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:

\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!

 

[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:

\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 independent 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!

 

[Astronuc]

I am trying to find a normal mode independent 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:

\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}

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.

 

[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:

a^\nu-\rho^\nu

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

Our equation reads:

\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.

 

[spock0149]

anyone care to take a stab?