Q.For a particular tube, among six harmonic frequencies below 1000 Hz, only four harmonic frequencies are given: 300 Hz, 600 Hz, 750 Hz and 900 Hz. What are the two other frequencies missing from this list?

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)

In both cases the frequency depends on the length L of the air column AND on the speed of sound v in air (which itself varies with temperature). Hence both factors matter.

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 $f_{closed} = \dfrac{v}{4L}$
• Open organ pipe (open at both ends): fundamental frequency $f_{open} = \dfrac{v}{2L}$

Ratio:

$\dfrac{f_{closed}}{f_{open}} = \dfrac{v/4L}{v/2L} = \dfrac{1}{2}$

So $f_{closed} : f_{open} = 1 : 2$.

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 organ pipe — which is why only odd harmonics ($f, 3f, 5f, \dots$) are produced as the water level (effective length) is varied to find resonance.

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

$f_{closed} = \frac{v}{\lambda_{closed}} = \frac{v}{4L}$

Taking the ratio for the same length $L$ and same speed of sound $v$:

$\frac{f_{open}}{f_{closed}} = \frac{v/2L}{v/4L} = \frac{4L}{2L} = 2$

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

Here v stands for the speed of sound in air; the answer is left in terms of v since no numerical value of v is supplied in this question, and 2.5v Hz matches option (b) exactly.

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

where f1 = v/4L is the fundamental frequency.

So the possible harmonics are f1, 3f1, 5f1, … — only odd harmonics are present, in the ratio 1:3:5.

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, ...)

Checking options:

(A) Odd harmonics only — this describes a closed pipe, not an open one

(B) Even harmonics only — incorrect, the fundamental (1st, odd) is always present

(C) Even as well as odd harmonics — correct for an open pipe

(D) None of these — not needed