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Pipe Open at Both Ends

There are displacement antinodes at both ends. This is just like a string free at both ends. We could therefore proceed from

s(x,t) = s_0 \cos(k_n x )\cos(\omega_n t)
\end{displaymath} (190)

\cos(k_n L) = \pm 1
\end{displaymath} (191)

but we could also remember that there are pressure nodes at both ends, which makes them like a string fixed at both ends again. Either way one will get the same frequencies one gets for the pipe closed at both ends above (as the cosine is $\pm 1$ for $k_n L = n\pi$) but the picture of the nodes is still different - be sure to draw displacement antinodes at the open ends!

Robert G. Brown 2004-04-12