Standing waves in open-open pipes alternate nodes and antinodes, with antinodes at both ends

Steps:

• Both open ends are displacement antinodes, so the pattern starts and ends with antinodes.
• Nodes occur between antinodes; three nodes divide the tube into four segments.
• Each segment is bounded by antinodes, requiring four antinodes total.
• This matches the nth harmonic formula: n nodes and n+1 antinodes, where n=3.

Why D is correct:

• The nth harmonic in an open-open pipe has n+1 antinodes by the wave equation cos(nπx/L), yielding 4 antinodes for n=3 (three nodes).

Why the others are wrong:

• A. 1: Impossible; minimum is 2 antinodes for fundamental (1 node).
• B. 2: Fits fundamental mode with 1 node, not 3 nodes.
• C. 3: Assumes equal nodes and antinodes, but open ends add one extra antinode.

Final answer: D