rho)\,T(t)[/tex]

[tex]\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[/tex]

which then give two equations.


[tex]\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,\frac{1}{Z(\rho)}\,=\, -\lambda^2[/tex]

and

[tex]\frac{1}{c^2}\frac{d^2 T(t)}{dt^2}\,\frac{1}{T(t)} = -\lambda^2[/tex]

The second equation can be written

[tex]\frac{d^2 T(t)}{dt^2}\,+\,\omega^2\,T(t)\,=\,0[/tex]

and the other equation can be written as

[tex]\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,\frac{1}{Z(\rho)}\,=\, -\lambda^2[/tex]

which becomes

[tex]\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,+\,{\rho}\lambda^2\,Z(\rho)\,=\,0[/tex]

The last equation should simplify to

[tex]\rho^2 Z''+\rho Z'+\rho^2 \lambda^2Z\,=\,0[/tex]

Then letting [tex] x\,=\lambda\rho[/tex]

One can write

[tex]x^2 Z''(x)\,+\,x Z'(x)\,+\,x^2Z\,=\,0[/tex]

which is Bessel's equation of order zero.

 

[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!

 

[spock0149]

Thanks again for the help.

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

[tex]a^\nu-\rho^\nu[/tex]

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

Our equation reads:

[tex]\rho^2 Z'' + \rho Z'+\rho\frac{\omega^2}{c^2}Z=0[/tex]

Let: [tex]Z=a^\nu-\rho^\nu[/tex]
So,
[tex]Z'=-\nu\rho^{\nu-1}[/tex]
and
[tex]Z''=-\nu^2\rho^{\nu-1}[/tex]

Plugging into our original equation we get:

[tex]-\nu^2\rho^2\rho^{\nu-2}-\rho\nu\rho^{\nu-1}+\rho^2\frac{\omega^2}{c^2}(a^\nu - \rho^\nu)=0[/tex]

...some lines of algebra...

[tex]\omega=\sqrt\frac{\rho^{\nu-2}c^2\nu(\nu+1)}{a^\nu-\rho^\nu}[/tex]

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

 

[spock0149]

anyone care to take a stab?