anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
goformit100
  • goformit100
"Welcome to OpenStudy. I can guide regarding this useful site; ask your doubts from me, for it you can message me. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."
anonymous
  • anonymous
So we have the function \(Y(r,\phi,t)\) and also we have: \[\frac{1}{v^2}\frac{d^2Y}{dt^2}=del^2[Y(r,\phi,t)]\]
anonymous
  • anonymous
Please explain.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Hmm actually this is much harder than I thought..I am not very well versed in differential equations...
anonymous
  • anonymous
Still, please help if possible.
anonymous
  • anonymous
Hmm well I don't know much. But I know that the wave equation can be simplified to: \[\frac{1}{v^2}\frac{d^2Y}{dt^2}=\nabla^2[Y(r,\phi,t)]\] \[\frac{1}{v^2}\frac{d^2Y}{dt^2}=\left(\frac{\partial^2 Y(r,\phi,t)}{\partial r^2},\frac{\partial^2 Y(r,\phi,t)}{\partial \phi^2},\frac{\partial^2 Y(r,\phi,t)}{\partial t^2}\right)\]
anonymous
  • anonymous
Thanks but can you explain because i have a test tommorow
anonymous
  • anonymous
I don't know how to solve this problem, but I can go a few more steps in the right direction.
anonymous
  • anonymous
I would actually be very interested and curious if you would @MrMoose
anonymous
  • anonymous
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: \[\frac{1}{v^2}\frac{d^2T(t)}{dt^2} = \nabla^2X(r,\phi)\]
anonymous
  • anonymous
Is it clear why this substitution works?
anonymous
  • anonymous
ie: the derivative with respect to time of a function of position is 0 and vice versa.
anonymous
  • anonymous
Note that the left hand side is a function of time only and the right hand side is a function of position only.
anonymous
  • anonymous
Therefore, in order to be equal at all points in time and space, they must be constants.
anonymous
  • anonymous
... I made a pretty massive typo in the last equation I wrote. It should be: \[\frac{1}{v^2T(t)}\frac{d^2T(t)}{dt^2} = \frac{\nabla^2X(r,\phi)}{X(r,\phi)}\]
anonymous
  • anonymous
Ask if this isn't clear.
anonymous
  • anonymous
And sorry for any confusion that caused
anonymous
  • anonymous
Thanks Obama.
anonymous
  • anonymous
Now, continuing on: \[\frac{d^2T(t)}{dt^2} = v^2T(t)C\] and \[\nabla^2X(r,\phi) = X(r,\phi)C\]
anonymous
  • anonymous
I can solve the time equation, but not the position equation.
anonymous
  • anonymous
Note that if we make the substitution: \[\omega^2 = -v^2C\] we get the harmonic oscillator equation: \[\frac{d^2T(t)}{dt^2} +\omega^2T(t) = 0\]
anonymous
  • anonymous
@MrMoose can you just explain me what you would do instead of torturing my brain?
anonymous
  • anonymous
Short answer: I have no idea and I am trying to progress as far as I can and figure it out as I go.
anonymous
  • anonymous
So anyway, we see that: \[Y(r,\phi,t) = X(r,\phi)(Acos(\omega t)+Bsin(\omega t) )\]
anonymous
  • anonymous
Or: \[X(r, \phi )Acos(\omega t + \delta)\]
anonymous
  • anonymous
By the way, what class is this for?
anonymous
  • anonymous
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.
anonymous
  • anonymous
I found out the solution to the problem
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Bessel_function
anonymous
  • anonymous
\[f_0 = \frac{2.4v}{2\pi R}\]
anonymous
  • anonymous
I think that this is the equation for the fundamental frequency
anonymous
  • anonymous
Oh now get it thanks bud.
anonymous
  • anonymous
The description of a radially symmetric wave on a struck drum is: \[Y(r,\phi,t) = J_0(\frac{\omega r}{v})Acos(\omega t)\]
anonymous
  • anonymous
where J_0 is the 0-order Bessel function

Looking for something else?

Not the answer you are looking for? Search for more explanations.