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An open pipe emits a fundamental frequency $n$ when it emits the 3rd harmonic, then the pipe can accommodate
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The correct answer is:
3 nodes, 4 anti-nodes
For third harmonic, $n=3$. So, for an open pipe, the propagation of waves can be shown as
$\therefore$ Nodes $=3$ and anti-nodes $=4$
$\therefore$ Nodes $=3$ and anti-nodes $=4$
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