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.

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.

Examples in context

Example 1. A Queensland Symphony Orchestra cellist at QPAC tunes the C string ($f_1 = 65.4 \text{ Hz}$) to a string length $0.69 \text{ m}$ fixed at both ends. Wave speed is $v = 2L f_1 = 2 \times 0.69 \times 65.4 = 90.3 \text{ m s}^{-1}$. The first overtone (2nd harmonic) sits at $130.8 \text{ Hz}$. The cellist adjusts tension to alter $v$ and so $f_1$, a direct application of the QCAA Unit 2 harmonics relation $f_n = nv/(2L)$.

Example 2. A Bundaberg sugar mill operator notices a vibration peak at $58 \text{ Hz}$ in a steam pipe open at both ends, length $2.9 \text{ m}$. For an open pipe, $f_n = nv/(2L)$; with $v = 340 \text{ m s}^{-1}$, $f_1 = 58.6 \text{ Hz}$, confirming the resonance is the fundamental. Engineers add a damping clamp at the midpoint (a pressure node) to suppress it. QCAA EA Unit 2 thematic items pair industrial stimulus with this kind of harmonic identification.

Try this

Q1. State the relationship between length and wavelength for the fundamental of a string fixed at both ends. [2 marks]

• Cue. $L = \lambda/2$, so $\lambda = 2L$.

Q2. A pipe closed at one end has length $0.85 \text{ m}$. Using $v = 340 \text{ m s}^{-1}$, calculate the first three resonant frequencies. [3 marks]

• Cue. Closed-pipe odd harmonics only; $f_1 = v/(4L) = 100 \text{ Hz}$; $f_3 = 300 \text{ Hz}$; $f_5 = 500 \text{ Hz}$.

Q3. A guitar string of length $0.65 \text{ m}$ has wave speed $260 \text{ m s}^{-1}$. (a) Calculate $f_1$ and $f_3$. (b) Explain why pressing the string at the $12$th fret (midpoint) doubles the fundamental frequency. (c) Discuss one factor that limits the precision of a QCAA IA2 resonance experiment. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

• Cue. (a) $f_1 = 200 \text{ Hz}$, $f_3 = 600 \text{ Hz}$; (b) halves $L$ so doubles $f_1$; (c) endpoint position uncertainty.