Q.An organ pipe closed at one end has a length of $0.5\ \text{m}$. Taking the speed of sound in air as $340\ \text{m s}^{-1}$, find the fundamental frequency, the first overtone, and the second overtone of the pipe.

In an open pipe, stationary waves form with antinodes at both open ends, giving allowed frequencies f_n = nv/2L for n = 1, 2, 3,... (all harmonics). For a closed pipe of length 0.7 m with v = 331 m/s, the fundamental frequency works out to about 118.2 Hz.

1. Stationary waves in an open pipe: When sound is produced in an air column enclosed in a pipe that is open at both ends, sound waves travel down the pipe and reflect at each open end (due to the pressure/density mismatch with the outside air). The incident and reflected waves of the same frequency travelling in opposite directions superpose to form a stationary (standing) wave. At each open end, the air is free to move, so a displacement antinode forms there, while an open pipe of length L can accommodate any pattern with antinodes at both ends and equally-spaced nodes in between. If there are $n$ half-wavelength loops fitted into the length L, $L = n\frac{\lambda_n}{2},\qquad n = 1, 2, 3, \dots$ so $\lambda_n = \frac{2L}{n}$ and the corresponding frequency (using $f = v/\lambda$, v = speed of sound) is $f_n = \frac{nv}{2L},\qquad n = 1, 2, 3,\dots$ For n = 1 this gives the fundamental (first harmonic) $f_1 = v/2L$; n = 2 gives the second harmonic (first overtone) $f_2 = 2f_1$; and so on. So an open organ pipe produces all harmonics (both even and odd multiples) of its fundamental frequency.
2. Fundamental frequency of the given closed organ pipe: …

An open pipe forms antinodes at both ends and supports all harmonics ($f_n = nv/2L$); a closed pipe has a node at the closed end and an antinode at the open end, supporting only odd harmonics ($f_n = nv/4L$) — for the given 70 cm closed pipe this gives a fundamental of about 118.2 Hz.

1. Stationary waves in an open organ pipe: When sound is produced in a pipe open at both ends, the incident and the wave reflected from each open end superpose to form a stationary (standing) wave inside the air column. Because both ends are open to the atmosphere, the pressure variation must be zero there, which corresponds to displacement antinodes at both open ends. The simplest possible standing wave pattern (the fundamental mode) that satisfies this — antinode at each end — has exactly one node in the middle. This means half a wavelength fits into the pipe length $L$: $L = \dfrac{\lambda_1}{2} \implies \lambda_1 = 2L \implies f_1 = \dfrac{v}{\lambda_1} = \dfrac{v}{2L}$ Higher modes (harmonics) are possible whenever an integer number of half-wavelengths fit into $L$, i.e. $L = p\left(\dfrac{\lambda_p}{2}\right)$ for $p = 1, 2, 3, \ldots$, giving: $f_p = \dfrac{pv}{2L} = p f_1, \quad p = 1, 2, 3, \ldots$ So an open pipe supports all harmonics (both odd and even integer multiples) of its fundamental frequency $f_1 = v/2L$.
2. Fundamental frequency of a closed organ pipe: …

In an open organ pipe, antinodes form at both open ends and all harmonics (fₙ = nv/2L) are produced. (Note: this topic is Waves, a Class-11/Physics Paper-I topic, and is not part of the current Physics-II chapter set — it is answered here for completeness since it appeared on this paper.)

Formation of stationary waves in an open organ pipe

An open organ pipe is open at both ends. When air is blown into it, sound waves travel down the pipe and reflect off the open ends. The incident and reflected waves (of the same frequency and speed, travelling in opposite directions) superpose to form a stationary (standing) wave. Since both ends are open, air molecules there are free to move maximally, so antinodes are formed at both open ends.

Derivation of harmonic frequencies:

For the simplest stationary-wave pattern (fundamental mode), there is one node in the middle and antinodes at both ends. The length of the pipe equals half a wavelength:

$L = \frac{\lambda_1}{2} \implies \lambda_1 = 2L$

Fundamental frequency:

$f_1 = \frac{v}{\lambda_1} = \frac{v}{2L}$

For higher modes, additional nodes/antinodes fit inside the pipe such that the length is an integer number of half-wavelengths:

$L = \frac{n\lambda_n}{2} \implies \lambda_n = \frac{2L}{n}, \quad n = 1, 2, 3, \dots$

So the frequency of the $n$-th harmonic is:

$f_n = \frac{nv}{2L} = n f_1$

…