Explain the formation of standing waves in strings (fixed at both ends) and in air columns (open and closed pipes), and solve problems involving the resonant frequencies of mechanical systems

What this dot point is asking

QCAA wants you to explain how a standing wave forms (superposition of two waves travelling in opposite directions with the same frequency), to identify nodes and antinodes, and to derive the resonant-frequency series for three standard systems: a string fixed at both ends, an open pipe, and a closed pipe.

How a standing wave forms

When a wave reflects off a boundary, the reflected wave travels back through the incident wave. The two waves superpose. At points where they are always in antiphase, the displacements cancel (a node). At points where they are always in phase, the displacements add to maximum (an antinode).

The result is a wave pattern with fixed nodes and antinodes that does not propagate. Energy is trapped in the standing-wave region. Only frequencies that fit the boundary conditions survive; others decay through destructive interference.

String fixed at both ends

Both ends are nodes (the string cannot move). The string supports a series of resonant modes:

The fundamental ($n = 1$) has $\lambda_1 = 2L$ and $f_1 = v / 2L$. Higher modes are integer multiples: $f_2 = 2 f_1$, $f_3 = 3 f_1$, and so on. The series contains all integers (a full harmonic series).

Open pipe (both ends open)

Both ends are antinodes (air can move freely). Same length-wavelength relation as the string:

All harmonics present. A flute is approximately an open pipe.

Closed pipe (one end closed)

The closed end is a node; the open end is an antinode. Quarter-wavelengths fit the length:

Only odd harmonics are present ($f_1, 3f_1, 5f_1, \ldots$). A clarinet is approximately a closed pipe at low register, which is why a clarinet sounds different from a flute of the same length playing the same fundamental.

Resonance

Resonance occurs when an applied periodic force has a frequency equal to one of the natural frequencies of the system. Energy is transferred efficiently into the standing wave and amplitude builds up over many cycles.

Examples: a tuning fork held above a tube of adjustable length will resonate when the air column matches a closed-pipe harmonic; pushing a child on a swing at the swing's natural frequency builds amplitude.

Worked example

An organ pipe is $1.20$ m long and closed at one end. Take the speed of sound as $343$ m s$^{-1}$. Find the fundamental frequency and the first three audible resonances.

Closed pipe: $f_n = n v / 4L$ with $n = 1, 3, 5$.

$f_1 = (1)(343) / (4 \times 1.20) = 71.5$ Hz.

$f_3 = 3 f_1 = 214$ Hz.

$f_5 = 5 f_1 = 357$ Hz.

The even harmonics ($n = 2, 4$) are absent.

Common traps

Confusing the closed-pipe series with the open-pipe series. Closed pipes only have odd harmonics. Drawing the antinode at the wrong end is the most common error.

Treating the string fundamental as $\lambda_1 = L$. A string fixed at both ends has nodes at the ends, so the fundamental has only one antinode and $\lambda_1 = 2L$, not $L$.

Forgetting end correction in air columns. Real open ends have antinodes slightly outside the physical pipe. QCAA problems use ideal pipes unless they specify a correction.

Treating resonance as one specific frequency. Resonance happens at every natural frequency, not just the fundamental.

How this appears in IA1 and EA

IA1. Often a stimulus showing the standing-wave pattern on a string or in a pipe and asking for either the harmonic number or the length given a frequency and speed.

EA Paper 1. Multiple choice on which harmonics are present in closed pipes versus open pipes.

EA Paper 2. A two-part question on a musical-instrument system, typically asking for wave speed and then a higher harmonic.

In one sentence

Standing waves form when two travelling waves of the same frequency superpose in opposite directions, producing fixed nodes and antinodes; a string fixed at both ends and an open pipe both have all-integer resonances $f_n = n v / 2L$, while a closed pipe has only odd resonances $f_n = n v / 4L$.