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AlephZero
#2
Dec6-12, 10:35 AM
Engineering
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Quote Quote by BomboshMan View Post
and that the boundary conditions for antinodes at x = 0 and x = L are

D(0,t) = 2a
D(L,t) = 2a
That looks wrong. I think it should be
##\partial D(0,t)/\partial x## = 0
##\partial D(1,t)/\partial x## = 0

which gives ##D(0,t) = D(1,t) = 2a \cos(\omega t)##.