QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 14.3NCERT Class 11 · Physics · Chapter 14

Standing (Stationary) Waves in Strings & Organ Pipes

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 4 min read
Formation of Standing Waves:

When two identical progressive waves of the same frequency, wavelength, and amplitude travel through a medium in mutually opposite directions with the same speed, their superposition gives rise to a wave pattern that does not propagate forward or backward. This is called a Standing Wave or Stationary Wave.

y1 = A sin(kx − ωt)  &  y2 = −A sin(kx + ωt) (reflection at rigid end)
y(x, t) = [ 2 A sin(kx) ] cos(ωt) = A(x) cos(ωt)

Here, the amplitude A(x) = 2A sin(kx) is a periodic function of position x, varying from 0 to 2A.

1. Nodes and Antinodes

Nodes (N)

  • Positions where amplitude of oscillation is strictly zero: sin(kx) = 0 ⇒ kx = nπ ⇒ x = n λ/2 (0, λ/2, λ, 3λ/2…).
  • Particles at nodes remain permanently at rest.
  • Strain and pressure variation are MAXIMUM at nodes!

Antinodes (A)

  • Positions where amplitude is maximum (2A): |sin(kx)| = 1 ⇒ x = (2n − 1) λ/4 (λ/4, 3λ/4, 5λ/4…).
  • Particles oscillate with maximum velocity vmax = 2ωA.
  • Strain and pressure variation are ZERO at antinodes!
Spatial Spacing Rules:
  • Distance between two consecutive nodes: λ / 2
  • Distance between two consecutive antinodes: λ / 2
  • Distance between an adjacent node and antinode: λ / 4
  • Energy: In a standing wave, energy is confined between nodes; there is zero net transmission of energy along the medium.

2. Normal Modes of Stretched Strings (Fixed at Both Ends)

Both ends (x = 0 and x = L) are fixed, so nodes must exist at both ends. Hence L = n (λn / 2):

λn = 2L / n  ⇒  fn = n (v / 2L) = (n / 2L) √(T / μ)    (where n = 1, 2, 3, …)
Harmonic / Mode Wavelength λ Frequency f Nodes (N) & Antinodes (A)
Fundamental (1st Harmonic) λ1 = 2L f1 = v / (2L) 2 Nodes, 1 Antinode
2nd Harmonic (1st Overtone) λ2 = L f2 = 2 f1 3 Nodes, 2 Antinodes
3rd Harmonic (2nd Overtone) λ3 = 2L / 3 f3 = 3 f1 4 Nodes, 3 Antinodes

Conclusion: In a stretched string, all harmonics (even and odd) are present in the ratio 1 : 2 : 3 : 4 : …

3. Organ Pipes: Open Pipe vs. Closed Pipe

Open Organ Pipe (Open at both ends)

  • Air at both ends is free to oscillate ⇒ Antinodes at both ends.
  • Length L = n (λ/2).
  • Frequencies: fn = n (v / 2L) = n f1 (n = 1, 2, 3…).
  • ALL harmonics (odd & even) are present: f1 : f2 : f3 = 1 : 2 : 3 : 4…
  • Sound quality is rich and pleasant.

Closed Organ Pipe (Closed at one end)

  • Air at closed end cannot move ⇒ Node at closed end; Antinode at open end.
  • Length L = (2n − 1) (λ/4).
  • Frequencies: fn = (2n − 1) [ v / (4L) ] (n = 1, 2, 3…).
  • ONLY ODD harmonics are present: f1 : f3 : f5 = 1 : 3 : 5 : 7…
  • Fundamental frequency of closed pipe is HALF that of an open pipe of the same length: fclosed = fopen / 2.

4. End Correction (Rayleigh's Correction) & Resonance Tube

Because the air particles slightly outside the open end also participate in oscillation, the actual antinode is formed at a small distance e outside the tube:

e ≈ 0.6 r   (where r is internal radius of the pipe)
  • For an open pipe: L' = L + 2e = L + 1.2 r
  • For a closed pipe: L' = L + e = L + 0.6 r

Resonance Tube Experiment:

Let l1 and l2 be the first two resonating lengths of air column for a tuning fork of frequency f:

l1 + e = λ/4  &  l2 + e = 3λ/4 ⇒ l2 − l1 = λ/2
v = 2 f (l2 − l1)    and    e = (l2 − 3 l1) / 2
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