Find u(x,t) of the String of Length L=π

  • Thread starter Thread starter asdf1
  • Start date Start date
  • Tags Tags
    Length String
asdf1
Messages
734
Reaction score
0
How do you find u(x,t) of the string of length L=pi when c^2=1, the initial velocity=0 and the initial deflection is 0.1x[(pi^2 )-(x^2)]?
 
Physics news on Phys.org
Since this is in the homework section, presumably you know something about the "wave equation" (although you don't actually say that you are trying to solve an equation- in fact, you don't even say what you mean by u(x,t)!). Do you know anything about either "characteristic equations" or "separation of variables"?
 
i know what separation of variables is but I'm not too sure about what you mean by "characteristic equations"... I've only heard of "characteristice equation" for the eigenvalues and eigenvectors...
 
My real point was that you have, in several posts now, just asked people to tell you how to do something without showing that you have tried anything yourself. As I pointed out before, you haven't even stated the problem clearly.

I really should have said "characteristic directions"- or just "characteristics" which is the direction of the eigenvectors.
Let u= x-ct, v= x+ct and rewrite the equation in those variables rather than x and t.
 
How do you formulate the problem in mathematical terms?
Set up the wave equation, the boundary conditions, the initial
conditions...
Separate variables.
expand the inital condition in apropriate eigenfunctions.
Thats all.
 
ok, i'll try that...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top