Standing waves and normal modes

Pluck a guitar string and something strange happens: the string does not just wiggle randomly, it settles into a fixed, glowing pattern — a shape that stays put while the string vibrates furiously inside it. Some points on the string barely move at all; others swing through the widest possible arc. Blow across the mouth of a bottle and the same thing happens to the air column trapped inside, except now you cannot see the pattern, you only hear it as a single clean musical note. Both are the same physics: a standing wave, built when a travelling wave reflects off a boundary and meets its own reflection coming back the other way. This chapter is where the "wave" chapter finally explains music. Every string instrument, every wind instrument, every organ pipe works because the medium — string or air column — is confined between two boundaries, and confinement forbids most wavelengths from surviving. Only a special, discrete ladder of wavelengths fits neatly between the boundaries; these survivors are the normal modes, or harmonics, and the lowest one is the fundamental. Everything else is silence, cancelled out by destructive interference on every single vibration. What makes this topic satisfying for CBSE exams is that it is almost entirely algebra once you accept one idea: superposition of two oppositely travelling waves of the same amplitude and frequency produces a wave whose shape stops moving. From there, three boundary conditions — string fixed at both ends, pipe closed at one end, pipe open at both ends — hand you three harmonic-frequency formulas, each provable from a simple picture of where the nodes and antinodes must sit. This section builds that picture from scratch, derives the standing-wave equation, and then works two numerical problems — a guitar-string harmonic and an organ-pipe harmonic — exactly the flavour of question that shows up in board and competitive exams alike.

Key takeaways

A standing wave forms when two identical waves travel in opposite directions and superpose; the resulting pattern does not travel — only its amplitude oscillates in time.
Superposing $y_1=A\sin(kx-\omega t)$ and $y_2=A\sin(kx+\omega t)$ gives $y = 2A\sin(kx)\cos(\omega t)$ — a fixed spatial shape $2A\sin(kx)$ multiplied by a time factor $\cos(\omega t)$ .
Nodes are points of permanent zero displacement ( $\sin kx = 0$ ); antinodes are points of maximum displacement ( $\sin kx = \pm1$ ); adjacent like-points are $\lambda/2$ apart.
A string fixed at both ends (both ends nodes) supports all integer harmonics: $f_n = \dfrac{n}{2L}\sqrt{T/\mu}$ , $n=1,2,3,\dots$
A pipe closed at one end (node at closed end, antinode at open end) supports only odd harmonics: $f_n = \dfrac{(2n-1)v}{4L}$ .
A pipe open at both ends (antinode at each end) supports all integer harmonics, structurally identical to the string: $f_n = \dfrac{nv}{2L}$ .
Key idea: superposing two oppositely travelling identical waves freezes the pattern into fixed nodes and antinodes, and forcing the boundary conditions of a string or pipe onto that pattern restricts vibration to a discrete ladder of normal-mode frequencies — all harmonics for a string or open pipe, only odd harmonics for a closed pipe.

The core insight

Two identical waves travelling in opposite directions superpose into a standing wave $y=2A\sin(kx)\cos(\omega t)$ whose fixed nodes and antinodes, forced to match a string's or pipe's boundary conditions, permit only a discrete ladder of normal-mode frequencies.

Where it fits in the syllabus

This sits inside the CBSE Class 11 Waves chapter (Unit: Oscillations and Waves), directly built on the principle of superposition and on reflection of waves at fixed and free ends — both of which you should have already met before this section. You are expected to derive the standing-wave displacement equation from two oppositely travelling waves, correctly place nodes and antinodes for the three standard boundary situations, and apply the three harmonic formulas above to strings and organ pipes. The wave equation as a differential equation, energy-flux derivations, and the Doppler effect are outside the current 2025-26 prescription and are not developed here — only the algebraic route through superposition, exactly as NCERT presents it.

The physical picture

A standing wave is not a new kind of wave physically — it is what you get for free whenever a travelling wave reflects back on itself, so that two waves of equal amplitude and frequency, moving in opposite directions, occupy the same string or air column at once. Take a rightward-moving wave and a leftward-moving wave of the same amplitude $A$ , wave number $k$ , and angular frequency $\omega$ :

y_1(x,t) = A\sin(kx-\omega t), \qquad y_2(x,t) = A\sin(kx+\omega t)

where: $A$ is the amplitude of each individual travelling wave, $k=2\pi/\lambda$ is the wave number, $\omega=2\pi f$ is the angular frequency, $x$ is position along the string or pipe, and $t$ is time. Physically, $y_1$ is the wave heading toward the far boundary and $y_2$ is exactly that wave reflected back. By the principle of superposition, the net displacement is simply the sum $y=y_1+y_2$ . Using the trigonometric identity for a sum of sines gives:

y(x,t) = 2A\sin(kx)\cos(\omega t)

where: the factor $2A\sin(kx)$ depends only on position $x$ — it is a fixed "envelope" shape that never moves along the string — and the factor $\cos(\omega t)$ depends only on time, meaning every point of the string oscillates up and down together, in step, but each with a different fixed amplitude given by $2A\sin(kx)$ . This is the single most important structural fact about a standing wave: unlike a travelling wave, where a fixed shape glides along $x$ as $t$ increases, here the shape is pinned in space and only its height breathes in and out with time. Points where $\sin(kx)=0$ have zero amplitude at all times — these are nodes. Points where $\sin(kx)=\pm1$ have the maximum possible amplitude $2A$ — these are antinodes. Because $\sin(kx)$ passes through zero every half-wavelength, nodes repeat every $\lambda/2$ along the string, and antinodes sit exactly midway between consecutive nodes.

Figure

A standing wave pattern on a horizontal string showing four nodes as filled dots on the centre line and three antinode loops alternating above and below the line, with a bracket marking the half wavelength distance between two adjacent nodes

Whether a boundary forces a node or an antinode depends on what the boundary is. A fixed end (a string tied to a wall, or the closed end of a pipe where air cannot move) cannot displace, so it must always be a node. A free end (a string tip that can slide freely, or the open end of a pipe where air is free to surge) is where displacement is largest, so it is always an antinode. These two rules are the entire content of the three harmonic-formula cases below — everything follows from asking "node or antinode at each end, and how many half-wavelengths fit in between."

Stretched string, fixed at both ends, length $L$ : both ends are nodes, so $L$ must hold a whole number of half-wavelengths, $L = n(\lambda_n/2)$ , giving $\lambda_n = 2L/n$ and

f_n = \frac{nv}{2L} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \quad n=1,2,3,\dots

where: $n$ is the harmonic number ( $n=1$ is the fundamental, or first harmonic), $L$ is the string length, $T$ is the tension, and $\mu$ is the linear mass density (mass per unit length). Every integer $n$ is allowed — a fixed-fixed string plays all harmonics.

Organ pipe closed at one end, length $L$ : the closed end is a node, the open end is an antinode; fitting a node at one end and an antinode at the other requires an odd number of quarter-wavelengths, $L=(2n-1)(\lambda_n/4)$ , giving

f_n = \frac{(2n-1)v}{4L}, \quad n=1,2,3,\dots

where: $v$ is the speed of sound in the enclosed air. The frequencies come in the ratio $1:3:5:\dots$ — even harmonics are structurally forbidden.

Organ pipe open at both ends, length $L$ : both ends are antinodes, which requires a whole number of half-wavelengths just like the fixed-fixed string, giving the same structural form,

f_n = \frac{nv}{2L}, \quad n=1,2,3,\dots

with all integer harmonics present, in the ratio $1:2:3:\dots$ It is worth dwelling on why a string fixed at both ends and a pipe open at both ends give the identical formula. Both situations demand the same type of boundary condition repeated at both ends — node-node for the string, antinode-antinode for the pipe — and requiring the same condition twice always forces a whole number of half-wavelengths into the length $L$ , regardless of whether that repeated condition happens to be "node" or "antinode." A closed pipe instead mixes the two condition types — node at one end, antinode at the other — and mixing them is exactly what forces an odd count of quarter-wavelengths, hence only odd harmonics. Recognising this symmetry is a fast way to avoid memorising three unrelated formulas: there are really only two shapes of boundary problem, "same condition twice" and "different conditions," and each has its own harmonic-counting rule.

Derivation AThe standing-wave equation from two oppositely travelling waves

The goal is to show, using only the trigonometric identity for a sum of two sine functions and the principle of superposition, that two identical waves travelling in opposite directions combine into a standing wave whose spatial and time dependence separate cleanly — a wave whose shape is pinned in space while only its height oscillates in time — and to read off from that separated form exactly where the permanent nodes and antinodes must lie. No calculus is needed; the whole derivation is algebraic identity manipulation on the two given wave functions.

Step 1. Write the two waves. Let a wave travelling in the $+x$ direction be $y_1(x,t)=A\sin(kx-\omega t)$ and its reflection, travelling in the $-x$ direction with the same amplitude, wave number, and angular frequency, be $y_2(x,t)=A\sin(kx+\omega t)$ . Both have amplitude $A$ , so the reflection has lost none of the original wave's strength — a reasonable idealisation at a rigid or fully open boundary.

Step 2. Apply the principle of superposition. Because waves add linearly on a string (for the small displacements considered in this chapter), the actual displacement at any point is simply $y(x,t) = y_1(x,t) + y_2(x,t) = A\sin(kx-\omega t) + A\sin(kx+\omega t)$ .

Step 3. Expand using the sine-sum identity. Using $\sin(P\mp Q) = \sin P\cos Q \mp \cos P \sin Q$ with $P=kx$ and $Q=\omega t$ , the two terms are $A(\sin kx\cos\omega t - \cos kx\sin\omega t)$ and $A(\sin kx\cos\omega t + \cos kx\sin\omega t)$ . Adding these, the $\cos kx \sin\omega t$ terms are equal and opposite, so they cancel exactly, and the $\sin kx\cos\omega t$ terms add.

Step 4. Collect the surviving terms. What remains is $y(x,t) = 2A\sin(kx)\cos(\omega t)$ . This is the standing-wave equation. Notice that $x$ and $t$ now appear in two separate factors rather than mixed together inside a single sine argument the way they were in $y_1$ or $y_2$ individually — this separation is the mathematical signature that the pattern no longer travels.

Step 5. Read off nodes and antinodes from the amplitude function $2A\sin(kx)$ . At positions where $\sin(kx)=0$ , that is $kx = 0,\pi,2\pi,\dots$ , the displacement is zero for every value of $t$ — these are the nodes. At positions where $\sin(kx)=\pm1$ , that is $kx=\pi/2, 3\pi/2,\dots$ , the displacement swings through the full range $\pm2A$ — these are the antinodes. Since consecutive zeros of $\sin(kx)$ are separated by $kx=\pi$ , i.e. $\Delta x = \pi/k = \lambda/2$ , adjacent nodes (and adjacent antinodes) are always $\lambda/2$ apart, with each antinode sitting exactly $\lambda/4$ from its neighbouring nodes.

Worked examples

Board level - fundamental and third harmonic of a fixed string

A string of length $1.0\ \text{m}$ , fixed at both ends, has linear mass density $\mu = 5\times10^{-4}\ \text{kg/m}$ and is under tension $T=80\ \text{N}$ . Find the fundamental frequency and the frequency of the third harmonic.

Step 1. Find the wave speed on the string using $v=\sqrt{T/\mu}$ :

v = \sqrt{\frac{80}{5\times10^{-4}}} = \sqrt{160000} = 400\ \text{m/s}

Step 2. Apply the fixed-fixed string formula $f_n = nv/2L$ with $n=1$ for the fundamental:

f_1 = \frac{1\times400}{2\times1.0} = 200\ \text{Hz}

Step 3. For the third harmonic, set $n=3$ . Because a string fixed at both ends supports every integer harmonic, $f_3$ is simply three times $f_1$ :

f_3 = \frac{3\times400}{2\times1.0} = 3\times200 = 600\ \text{Hz}

The fundamental frequency is 200 Hz and the third harmonic is 600 Hz.

Competitive level - closed pipe harmonics and a missing frequency check

A closed organ pipe of length $0.5\ \text{m}$ contains air in which sound travels at $v=340\ \text{m/s}$ . Find the fundamental frequency and the next two allowed harmonics, and determine whether a tone of $680\ \text{Hz}$ could be produced by this pipe.

Step 1. Apply the closed-pipe formula $f_n=(2n-1)v/4L$ with $n=1$ for the fundamental:

f_1 = \frac{(2\times1-1)\times340}{4\times0.5} = \frac{340}{2} = 170\ \text{Hz}

Step 2. A closed pipe allows only odd harmonics, in the ratio $1:3:5:\dots$ , so the next two allowed frequencies come from $n=2$ and $n=3$ , which correspond to the 3rd and 5th harmonics:

f_3 = 3\times f_1 = 3\times170 = 510\ \text{Hz}, \qquad f_5 = 5\times f_1 = 5\times170 = 850\ \text{Hz}

Step 3. Check whether $680\ \text{Hz}$ is possible. Dividing, $680/170 = 4$ , so $680\ \text{Hz}$ would be the fourth harmonic — an even multiple of the fundamental. Since a closed pipe forbids even harmonics (the closed end must remain a node and the open end an antinode, which only an odd number of quarter-wavelengths can satisfy), 680 Hz cannot be produced by this pipe; the nearest allowed tones are 510 Hz and 850 Hz.

The fundamental is 170 Hz, the next two allowed harmonics are 510 Hz and 850 Hz, and 680 Hz is forbidden because it is an even harmonic. Timing note: In an exam, do not compute $f_2, f_3, f_4,\dots$ one by one and then discard the even ones — that wastes minutes. Recognise immediately that a closed pipe's allowed set is $f_1, 3f_1, 5f_1,\dots$ and jump straight to the odd multiples; use the ratio test ( $f/f_1$ must be an odd integer) to check any given frequency in a single division.

Common traps

Applying the open-pipe/fixed-string formula ( $f_n=nv/2L$ , all harmonics) to a closed pipe, or the closed-pipe formula (odd harmonics only) to an open pipe. Always identify the boundary conditions first — a closed end is a node, an open end or free end is an antinode — before choosing a formula.
Confusing which end of a closed pipe is the node. The closed end (where air cannot surge) is the node; the open end (where air is free) is the antinode — reversing this flips the whole harmonic series.
Writing $\lambda_n = L/n$ instead of $\lambda_n = 2L/n$ for a fixed-fixed string or open-open pipe. The length $L$ contains a whole number of half-wavelengths, because consecutive nodes (or consecutive antinodes) are $\lambda/2$ apart, not whole wavelengths.
Treating an odd multiple of the fundamental as automatically "the second harmonic" or similar. For a closed pipe, the harmonic number equals $2n-1$ , so the third allowed tone is the fifth harmonic, not the third — keep the harmonic number and the sequence position distinct.
Assuming a standing wave carries energy along the string or pipe the way a travelling wave does. It does not; energy is trapped and simply oscillates between kinetic and potential form within each segment between adjacent nodes.

You should be able to…

Derive $y=2A\sin(kx)\cos(\omega t)$ by superposing two oppositely travelling waves of equal amplitude, and explain why the spatial and time parts separate.
Locate nodes and antinodes for a given standing wave and state that adjacent like-points are $\lambda/2$ apart.
Decide, from a picture of the boundaries, whether each end is a node or an antinode, and hence which harmonic formula applies.
Apply $f_n=\dfrac{n}{2L}\sqrt{T/\mu}$ for a string fixed at both ends and compute the fundamental and any higher harmonic.
Apply $f_n=\dfrac{(2n-1)v}{4L}$ for a closed pipe and $f_n=\dfrac{nv}{2L}$ for an open pipe, and explain in one sentence why closed pipes skip even harmonics.
Check quickly whether a given frequency belongs to a closed pipe's allowed set using the odd-integer ratio test.

Ask the tutor: Want the AI tutor to sketch, side by side, exactly which harmonics survive and which are missing for a fixed string versus a closed pipe versus an open pipe, so the pattern of "all integers" versus "odd only" becomes visually obvious?

Why does a closed pipe skip every other harmonic but an open pipe does not?

Because the two ends of a closed pipe demand different things: the closed end must be a node (air cannot move there) while the open end must be an antinode (air is free to surge). Fitting a node at one end and an antinode at the other only works if the pipe length holds an odd number of quarter-wavelengths — one quarter-wave, three quarter-waves, five, and so on — which is exactly why only odd multiples of the fundamental survive. An open pipe demands an antinode at both ends, the same condition repeated, which allows any whole number of half-wavelengths and therefore every integer harmonic.

Does a standing wave actually carry energy from one end of the string to the other?

No, not in the steady sense that a travelling wave carries energy forward. In a standing wave the energy is trapped between adjacent nodes, sloshing back and forth between kinetic energy (maximum when the string is flat and moving fastest through the antinode) and elastic potential energy (maximum when the string is stretched to its extreme displacement). No net energy crosses a node, since a node never moves at all — it is a permanent zero-displacement point, so it can do no work on the segment beside it.

If I make a guitar string tighter, what happens to all its harmonics?

Tightening the string increases the tension $T$ , which increases the wave speed $v=\sqrt{T/\mu}$ , and every harmonic frequency $f_n = nv/2L$ scales directly with $v$ . So every harmonic rises together, all in the same proportion — the fundamental gets higher-pitched and so does the third harmonic, the fifth, and so on, keeping their ratio 1:2:3 unchanged. This is exactly why turning a tuning peg raises the pitch of a guitar string smoothly.

Is the Doppler effect part of this topic?

No — the Doppler effect (the frequency shift you hear when a source or listener moves) is not part of the current CBSE Class 11 prescription and is studied only later, in higher classes. This section stays entirely within standing waves and normal modes: superposition, reflection, nodes and antinodes, and the harmonic formulas for strings and pipes.