How a standing wave forms
A standing wave is not a new kind of wave; it is what happens when two travelling waves of equal amplitude and wavelength move through the same medium in opposite directions and add up by the principle of superposition. The natural setting is a bounded medium — a string fixed at both ends, or an air column in a pipe — where a wave launched in one direction reflects from a boundary, travels back, reflects again, and so on. After the transients die away, a steady pattern locks into place.
NCERT sets up exactly this in Section 14.6.1: a wave travelling along $+x$ and its reflection travelling along $-x$, with equal amplitude and wavelength. Because the medium is bounded at two ends, only certain wavelengths can survive — the rest interfere destructively with themselves and vanish.
Two harmonic waves of the same amplitude moving oppositely add to a stationary pattern. Red dots mark nodes — points that never move at any instant.
The standing-wave equation
Take the two component waves with phase constant zero, as NCERT does:
$$y_1(x,t) = A\sin(kx - \omega t), \qquad y_2(x,t) = A\sin(kx + \omega t)$$
Adding them and using $\sin(A-B) + \sin(A+B) = 2\sin A\cos B$ gives the defining result:
$$y(x,t) = 2A\,\sin(kx)\,\cos(\omega t)$$
The single most important thing about this expression is that $kx$ and $\omega t$ appear separately, not bundled as $(kx - \omega t)$. A travelling wave depends on $x$ and $t$ only through that combination; a standing wave does not. The factor $2A\sin(kx)$ is a position-dependent amplitude, and $\cos(\omega t)$ is a common time oscillation. Every element of the string therefore vibrates with the same angular frequency $\omega$ and with zero phase difference from its neighbours, but with an amplitude that depends on where it sits.
Nodes, antinodes & no energy transport
The amplitude $2A\sin(kx)$ is zero wherever $\sin(kx) = 0$, i.e. $kx = n\pi$. Using $k = 2\pi/\lambda$, these nodes sit at $x = n\lambda/2$ — points of permanently zero displacement. The amplitude is greatest where $\sin(kx) = \pm 1$, i.e. $kx = (n + \tfrac12)\pi$; these antinodes swing between $+2A$ and $-2A$ and lie exactly halfway between adjacent nodes.
| Feature | Node | Antinode |
|---|---|---|
| Amplitude | Zero (no motion) | Maximum, $2A$ |
| Condition | sin kx = 0 → $x = n\lambda/2$ | sin kx = ±1 → $x=(n+\tfrac12)\tfrac{\lambda}{2}$ |
| Spacing (like to like) | $\lambda/2$ apart | $\lambda/2$ apart |
| Node-to-nearest-antinode | $\lambda/4$ | |
Node spacing is λ/2 — not λ
Successive nodes are separated by $\lambda/2$, and so are successive antinodes. Students who write λ for the gap halve their wavelength and wreck every downstream frequency. A node and the antinode beside it are only $\lambda/4$ apart.
Remember: node→node = λ/2, antinode→antinode = λ/2, node→antinode = λ/4.
Because the pattern is built from two equal waves carrying equal and opposite energy fluxes, the two transports cancel. The wave pattern moves neither left nor right; it merely pulses in place. Consequently a standing wave transports no net energy across any point — energy is trapped within each loop, shuttling between kinetic form (when the string streaks through the axis) and potential form (at maximum displacement). This is the deepest contrast with a progressive wave, which carries energy steadily forward.
Standing waves carry no energy
A common assertion-reason item claims a standing wave transports energy "because it has amplitude". It does not. Each loop stores and recycles energy locally; there is no average flow past a node. Treat any statement about "energy flowing along a standing wave" as false.
Normal modes: string fixed at both ends
The boundary conditions decide which wavelengths survive. For a string of length $L$ clamped at both ends, both ends must be nodes. Taking one end at $x = 0$ (already a node) and demanding $x = L$ also be a node forces
$$L = n\,\frac{\lambda}{2} \quad\Longrightarrow\quad \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3,\dots$$
Since the wave speed on the string is set by the medium as $v = \sqrt{T/\mu}$ (tension $T$, linear mass density $\mu$), the allowed frequencies are
$$f_n = \frac{v}{\lambda_n} = \frac{n\,v}{2L} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \qquad n = 1, 2, 3,\dots$$
The lowest, $n=1$, is the fundamental (first harmonic), $f_1 = v/2L$. Every higher mode is an integer multiple of it, so a string fixed at both ends produces all harmonics — first, second, third, and so on. This integer ladder is what gives bowed and plucked strings their rich, full timbre.
Both ends are nodes (red). The nth harmonic fits n half-wavelengths — n loops — into length L. Matches NCERT Fig. 14.13.
The $v = \sqrt{T/\mu}$ that drives every string frequency is derived in Speed of a Travelling Wave.
Normal modes: pipes (open & closed)
Air columns work on the same logic, with one twist: an open end is a displacement antinode (air is free to move) while a closed end is a displacement node (air cannot move past the rigid wall). The boundary conditions therefore differ between the two pipe types, and that single difference reshapes the whole harmonic series.
Pipe open at both ends
With an antinode at each end, the column behaves like the string in reverse — antinodes where the string had nodes — but the counting works out identically. The allowed wavelengths are $\lambda_n = 2L/n$ and the frequencies are
$$f_n = \frac{n\,v}{2L}, \qquad n = 1, 2, 3,\dots \quad \text{(all harmonics)}$$
Fundamental $f_1 = v/2L$, and an open pipe sounds all harmonics, just like a string fixed at both ends.
Pipe closed at one end
Now one end is a node (closed) and the other an antinode (open). NCERT takes the closed end at $x=0$ and requires the open end $x=L$ to be an antinode, giving $L = (n+\tfrac12)\lambda$ for $n = 0, 1, 2,\dots$. Rewriting with $n = 1, 2, 3,\dots$ for the harmonic count yields the standard form:
$$\lambda_n = \frac{4L}{2n-1}, \qquad f_n = \frac{(2n-1)v}{4L}, \qquad n = 1, 2, 3,\dots$$
The fundamental is $f_1 = v/4L$ — half the open-pipe fundamental of the same length. Crucially, the series runs $v/4L, 3v/4L, 5v/4L,\dots$ — odd harmonics only. The even harmonics (2f, 4f) are physically impossible because they would demand a node and an antinode at the same closed end.
Thick black bar = rigid closed end (node). Open ends are antinodes. The closed pipe skips the even harmonics. Matches NCERT Figs. 14.14–14.15.
Closed pipe: odd harmonics only, fundamental v/4L
The two halves of this trap go together. A pipe closed at one end has fundamental $v/4L$ (one quarter wavelength fits the length), and it can sound only $v/4L, 3v/4L, 5v/4L,\dots$ Asking for "the second harmonic of a closed pipe" is a deliberate trick — there isn't one; the next mode above the fundamental is the third harmonic at $3f_1$.
Open pipe → 1, 2, 3, 4… (all). Closed pipe → 1, 3, 5… (odd only). Closed f₁ = ½ × open f₁ for equal length.
A pipe 30.0 cm long, open at both ends, resonates with a 1.1 kHz source. Which harmonic is it? (Take v = 330 m/s.)
For an open pipe $f_n = nv/2L = n(330)/(2\times0.30) = 550\,n$ Hz. Setting $550n = 1100$ gives $n = 2$ — the second harmonic resonates. If one end were closed, only odd multiples of $v/4L = 275$ Hz are allowed (275, 825, 1375…); 1100 Hz is not among them, so the closed pipe would not resonate with the same source.
Master comparison: string vs pipes
The three bounded systems differ only in their boundary conditions, yet that fully determines their wavelengths, frequencies and which harmonics survive. This single table is the highest-yield object on the page.
| System | Boundary ends | $\lambda_n$ | $f_n$ | Fundamental | Harmonics present |
|---|---|---|---|---|---|
| String fixed both ends | Node — Node | $\dfrac{2L}{n}$ | $\dfrac{nv}{2L}$ | $\dfrac{v}{2L}$ | All (1, 2, 3, 4…) |
| Pipe open both ends | Antinode — Antinode | $\dfrac{2L}{n}$ | $\dfrac{nv}{2L}$ | $\dfrac{v}{2L}$ | All (1, 2, 3, 4…) |
| Pipe closed one end | Node — Antinode | $\dfrac{4L}{2n-1}$ | $\dfrac{(2n-1)v}{4L}$ | $\dfrac{v}{4L}$ | Odd only (1, 3, 5…) |
For the string, $v = \sqrt{T/\mu}$; for pipes, $v$ is the speed of sound in air. Note that for equal length the open pipe and the both-ends-fixed string share an identical frequency ladder, while the closed pipe is the outlier — half the fundamental and odd harmonics only.
End correction
The idealised treatment places the displacement antinode exactly at the open end. In reality the air just outside the mouth also participates, so the antinode sits a small distance $e$ beyond the geometric end. The effective length is $L + e$ (or $L + 2e$ for a pipe open at both ends), with $e \approx 0.6\,r$ for a tube of radius $r$. The practical consequence is that measured resonant frequencies run slightly below the ideal $v/2L$ or $v/4L$.
Resonance-tube experiments sidestep the unknown $e$ by taking the difference of two successive resonance lengths: the gap equals $\lambda/2$, and the end correction cancels — exactly the method behind the NEET 2018 resonance-tube problem below.
Standing waves & normal modes in one screen
- Standing wave = two identical opposite-travelling waves superposed: $y = 2A\sin(kx)\cos(\omega t)$ — $kx$ and $\omega t$ are separate.
- Nodes (zero amplitude) and antinodes (max $2A$) are fixed; node→node = antinode→antinode = $\lambda/2$; node→antinode = $\lambda/4$.
- A standing wave transports no net energy; energy oscillates within each loop.
- String fixed both ends & open pipe: $f_n = nv/2L$, fundamental $v/2L$, all harmonics. String speed $v = \sqrt{T/\mu}$.
- Closed pipe: $f_n = (2n-1)v/4L$, fundamental $v/4L$, odd harmonics only — open : closed fundamental = 2 : 1 at equal length.
- End correction shifts the antinode just beyond the open mouth, lowering observed frequencies slightly.