Q.The following figure shows vibrating modes of an air column. Find the ratio of frequencies of the two modes.

Organ-pipe frequency depends on both air-column length and sound speed. Answer (C).

For an air column, the fundamental frequency is:

• open pipe: f = v/(2L)
• closed pipe: f = v/(4L) …

For the same length, the fundamental frequency of a closed pipe is half that of an open pipe — ratio 1:2.

For a pipe of length $L$ with speed of sound $v$:

• Closed organ pipe (closed at one end): fundamental frequency $f_{closed} = \dfrac{v}{4L}$ …

A resonance tube behaves as a closed organ pipe.

In the resonance tube experiment, a vertical tube is partly filled with water; the water surface acts as a rigid, closed boundary (a displacement node forms there) while the top of the tube is open to the atmosphere (a displacement antinode forms there). This is exactly the boundary condition of a pipe closed at one end and open at the other — i.e. a closed or …

Comparing $f_{open} = v/2L$ with $f_{closed} = v/4L$ for the same length L gives a ratio of 2:1.

In an open organ pipe (open at both ends), antinodes form at both ends, and the fundamental mode fits half a wavelength into the pipe length $L$: $L = \lambda_{open}/2$, so $\lambda_{open} = 2L$, giving fundamental frequency

$f_{open} = \frac{v}{\lambda_{open}} = \frac{v}{2L}$

In a closed organ pipe (closed at one end, open at the other), a node forms at the closed end and an antinode at the open end, and the fundamental mode fits a quarter wavelength into the length $L$: $L = \lambda_{closed}/4$, so $\lambda_{closed} = 4L$, giving fundamental frequency …

A closed organ pipe has a node at the closed end and an antinode at the open end, giving fundamental frequency f = v/(4L) = 2.5v Hz for L = 0.1 m.

A closed organ pipe (closed at one end, open at the other) vibrates with a node at the closed end and an antinode at the open end. The fundamental (first) mode fits a quarter wavelength inside the pipe length L:

L = lambda/4, so lambda = 4L

The fundamental frequency is

f = v/lambda = v/(4L)

Given L = 10 cm = 0.1 m,

f = v/(4 x 0.1) = v/0.4 = 2.5v Hz

…

A pipe closed at one end must have a node at the closed end and an antinode at the open end, which allows only odd-integer multiples of the fundamental frequency — the ratio 1:3:5.

In a closed organ pipe (closed at one end, open at the other), the closed end must always be a displacement node and the open end an antinode. This boundary condition permits only standing waves whose length corresponds to an odd number of quarter-wavelengths fitting the pipe length L:

L = (2n-1)λ/4, for n = 1, 2, 3, …

The allowed frequencies are then:

f_n = (2n-1)v/4L = (2n-1) f1

…

An open organ pipe (open at both ends) has antinodes at both ends and supports all harmonics — both even and odd multiples of the fundamental frequency.

An organ pipe open at both ends has displacement antinodes at both open ends. The allowed modes of vibration must fit an integer number of half-wavelengths between the two ends:

L = n (lambda_n / 2), n = 1, 2, 3, ...

So the frequencies of the harmonics are:

f_n = n v / (2L) = n f_1

where f_1 = v/2L is the fundamental frequency.

This means the open pipe supports f_1, 2f_1, 3f_1, 4f_1, ... — i.e., ALL integer harmonics (both odd: 1st, 3rd, 5th... and even: 2nd, 4th, 6th...) are present.

(This is in contrast to a pipe closed at one end, which has a node at the closed end and an antinode at the open end, and supports only odd harmonics: f_1, 3f_1, 5f_1, ...)

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