# Why are there both antinodes at both ends of the tube? [duplicate]

I learned stationary/standing waves the other day. For stationary waves in open tubes, the textbook says both ends must have an antinode. Can anyone tell me why? (shown as figure)

And also, when playing with the instruments like guitar, what's the number of harmonic on the string, i.e. how many antinodes and nodes are there on the string?

It's a pressure node as the pressure at the open ends of tube is atmospheric pressure. It's then automatically a displacement antinode because the equation of motion
$$\rho_o \partial_t v = - \partial_x P$$ means that a standing wave $$P(x,t)= A\sin(kx) \sin (\omega t)$$ with a pressure node at $$x=0$$ makes the velocity obey $$\rho_0 \omega^2 v(x,t)= \partial^2_{xt} P= A k\omega \cos(kx) \cos (\omega t),$$ which has a velocity (and hence a displacement) maximum at $$x=0$$.

The air molecules are free to move at the open end of a tube, so there's an antinode. At a closed end, there must be a node as the air molecules don't move there.

That's similar to a wave in a string, at the fixed ends the string can't move and there are nodes there. The first harmonic for a string (the fundamental), has two nodes at the ends and one antinode in the middle. The second harmonic has two antinodes etc...