PhysicsISC Class 11 · Waves

Standing Waves: Strings, Organ Pipes, Resonance and Beats

Updated 17 October 2026 · Fact-checked

Superposition of two identical waves travelling in opposite directions forms a standing wave with nodes λ/2 apart. A string fixed at both ends has f = nv/2L (all harmonics), a closed pipe has f = (2n − 1)v/4L (odd harmonics only) and an open pipe has f = nv/2L. Two close frequencies give beats at |f₁ − f₂|.

Closed ends and fixed ends are nodes, open ends are antinodes; a closed pipe holds only a quarter wavelength, so its fundamental v/4L is half that of an open pi

Illustration drawn for QuestionBuddy.

Understand Standing Waves, Organ Pipes and Beats

Standing waves. A wave and its reflection from a boundary superpose to give a stationary pattern in which some points (nodes) never move, and some points (antinodes) vibrate with maximum amplitude. Adjacent nodes (or adjacent antinodes) are λ/2 apart, and a node and the next antinode λ/4 apart. No energy is transmitted along the medium. At a fixed (rigid) end there is a node; at a free end of a string or an open end of a pipe there is an antinode; at a closed end of a pipe there is a node.

String fixed at both ends. L = nλ/2, so f_n = nv/2L = (n/2L)√(T/μ), n = 1, 2, 3... The fundamental is f₁ = v/2L. In a sonometer, f ∝ 1/L, f ∝ √T and f ∝ 1/√μ.

Closed pipe. One end closed (node), one end open (antinode): L = (2n − 1)λ/4, f = (2n − 1)v/4L: fundamental v/4L, followed by 3v/4L, 5v/4L: odd harmonics only.

Open pipe. Both ends open (antinodes): L = nλ/2, f = nv/2L: all harmonics, with the fundamental v/2L, twice that of a closed pipe of the same length.

Resonance tube. A vibrating fork over a tube of adjustable air column resonates at the first and second resonating lengths; L₂ − L₁ = λ/2, so v = 2f(L₂ − L₁), independent of end correction.

Beats. Two sound waves of nearly equal frequencies f₁ and f₂ give periodic variation of loudness at the beat frequency |f₁ − f₂|. A beat is one rise and fall of loudness. Beats help to tune instruments and to find an unknown frequency (loading a fork with wax lowers its frequency).

Standing Waves, Organ Pipes and Beats: formulas and key facts

String: $f_{n} = n v /2 L ∣ f_{n} = (n /2 L) \sqrt (T / μ)$; n = 1, 2, 3...
Closed pipe: $f = (2 n - 1) v /4 L$; Odd harmonics.
Open pipe: $f = n v /2 L$; All harmonics.
Beats: $f_{b} e a t = ∣ f_{1} - f_{2} ∣$; Beats per second.

Standing Waves, Organ Pipes and Beats in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A guitar-like wire

Suppose a wire of length 0.8 m is fixed at both ends, and waves travel on it at 200 m s⁻¹.

Find the frequencies of the first three harmonics and the distance between adjacent nodes in the second harmonic.

Try it first, then show the solutionSolution

f₁ = v/2L = 200/1.6 = 125 Hz.
f₂ = 250 Hz and f₃ = 375 Hz.
For the second harmonic λ = L = 0.8 m, so adjacent nodes are λ/2 = 0.4 m apart.

Answer: 125 Hz, 250 Hz, 375 Hz; nodes 0.4 m apart.

Memory hook: Both ends fixed: all harmonics are possible.

Tricks to remember Standing Waves, Organ Pipes and Beats

Short memory aids. Say each one aloud once, then use it on a question the same day.

Closed means odd
The closed pipe has a node at one end and an antinode at the other: only odd multiples of the fundamental.
Use it when: Harmonics of pipes.
Open is double the closed
For the same length, the open pipe fundamental is twice that of the closed pipe.
Use it when: Comparing pipes.

Standing Waves, Organ Pipes and Beats: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Harmonics of a closed pipe · 1 mark · MCQBoard level

The resonant frequencies of an air column closed at one end form which series?
(a) only odd harmonics
(b) only even harmonics
(c) all harmonics
(d) no harmonics

Try it first, then show the solutionSolution

A node at the closed end and an antinode at the open end allow only odd multiples of the fundamental.

Answer: (a) only odd harmonics

Example 2: Vibrating wire · 3 marks · Short answerBoard level

A wire 0.8 m long, fixed at both ends, carries waves of speed 200 m s⁻¹. Find the frequencies of the fundamental and the next two harmonics.

Try it first, then show the solutionSolution

f₁ = v/2L = 200/(2 × 0.8). (1 mark)
= 125 Hz. (0.5 mark)
f₂ = 2f₁ = 250 Hz. (0.75 mark)
f₃ = 3f₁ = 375 Hz. (0.75 mark)

Answer: 125 Hz, 250 Hz and 375 Hz.

Example 3: Closed and open pipes · 4 marks · Short answerHarder board level

Find the first three resonant frequencies of a closed pipe 0.85 m long, and the fundamental of an open pipe of the same length. (v = 340 m s⁻¹)

Try it first, then show the solutionSolution

Closed: f₁ = v/4L = 340/3.4 = 100 Hz. (1 mark)
Next: 3f₁ = 300 Hz and 5f₁ = 500 Hz. (1 mark)
Open: f₁ = v/2L = 340/1.7. (1 mark)
= 200 Hz, twice the closed-pipe fundamental. (1 mark)

Answer: Closed: 100, 300, 500 Hz. Open fundamental: 200 Hz.

Example 4: Resonance tube · 4 marks · Short answerHarder board level

A tuning fork of frequency 512 Hz gives resonance with an air column at lengths 16.0 cm and 50.0 cm. Find the speed of sound and the end correction.

Try it first, then show the solutionSolution

L₂ − L₁ = λ/2 = 34.0 cm, so λ = 0.68 m. (1 mark)
v = fλ = 512 × 0.68 = 348 m s⁻¹. (1 mark)
L₁ + e = λ/4 = 17.0 cm. (1 mark)
e = 17.0 − 16.0 = 1.0 cm. (1 mark)

Answer: v = 348 m s⁻¹; end correction 1.0 cm.

Example 5: Beats and a sonometer · 5 marks · Long answerHOTS / competency

A tuning fork of unknown frequency gives 4 beats per second with a 256 Hz fork. When the unknown fork is loaded with a little wax, the beat frequency rises to 6 per second. Find the unknown frequency. A sonometer wire 0.5 m long has a fundamental of 200 Hz; find the length that resonates with the unknown fork at the same tension.

Try it first, then show the solutionSolution

Unknown frequency is 252 Hz or 260 Hz. (1 mark)
Loading lowers the frequency. If f = 260 Hz, it would move closer to 256 Hz and the beats would fall; the beats rose, so f = 252 Hz. (2 marks)
Sonometer: f ∝ 1/L (same T and μ). (1 mark)
L = 0.5 × 200/252 = 0.397 m. (1 mark)

Answer: Unknown frequency 252 Hz; length 0.397 m.

Where students lose marks in Standing Waves, Organ Pipes and Beats

The slips examiners see most often, and what to write instead.

Mistake: Using even harmonics for a closed pipe
Writing 2f, 4f for a closed pipe.
Do this: A closed pipe has f, 3f, 5f only.
Usually costs: 2 marks.
Mistake: Nodes at the open end
Marking a node at the open end of a pipe.
Do this: Open end: antinode. Closed end: node.
Usually costs: 1 to 2 marks.
Mistake: Sign of the beat change
Concluding the wrong fork frequency after loading.
Do this: Loading lowers a fork's frequency; use how the beats change to decide.
Usually costs: 2 marks.

Exam tips

Draw the displacement pattern of nodes and antinodes in the tube or string.
State the end conditions (node or antinode) before writing the formula.
For a resonance tube use L₂ − L₁ = λ/2 to remove the end correction.
Beats per second = |f₁ − f₂|.

Practice questions on Standing Waves, Organ Pipes and Beats

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The distance between two consecutive nodes in a standing wave of wavelength λ is:
(a) λ
(b) λ/2
(c) λ/4
(d) 2λ
Hint
Nodes are half a wavelength apart.
Answer
(b) λ/2
Q2 · 2 marks · Short answerBoard level
The wavelength of a standing wave on a string is 0.6 m. How far apart are adjacent nodes?
Hint
Distance = λ/2.
Answer
0.3 m
Q3 · 3 marks · Short answerHarder board level
A flute-like tube open at both ends is 0.6 m long. Find its three lowest resonant frequencies (v = 340 m s⁻¹).
Hint
f_n = nv/2L.
Answer
f₁ = 340/1.2 = 283 Hz; f₂ = 567 Hz; f₃ = 850 Hz.
Q4 · 3 marks · Short answerHarder board level
Two tuning forks of frequencies 440 Hz and 436 Hz sound together. Find the beat frequency, the number of beats in 10 s, and the frequency heard.
Hint
Beat frequency = difference; heard frequency = mean.
Answer
4 beats per second; 40 beats in 10 s; the pitch heard is the mean, 438 Hz.
Q5 · 3 marks · Short answerHarder board level
In a sonometer the length of the wire is doubled and the tension is made four times. How does the fundamental frequency change? What if the mass per unit length is also made four times?
Hint
f ∝ √T/(L√μ).
Answer
f ∝ √4/2 = 1, so it is unchanged. With μ four times as large, f is halved.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a closed pipe is 0.425 m long (v = 340 m s⁻¹).
(i) Find its fundamental frequency and the next two resonances. (1 mark)
(ii) Find the fundamental of an open pipe of the same length. (1 mark)
(iii) How many nodes (including the closed end) are there when the closed pipe sounds at 600 Hz? (1 mark)
(iv) What is the fundamental on a hotter day when v = 350 m s⁻¹? (1 mark)
Hint
The 600 Hz sound is the third harmonic of the closed pipe.
Answer
(i) f₁ = 340/1.7 = 200 Hz, then 600 Hz and 1000 Hz. (ii) 340/0.85 = 400 Hz. (iii) 600 Hz is the third harmonic, L = 3λ/4, so there are 2 nodes (including the closed end) and 2 antinodes (including the open end). (iv) f₁ = 350/1.7 = 206 Hz.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Frequencies of a closed pipe?Show answer
v/4L, 3v/4L, 5v/4L...
2.Frequencies of an open pipe?Show answer
v/2L, 2v/2L, 3v/2L...
3.Beat frequency?Show answer
|f₁ − f₂|.
4.How far is an antinode from its neighbouring node?Show answer
λ/4.

Standing Waves, Organ Pipes and Beats: frequently asked questions

Do standing waves transport energy?

No net energy is transmitted along the medium: the two travelling waves carry equal energy in opposite directions.

Why do we use the difference L₂ − L₁ in the resonance tube?

The antinode forms slightly beyond the open end (end correction e). In the difference the end correction cancels, leaving λ/2 exactly.

What limits the loudness of beats?

Beats are best heard when the two waves have nearly equal amplitudes; the beat frequency should be small enough (below about 10 per second) for the ear to follow.