00:01
In this video, we're going to go through the answer to question number 20, where it has to solve the vibrating string problem for a given function f and g, which are given here.
00:12
Okay, so the general solution, like the sort of the eigenmode solution for this problem, was derived in the notes, and that's u .n.
00:26
Is equal to a constant a .n.
00:29
Times cos, alpha, n pi over l, but l is pi in this case times t plus b n cos sorry b n sine alpha and t times by sign x.
00:57
Okay, first boundary condition u of x 0 equals what's going to be the sum of all these eigen modes and is equal to at zero to infinity.
01:24
But evaluated at t equals zero is going to get rid of it's going to mean that all of these signs are equal to zero.
01:35
So it's going to be a n.
01:38
Cos alpha n t, sine n x.
01:48
But this, the cause of alpha nt evaluated at t is equal zero is just one.
01:55
So let me just leave it as an n times sine n x and this is going to be equal to f of x which is zero so therefore all the a n's are equal to zero for any n...