Q.A pipe $20\ \text{cm}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430\ \text{Hz}$ source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is $340\ \text{m s}^{-1}$).

Concept: Acoustic Resonance Harmonics -- a pipe closed at one end supports only the odd harmonics of $f_1 = \dfrac{v}{4L}$; a pipe open at both ends supports all harmonics of $f_1' = \dfrac{v}{2L}$.

Step 1 -- Closed-pipe fundamental:

Step 2 -- Compare with the source: The source is $430\ \text{Hz}$, very close to the calculated fundamental $425\ \text{Hz}$ (a difference of only about $1\%$, well within the experimental tolerance expected of a real resonance-tube measurement). So the closed pipe is taken to be resonantly excited by the source in its fundamental (first harmonic) mode.

Step 3 -- Open-pipe fundamental: If both ends were open instead,

$f_1' = \frac{v}{2L} = \frac{340}{0.40} = 850\ \text{Hz}$ …

Why harmonics matter in pipes

When a sound source is placed near a pipe, the air column inside can vibrate strongly only at certain natural frequencies — its resonant modes (harmonics). The source frequency must match one of these modes for resonance to occur.

The key difference between a pipe closed at one end and a pipe open at both ends is the boundary condition for the pressure wave:

• Closed end: air molecules cannot move — a displacement node (pressure antinode).
• Open end: air molecules move freely — a displacement antinode (pressure node).

This changes which harmonics are possible.

Step-by-step solution

1. Find the fundamental frequency of the closed pipe.

Length $L = 20\ \text{cm} = 0.20\ \text{m}$, speed of sound $v = 340\ \text{m/s}$.

For the first (fundamental) mode of a closed pipe, the air column length equals one-quarter of the wavelength:

$L = \frac{\lambda_1}{4} \quad\Rightarrow\quad \lambda_1 = 4L = 0.80\ \text{m}$.

Fundamental frequency:

$f_1 = \frac{v}{\lambda_1} = \frac{340}{0.80} = 425\ \text{Hz}$.

2. Check which harmonic matches $430\ \text{Hz}$.

The allowed frequencies are $f_n = n \times 425\ \text{Hz}$ for $n = 1, 3, 5, \dots$.

• $n=1$: $425\ \text{Hz}$ — close to $430\ \text{Hz}$, but not exact.
• $n=3$: $1275\ \text{Hz}$ — too high.
• $n=5$: $2125\ \text{Hz}$ — even higher.

The source is $430\ \text{Hz}$, which is only $5\ \text{Hz}$ above the fundamental. In practice, a source near the fundamental will excite it because real pipes have a small bandwidth — the air column resonates over a narrow range around each harmonic. So the first harmonic (fundamental mode) is excited. …

Step 1: For the pipe closed at one end, the fundamental is $f_1=\dfrac{v}{4L}=\dfrac{340}{4\times0.20}=425\ \text{Hz}$; a closed pipe supports only odd harmonics $nf_1$ ($n=1,3,5,\dots$).

Step 2: Compare the source (430 Hz) to the allowed frequencies: it's very close to $f_1=425\ \text{Hz}$ (within a real pipe's resonance bandwidth) and nowhere near the next allowed harmonic $3f_1=1275\ \text{Hz}$ — so the source resonantly excites the first harmonic (fundamental). …