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PhysicsMediumClass 11

Standing Waves — Nodes, antinodes, harmonics

Waves & Sound

7

JEE Qs

8%

Hard

75

min

Master the boundary conditions for different systems (strings, open/closed pipes) to correctly determine allowed wavelengths and resonant frequencies.

🧮 Key Formulas

Equation of a standing wave: y(x,t) = 2A sin(kx) cos(ωt)
Condition for nodes: kx = nπ (n = 0, 1, 2, ...)
Condition for antinodes: kx = (n + 1/2)π (n = 0, 1, 2, ...)
Distance between consecutive nodes/antinodes = λ/2
Distance between a node and an adjacent antinode = λ/4
Wave speed: v = fλ
Wave speed in a string: v = sqrt(T/μ) (where T is tension, μ is linear mass density)
Wave speed in gas (sound): v = sqrt(γP/ρ) or v = sqrt(B/ρ) (where γ is adiabatic index, P is pressure, ρ is density, B is bulk modulus)
Frequencies for string fixed at both ends / open organ pipe: f_n = n * (v / 2L) (n = 1, 2, 3, ...)
Frequencies for closed organ pipe: f_n = (2n - 1) * (v / 4L) (n = 1, 2, 3, ...), only odd harmonics present

✅ Key Points for JEE

  • 1Standing waves are formed by the superposition of two identical waves traveling in opposite directions, resulting in a stationary pattern of nodes and antinodes.
  • 2Nodes are points of zero displacement amplitude, while antinodes are points of maximum displacement amplitude; energy is localized and not transported in a standing wave.
  • 3Boundary conditions are crucial: a fixed end or displacement minimum is a node, a free end or displacement maximum is an antinode.
  • 4For strings fixed at both ends and open organ pipes, all harmonics (integral multiples of the fundamental frequency) are present. For closed organ pipes, only odd harmonics are present.
  • 5The wave speed (v) is determined solely by the properties of the medium (e.g., tension and mass density for strings, elasticity and density for sound in air), not by the frequency or wavelength of the standing wave.

⚠️ Common Mistakes

  • Confusing displacement nodes/antinodes with pressure nodes/antinodes, especially for sound waves in pipes (displacement node is a pressure antinode and vice-versa).
  • Incorrectly applying boundary conditions or formulas for strings versus open/closed organ pipes, leading to wrong allowed frequencies or harmonics.
  • Misinterpreting or miscounting harmonics versus overtones (e.g., the 3rd harmonic is the 2nd overtone).

📝 Practice Questions

See all

Q27.Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below ⎪ ⎪ 2025 (28 Jan Shift 1) JEE Main Previous Year Paper (1) Both A and R are true but R is NOT the correct (2) A is true but R is false explanation of A (3) A is false but R is true (4) Both A and R are true and R is the correct explanation of A

2025·Assertion ReasoningMedium

Q40.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170. Reason (R): Nuclear force is long range. In the light of the above statements, choose the correct answer from the options given below : (1) (A) is true but (R) is false (2) (A) is false but (R) is true (3) Both (A) and (R) are true and (R) is the correct (4) Both (A) and (R) are true but (R) is NOT the explanation of (A) correct explanation of (A)

2025·MCQEasy

Q36.A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities ρ1 and ρ2′ , respectively. The frequency of 9th harmonic of closed tube is identical with 4th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is ρ1 : ρ2 = 1 : 16, then the length of the open tube is : (1) 15 7 cm (2) 207 cm (3) 15 9 cm (4) 209 cm

2025·MCQMedium

Q24.A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is ( a−1a ) then the value of a is ________ → 2^i+6^j+8^k 2^i+^j+^k . The

2024·NumericalMedium

Q24.Two open organ pipes of lengths 60 cm and 90 cm resonate at 6th and 5th harmonics respectively. The difference of frequencies for the given modes is _______ Hz . (Velocity of sound in air = 333 m/s )

2024·NumericalMedium

Q2. The equation of stationary wave is : y = 2a sin ( 2πntλ ) cos ( 2πxλ ). Which of the following is NOT correct : (1) The dimensions of n/λ is [T] (2) The dimensions of n is [LT−1] (3) The dimensions of x is [L] (4) The dimensions of nt is [L]

2024·MCQMedium

NCERT Chapters

  • Class 11 Physics Ch 15: Waves