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For the same number of nodes in a standing wave, the ratio of the number of nodes formed in an open organ pipe to that formed in a closed organ pipe, when both vibrate in their fundamental mode, is:

R
Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
1:1
2
2:1
3
1:2
4
3:1
Correct Answer
1:1
Solution
1

Open pipe fundamental: antinodes at both ends → 1 node at centre

Closed pipe fundamental: node at closed end, antinode at open end → 1 node total

2

Both have 1 node in fundamental mode → Ratio = 1:1

Answer: 1:1

Fundamental mode: Open pipe = 1 node (centre) | Closed pipe = 1 node (closed end)
Ratio nodes = 1:1
Theory: Waves / Sound
1. Standing Waves in Pipes

Standing waves form by superposition of incident and reflected waves. Pressure and displacement nodes/antinodes: at an open end → displacement antinode, pressure node. At a closed end → displacement node, pressure antinode. Open pipe (length L): all harmonics: $f_n = nv/2L$ for $n = 1, 2, 3...$. nth harmonic: n antinodes at... wait: actually nth harmonic has (n+1) antinodes and n nodes (internal). Fundamental (n=1): 2 antinodes (ends) + 1 node (centre). Closed pipe: only odd harmonics: $f_n = nv/4L$ for $n = 1, 3, 5...$. Fundamental: 1 antinode (open end) + 1 node (closed end).

2. Sound Wave Characteristics

Speed of sound: $v = \sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M}$. In air at 0°C: $v \approx 331$ m/s; at temperature $T$ K: $v \approx 331\sqrt{T/273}$. Speed increases with temperature, humidity. Frequency (pitch): $f$ = number of oscillations per second (Hz). Wavelength: $\lambda = v/f$. Intensity: $I \propto A^2 f^2$; Decibel: $\beta = 10\log(I/I_0)$ where $I_0 = 10^{-12}$ W/m².

3. Beats

Two waves of slightly different frequencies $f_1$ and $f_2$ superpose → beat frequency $f_{beat} = |f_1 - f_2|$. Maximum amplitude when waves in phase, minimum when out of phase. Used for tuning musical instruments: if beats heard → frequencies not equal; reduce beats until zero. Doppler effect: apparent frequency changes when source or observer moves. $f_{obs} = f_s\left(\frac{v \pm v_o}{v \mp v_s}\right)$ (+ for approach, − for recession).

4. Resonance and Musical Instruments

Resonance: system vibrates at maximum amplitude when driving frequency matches natural frequency. In a pipe, resonance occurs at frequencies satisfying boundary conditions. Guitar string: $f_n = nv/2L = \frac{n}{2L}\sqrt{T/\mu}$ where $T$ = tension, $\mu$ = linear mass density. Higher tension, shorter string, lighter string → higher frequency (pitch). Harmonics in musical instruments create timbre (quality of sound). Overtones are frequencies above fundamental.

Frequently Asked Questions
1. How many nodes does an open pipe have in fundamental mode? ⌄
Open pipe fundamental: wavelength λ = 2L. Standing wave has antinodes at both open ends and ONE node at the centre of the pipe. Number of nodes = 1.
2. How many nodes does a closed pipe have in fundamental mode? ⌄
Closed pipe fundamental: wavelength λ = 4L. Standing wave: node at closed end, antinode at open end. Number of nodes = 1 (at the closed end). Both open and closed pipes have exactly 1 node in fundamental mode.
3. What harmonics do open and closed pipes support? ⌄
Open pipe: all harmonics (f, 2f, 3f, 4f...). In nth harmonic: n nodes and (n+1) antinodes. Closed pipe: only odd harmonics (f, 3f, 5f, 7f...). In nth harmonic (odd): n nodes and n antinodes. Fundamental frequency of open pipe = v/2L; closed pipe = v/4L (half the open pipe frequency for same length).
4. What is the frequency relationship between open and closed pipes of same length? ⌄
Open pipe fundamental: $f_o = v/2L$. Closed pipe fundamental: $f_c = v/4L = f_o/2$. For same frequency, closed pipe must be twice as long as open pipe. Open pipe has all harmonics; closed pipe misses even harmonics.
5. What determines nodes and antinodes in pipes? ⌄
Boundary conditions: Open end: must be antinode (maximum displacement, pressure node). Closed end: must be node (zero displacement, pressure antinode). Open pipe: both ends open → antinodes at both ends → integer number of half-wavelengths fit. Closed pipe: one closed end → node at closed end, antinode at open → odd number of quarter-wavelengths fit.
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