Physics · Waves · NEET
A node is a fixed point in a standing wave where the two overlapping waves always cancel, so the displacement is always zero (No motion). An antinode is a fixed point where they always add, so the particle vibrates with the largest amplitude (All motion). Both stay in the same place, which is why standing waves look like they are standing still.
The displacement pattern is y = 2a sin(kx) cos(wt). Nodes need sin(kx) = 0, which happens at kx = 0, pi, 2pi..., i.e. x = 0, lambda/2, lambda... So nodes repeat every lambda/2. Many students wrongly write lambda because they picture a travelling wave; in a standing wave one full wavelength holds TWO nodes and TWO antinodes.
Antinodes lie exactly halfway between nodes, so a node to its nearest antinode is lambda/4. Sequence along the string: node, then lambda/4 to an antinode, then another lambda/4 to the next node. This lambda/4 gap is a very common NEET calculation.
For a sound (longitudinal) standing wave, a displacement node is a pressure antinode. Where particles do not move (displacement node), the air is squeezed most, so pressure change is largest. Where particles swing most (displacement antinode, like an open pipe end), pressure change is smallest. Displacement and pressure are always out of step by lambda/4.
First fix the ends using boundary conditions: a fixed end or a closed pipe end is a node, a free end or an open pipe end is an antinode. Then draw the loops. For a string fixed at both ends or a pipe open at both ends, the p-th harmonic has p+1 nodes... but be careful, NEET usually counts the interior nodes, so read the question. For an open pipe the number of displacement nodes equals the harmonic number; for a closed pipe only odd harmonics exist.
For sound waves, if the number of nodes for the 5th harmonic of an open-ended pipe is n and that for the 9th harmonic of the same pipe with one of its ends closed is m, the ratio n/m is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. In a pure standing wave their positions are fixed in space, unlike a travelling wave where the pattern moves. That is the defining feature of a standing (stationary) wave.
Taking a fixed end at x = 0: nodes are at x = n(lambda/2) and antinodes at x = (n + 1/2)(lambda/2), where n = 0, 1, 2, 3... Consecutive nodes and consecutive antinodes are both lambda/2 apart.
A rigid or closed boundary (fixed string end, closed pipe end, water surface in a resonance tube) is always a displacement node. A free or open end is a displacement antinode. Fixing the ends first is the key step in locating all the other nodes.
The boundary conditions force the ends to be nodes or antinodes, so only wavelengths that fit a whole number of loops (each of length lambda/2) survive. This gives the discrete normal modes and harmonics, unlike a travelling wave that accepts any frequency.
