Question

Consider a circular drum head or membrane, which may vibrate but is fixed everywhere along its edge. If Y(r, phi, t) describes the motion of the membrane, and the wave equation tells us that (1/v^2) d^2Y/dt^2 = del^2(Y(r, phi)), where v is some constant, then use the method of separation of variables to find the general description of its motion. What is the lowest frequency of vibration?

Answer 1

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Answer 2

So we have the function \(Y(r,\phi,t)\) and also we have:
\[\frac{1}{v^2}\frac{d^2Y}{dt^2}=del^2[Y(r,\phi,t)]\]

Answer 3

Please explain.

Answer 4

Hmm actually this is much harder than I thought..I am not very well versed in differential equations...

Answer 5

Still, please help if possible.

Answer 6

Hmm well I don't know much. But I know that the wave equation can be simplified to:
\[\frac{1}{v^2}\frac{d^2Y}{dt^2}=\nabla^2[Y(r,\phi,t)]\]
\[\frac{1}{v^2}\frac{d^2Y}{dt^2}=\left(\frac{\partial^2 Y(r,\phi,t)}{\partial r^2},\frac{\partial^2 Y(r,\phi,t)}{\partial \phi^2},\frac{\partial^2 Y(r,\phi,t)}{\partial t^2}\right)\]

Answer 7

Thanks but can you explain because i have a test tommorow

Answer 8

I don't know how to solve this problem, but I can go a few more steps in the right direction.

Answer 9

I would actually be very interested and curious if you would @MrMoose

Answer 10

The problem commands us to use the separation of variables method. Therefore, we assume:
\[Y(r,\phi,t) = X(r,\phi)T(t)\]
substituting in:
\[\frac{1}{v^2}\frac{d^2T(t)}{dt^2} = \nabla^2X(r,\phi)\]

Answer 11

Is it clear why this substitution works?

Answer 12

ie: the derivative with respect to time of a function of position is 0 and vice versa.

Answer 13

Note that the left hand side is a function of time only and the right hand side is a function of position only.

Answer 14

Therefore, in order to be equal at all points in time and space, they must be constants.

Answer 15

... I made a pretty massive typo in the last equation I wrote. It should be:

\[\frac{1}{v^2T(t)}\frac{d^2T(t)}{dt^2} = \frac{\nabla^2X(r,\phi)}{X(r,\phi)}\]

Answer 16

Ask if this isn't clear.

Answer 17

And sorry for any confusion that caused

Answer 18

Thanks Obama.

Answer 19

Now, continuing on:
\[\frac{d^2T(t)}{dt^2} = v^2T(t)C\]
and
\[\nabla^2X(r,\phi) = X(r,\phi)C\]

Answer 20

I can solve the time equation, but not the position equation.

Answer 21

Note that if we make the substitution:
\[\omega^2 = -v^2C\]
we get the harmonic oscillator equation:
\[\frac{d^2T(t)}{dt^2} +\omega^2T(t) = 0\]

Answer 22

@MrMoose can you just explain me what you would do instead of torturing my brain?

Answer 23

Short answer: I have no idea and I am trying to progress as far as I can and figure it out as I go.

Answer 24

So anyway, we see that:
\[Y(r,\phi,t) = X(r,\phi)(Acos(\omega t)+Bsin(\omega t) )\]

Answer 25

Or:
\[X(r, \phi )Acos(\omega t + \delta)\]

Answer 26

By the way, what class is this for?

Answer 27

well im 13 years old but i can do some calc, this is just for fun and i want to prove to some smartass senior in a test that im smarter than him. even without this i would beat him but still.

Answer 28

I found out the solution to the problem

Answer 29

http://en.wikipedia.org/wiki/Bessel_function

Answer 30

\[f_0 = \frac{2.4v}{2\pi R}\]

Answer 31

I think that this is the equation for the fundamental frequency

Answer 32

Oh now get it thanks bud.

Answer 33

The description of a radially symmetric wave on a struck drum is:
\[Y(r,\phi,t) = J_0(\frac{\omega r}{v})Acos(\omega t)\]

Answer 34

where J_0 is the 0-order Bessel function