PhysicsCBSE Class 11 · Waves

Standing Waves, Organ Pipes and Beats Class 11

Updated 17 October 2026 · Fact-checked

A string fixed at both ends vibrates at fn = nv/2L (n = 1, 2, 3 …); an open pipe gives the same set with the speed of sound, and a closed pipe gives only odd harmonics fn = (2n − 1)v/4L. Two waves of nearby frequencies f₁ and f₂ produce beats at |f₁ − f₂| per second. Adjacent nodes are λ/2 apart, and nodes are at points of zero amplitude.

Compare node and antinode positions: string and open pipe fit λ/2 in L, but a closed pipe fits only λ/4, so its fundamental is half as high.

Illustration drawn for QuestionBuddy.

Understand Standing Waves, Organ Pipes and Beats

Standing waves. When two identical waves travel in opposite directions, their superposition is $y = 2 A sin k x cos ω t$ : every point oscillates with the same phase but an amplitude $2 A sin k x$ that depends on position. Nodes (zero amplitude) occur at $x = nλ /2$ and antinodes (amplitude $2 A$ ) midway between them at $(2 n + 1) λ /4$ . Adjacent nodes (or antinodes) are $λ /2$ apart. No energy is carried along a standing wave.

**String fixed at both ends (length $L$ ).** Nodes at both ends, so $L = nλ /2$ :
$ $f_{n} = \frac{n v}{2 L} = \frac{n}{2 L} \frac{T}{μ} (n = 1, 2, 3, \dots) .$ $
The $n = 1$ mode is the fundamental (first harmonic); the others are overtones. All harmonics are present. With tension $T$ varied, $f \propto T$ (sonometer).

Open organ pipe (open at both ends: antinodes at both ends): $f_{n} = \frac{n v}{2 L}$ , all harmonics ( $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$ ).

Closed pipe (closed at one end: node at the closed end, antinode at the open end): $f = \frac{( 2 n - 1 ) v}{4 L}$ , only odd harmonics ( $1 : 3 : 5$ ), with fundamental $v /4 L$ (half the open pipe's fundamental of the same length). A closed pipe of length $L$ has the same fundamental as an open pipe of length $2 L$ . (The air column resonates slightly beyond the open end; this end effect is ignored here.) Two successive resonant lengths of an air column, $L_{2} - L_{1} = λ /2$ , give $v = 2 f (L_{2} - L_{1})$ .

Beats. Two waves of slightly different frequencies $f_{1}$ and $f_{2}$ produce a resultant whose amplitude rises and falls at the beat frequency $∣ f_{1} - f_{2} ∣$ . The sound is loud and soft alternately. Loading a source (e.g. wax on a fork's prong) lowers its frequency; filing raises it. Observing whether the beat rate rises or falls tells which of two frequencies is larger.

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

String / open pipe: $f_{n} = \frac{n v}{2 L}$; n = 1, 2, 3, …
Closed pipe: $f_{n} = \frac{( 2 n - 1 ) v}{4 L}$; Odd harmonics only.
String speed: $v = \frac{T}{μ}$; Sonometer.
Standing wave: $y = 2 A sin k x cos ω t$; Nodes at $x = nλ /2$ .
Beat frequency: $f_{b} = ∣ f_{1} - f_{2} ∣$; Beats per second.
Air column: $v = 2 f (L_{2} - L_{1})$; Successive resonances.

How to solve Standing Waves, Organ Pipes and Beats questions

Set the boundary conditions (node or antinode at each end), find the allowed λ, then use f = v/λ.

1Mark nodes and antinodes at the two ends: string (node, node), open pipe (antinode, antinode), closed pipe (node, antinode).
2Write the allowed wavelengths: 2L/n (string/open), 4L/(2n − 1) (closed).
3Compute v (√(T/μ) for strings, 340 m/s or the given value for air).
4Use f = v/λ for each harmonic required.
5For beats, take |f₁ − f₂| and use the change in beat rate to decide which is larger.
6Check that the frequencies you list are positive and ascending.

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 string

Suppose a guitar string has fundamental 330 Hz when its vibrating length is 0.64 m.

Find the wave speed on the string and the frequency of its third harmonic.

Try it first, then show the solutionSolution

v = 2Lf₁ = 2 × 0.64 × 330 = 422 m/s.
f₃ = 3 × 330 = 990 Hz.

Answer: 422 m/s; 990 Hz.

Memory hook: Harmonics are whole-number multiples.

A bottle

Suppose air in a bottle neck behaves as a closed pipe 17 cm long (v = 340 m/s).

Find its fundamental frequency.

Try it first, then show the solutionSolution

f₁ = v/4L = 340/(4 × 0.17) = 500 Hz.

Answer: 500 Hz.

Memory hook: A blown bottle sounds the closed-pipe note.

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.

Open = all, closed = odd
Open pipes have all harmonics; closed pipes only odd ones.
Use it when: Pipe questions.
Nodes are λ/2 apart
Distance between nodes gives λ.
Use it when: Standing wave spacing.
Wax lowers, file raises
Loading lowers frequency; filing raises it.
Use it when: Beats logic.
Closed is half
Fundamental of a closed pipe is half that of an open pipe of the same length.
Use it when: Quick comparisons.

Standing Waves, Organ Pipes and Beats: solved examples

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

Example 1: String harmonics · 3 marks · Short answerHarder board level

A string of linear mass density 5.0 × 10⁻³ kg/m under a tension of 180 N is fixed at both ends 0.50 m apart. Find the wave speed and the first three harmonic frequencies.

Try it first, then show the solutionSolution

v = √(T/μ) = √(180/0.005) = 190 m/s. (1)
f₁ = v/2L = 190/(1.0) = 190 Hz. (1)
f₂ = 380 Hz, f₃ = 570 Hz. (1)

Answer: 190 m/s; 190, 380, 570 Hz.

Example 2: Closed pipe · 2 marks · Very short answerBoard level

Find the first three resonant frequencies of a pipe 0.85 m long, closed at one end (v = 340 m/s).

Try it first, then show the solutionSolution

f₁ = v/4L = 340/(4 × 0.85) = 100 Hz. (1)
Next harmonics are odd multiples: 300 Hz and 500 Hz. (1)

Answer: 100, 300, 500 Hz.

Example 3: Speed from successive resonances · 3 marks · Short answerHarder board level

A tuning fork of frequency 512 Hz makes an air column in a tube (closed at one end) resonate at lengths 15.0 cm and 48.0 cm, successive resonances. Find the speed of sound, and explain why 15.0 cm is not a quarter wavelength.

Try it first, then show the solutionSolution

L₂ − L₁ = λ/2 = 33.0 cm, so λ = 66.0 cm. (1)
v = fλ = 512 × 0.660 = 338 m/s. (1)
λ/4 = 16.5 cm is larger than 15.0 cm, because the antinode forms slightly outside the open end of the tube. (1)

Answer: 338 m/s.

Example 4: Beats and unknown frequency · 4 marks · Long answerHarder board level

A fork of frequency 512 Hz gives 5 beats per second with a second fork. Loading the second fork with a little wax reduces the beat rate to 3 per second. Find the frequency of the second fork before and after loading, and the time between beats initially.

Try it first, then show the solutionSolution

Initially the unknown frequency is 512 ± 5, i.e. 517 or 507 Hz. (1)
Loading lowers the frequency; the beat rate falls, so the unknown fork was above 512 Hz: 517 Hz. (1)
After loading: 515 Hz (3 beats per second with 512 Hz). (1)
Time between beats initially = 1/5 = 0.2 s. (1)

Answer: 517 Hz before; 515 Hz after; 0.2 s between beats.

Example 5: Sonometer and beats · 4 marks · Long answerHOTS / competency

A sonometer wire of length 0.50 m and mass per unit length 4.0 × 10⁻³ kg/m vibrates in its second harmonic in resonance with a 320 Hz fork. (a) Find the speed of the wave and the tension. (b) The tension is raised by 2%. Find the new second-harmonic frequency and the beat frequency heard with the fork, and the time between beats.

Try it first, then show the solutionSolution

(a) f₂ = 2 × v/2L = v/L, so v = 320 × 0.50 = 160 m/s. (1)
T = μv² = 4.0 × 10⁻³ × 25600 = 102.4 N. (1)
(b) f ∝ √T: a 2% rise in T gives about a 1% rise in f, 320 × 1.01 = 323.2 Hz. (1)
Beat frequency = 3.2 Hz, so beats occur every 0.31 s. (1)

Answer: (a) 160 m/s; 102.4 N; (b) 323.2 Hz; 3.2 beats per second.

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

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

Mistake: Using f = v/2L for a closed pipe.
Confusing the pipe types.
Do this: Closed pipe: f = v/4L with odd harmonics only.
Usually costs: The answer mark.
Mistake: Calling the harmonic number n the overtone number.
Terminology.
Do this: First overtone = second harmonic for string and open pipe.
Usually costs: 1 mark.
Mistake: Getting the sign of the beat change wrong.
Not applying the loading/filing rule.
Do this: Wax lowers; file raises; compare the beat rate before and after.
Usually costs: The deduction mark.
Mistake: Forgetting to convert cm to m.
Unit slip.
Do this: Express lengths in metres when v is in m/s.
Usually costs: The answer mark.

Exam tips

Sketch the string or pipe with nodes and antinodes before writing the formula.
Name which harmonics are present (all, or odd only).
For beats, give both possibilities and the reasoning that selects one.
Show f_n = n f₁ for strings and open pipes.
State the end-effect assumption if lengths are compared.

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
Two consecutive nodes of a standing wave on a string are 0.40 m apart and the frequency is 150 Hz. The speed of the wave is
(a) 60 m/s (b) 120 m/s (c) 240 m/s (d) 300 m/s
Hint
λ = 2 × distance between nodes.
Answer
(b) 120 m/s: λ = 0.80 m, v = 150 × 0.80.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Beats are produced when two waves of slightly different frequencies are superposed.
Reason (R): The amplitude of the resultant wave rises and falls periodically at a rate equal to the difference of the two frequencies.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Phase difference grows with time.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the fundamental and the first overtone of a pipe 0.34 m long and open at both ends (v = 340 m/s).
Hint
f_n = nv/2L.
Answer
f₁ = 340/0.68 = 500 Hz; f₂ = 1000 Hz.
Q4 · 3 marks · Short answerHarder board level
A 300 Hz fork gives 3 beats per second with fork B and 5 beats per second with fork C. The beat frequency between B and C is 2 per second. Find the possible frequencies of B and C.
Hint
B is 297 or 303; C is 295 or 305.
Answer
B ∈ {297, 303}, C ∈ {295, 305}. Beats of 2 between B and C require both below or both above 300 Hz: (B, C) = (297, 295) or (303, 305).
Q5 · 4 marks · Long answerHOTS / competency
A standing wave on a string is y = 0.004 sin(15x) cos(300t) (SI). (a) Find the amplitudes, wavelength, frequency and speed of the component travelling waves. (b) Find the distance between nodes. (c) The string is 0.628 m long and fixed at both ends; which harmonic is this? (d) Find the maximum speed of a particle at an antinode.
Hint
Compare with y = 2A sin kx cos ωt.
Answer
(a) 2A = 4 mm so each wave has 2 mm; k = 15 so λ = 0.419 m; ω = 300 so f = 47.7 Hz; v = ω/k = 20 m/s. (b) λ/2 = 0.209 m. (c) 0.628/0.209 = 3 loops: the third harmonic. (d) v = ωA_max = 300 × 0.004 = 1.2 m/s.

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.Distance between adjacent nodes?Show answer
λ/2.
2.Closed pipe fundamental?Show answer
v/4L.
3.Open pipe harmonics?Show answer
All integers.
4.Beat frequency?Show answer
|f₁ − f₂|.
5.Loading a fork does what?Show answer
Lowers its frequency.
6.String fundamental?Show answer
v/2L.

Standing Waves, Organ Pipes and Beats: frequently asked questions

Why does a closed pipe have only odd harmonics?

The closed end must be a node and the open end an antinode, which allows only odd multiples of a quarter wavelength.

Do standing waves carry energy?

No net energy flows along a standing wave; energy oscillates between kinetic and potential in each loop.

What is the maximum beat frequency we can hear distinctly?

About 10 to 15 per second; faster variations are heard as a rough or separate tone.

What is the first overtone?

The next harmonic after the fundamental: n = 2 for strings and open pipes, the third harmonic for a closed pipe.

How are beats used in practice?

To tune instruments: the beats vanish when the two notes are in unison.