Superposition and stationary waves

What the exam asks

• Explain how a stationary wave forms: reflection, superposition, nodes and antinodes.
• Compare stationary and progressive waves for energy, amplitude, phase and frequency.
• Draw and use harmonic patterns for strings and for pipes (open at both ends, or closed at one end).
• Use $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ and describe the string practical.
• Find a wavelength from node positions with microwaves or sound, and the speed of sound from a resonance tube.

Core ideas

How a stationary wave forms

1. A progressive wave travels along the string (or the air in a pipe) and reflects at the end.
2. The incident and reflected waves have the same frequency, wavelength and speed (and similar amplitudes) and travel in opposite directions.
3. They superpose. At some points the two waves always arrive in antiphase and cancel: these are nodes, with zero amplitude. Midway between them the waves always arrive in phase and reinforce: these are antinodes, with maximum amplitude.
4. A large, steady pattern only builds up at resonant frequencies, when a whole number of half-wavelengths fits the length and matches the conditions at the ends.

Adjacent nodes (or adjacent antinodes) are $\lambda/2$ apart. A node and the next antinode are $\lambda/4$ apart.

Stationary versus progressive

Strings: both ends are nodes

$\mu$ is the mass per unit length in kg m⁻¹. For a wire of density $\rho$ and diameter $d$, $\mu = \rho A = \rho\pi d^2/4$.

So $f \propto 1/L$, $f \propto \sqrt{T}$ and . Every harmonic () is possible.

Air columns

These are displacement nodes and antinodes; a displacement node is where the pressure varies most. The antinode at an open end actually forms slightly beyond the end, so the effective length is $L + e$, where the end correction $e$ is about $0.6r$. Cambridge and OCR B treat it as negligible; other boards may raise it in resonance-tube questions, so understand it qualitatively.

The string practical (AQA RP1, Edexcel and IAL core practical, OCR PAG5)

• A string runs from a mechanical vibrator, driven by a signal generator, over a pulley to a mass hanger, so $T = mg$.
• Adjust the frequency until the first harmonic (one loop) has maximum amplitude. Read $f$ from the signal generator, or check it with an oscilloscope if the dial is poorly calibrated.
• Measure $L$ from the vibrator to the pulley (node to node) with a metre rule. Find $\mu$ by weighing a long, measured length of string on a top-pan balance.
• Change one variable at a time. Straight-line graphs: against (gradient ), against (gradient ), against .

Microwaves and sound

A transmitter faces a metal plate, and a probe moved along the line between them finds maxima and minima. Adjacent minima are $\lambda/2$ apart. Count intervals, not minima: from the 1st to the 11th minimum is 10 half-wavelengths. The same method works with a loudspeaker, a reflecting board and a microphone.

Resonance tube: hold a tuning fork (or small loudspeaker) over a tube whose air column can be lengthened, for example by raising it out of water. The first resonance occurs at $L_1 + e = \lambda/4$ and the second at $L_2 + e = 3\lambda/4$. Subtracting removes the end correction: and .

Worked examples

Exam technique

• A formation explanation needs four linked ideas: reflection gives two waves; they have the same frequency and travel in opposite directions; they superpose; nodes form where they are always in antiphase and antinodes where they are always in phase. Write “superpose” or “interfere”, not just “the waves meet”.
• Sketch the harmonic before writing an equation: mark the nodes (N) and antinodes (A) and count the quarter- or half-wavelengths.
• For scaling questions, write the proportionality ($f \propto \sqrt{T}/L$) and multiply the factors. No numbers are needed.
• For graph questions, rearrange into the form $$, identify the gradient in symbols, then substitute.

Common mistakes

Quick recap

• Two waves of the same frequency travelling in opposite directions superpose to form a stationary wave.
• Nodes have zero amplitude and antinodes maximum amplitude; node to node is $\lambda/2$.
• There is no net energy transfer. The amplitude varies with position; the phase is the same within a loop and reverses across a node.
• String: $f_n = \dfrac{n}{2L}\sqrt{T/\mu}$, with every harmonic present.