Conditions for an antinode?


by BomboshMan
Tags: antinodes, boundary, conditions, standing waves
BomboshMan
BomboshMan is offline
#1
Dec6-12, 09:43 AM
P: 16
Hi,

If I have two sinusoidal waves with the same frequency, wavelength and amplitude, travelling along the same line in opposite directions, the net displacement of the resulting standing wave wave is given by

D(x,t) = 2a*sin(kx)*cos(ωt)

the boundary conditions for standing waves on a string from x=0 to x=L are

D(0,t) = 0 (which satisfies the above equation at all times), and
D(L,t) = 0, which satisfies the equation when L = 0.5mλ , m=1,2,3... (or λ = 2L/m)

This I understand because the displacement of a node at any time must = 0.

Now I've been told by my physics lecturer (who's not very good), that for antinodes at x = 0 and x = L, e.g. standing waves in a pool, the equation for disturbance is

D(x,t) = 2a*cos(kx)*cos(ωt),

and that the boundary conditions for antinodes at x = 0 and x = L are

D(0,t) = 2a
D(L,t) = 2a

which I get where this is coming from (same sort of thing as boundary conditions for nodes), but wouldn't this suggest that the displacement of x = 0 and x = L are 2a at all times? Surely it only makes sense to say the max displacement of x = 0 and x = L is 2a

I may just be getting confused, but if someone could shed some light on this it would help a lot!

Thanks,

Matt
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AlephZero
AlephZero is offline
#2
Dec6-12, 10:35 AM
Engineering
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P: 6,356
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)##.


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