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[FAQ]

In modal analyses, the naturally-occurring boundary condition is a rigid wall, which is the same for harmonic or transient analyses as well. There is one exception for handling boundary conditions – in modal analyses, special care has to be taken when one has an anechoic termination or open condition. Let’s say one has a pipe with one end open (say one is looking into something similar to this case): http://labman.phys.utk.edu/phys221core/modules/m12/Standing%20sound%20waves.html In the figures for the section “Tubes with one open and one closed end” above, we see that there is a ‘node’ (vs. antinode) on the left, which represents a rigid wall. The other end is open. In this case, if one constrain pressure=0 on the left (which is actually a rigid wall), one will get the frequencies for this case. The reason why this works is because, in a modal analysis, we are solving for frequencies, so by applying a boundary condition such that we are constraining a ‘node’ (vs. antinode), we are solving for those specific frequencies. Normally, we would not do this, but one can think of it as solving for a specific set of frequencies by imposing a known ‘node’ (vs. antinode). One can solve this same situation without imposing a boundary condition, but would need to model the air around the pipe as well. If one does so, they can solve for the same frequencies, but it will require longer computational time since they are modeling the air around the pipe as well – also, one may get other non-pipe modes included in your solution. Thus, constraining the closed end to pressure=0 to enforce a ‘node’ (vs. antinode) is a simpler approach. This does not mean that one set pressure=0 for a rigid wall. Instead, think of it as enforcing a ‘node’ (vs. antinode) in a specific location.