Superposition and stationary wavesAQA A-Level Physics: Revision notes

When two or more waves meet at a point, the resultant displacement is the vector sum of the displacements that each wave would produce there on its own. This is the principle of superposition.

Displacements have direction, so a crest (+) and a trough (−) partly or completely cancel. Where the waves arrive in phase the result is constructive interference with displacement as large as possible; where they arrive in antiphase the result is destructive interference.

Worked example. Crests of 0.040 m and 0.030 m overlap: resultant $0.040 + 0.030 = 0.070$ m. A 0.040 m crest and a 0.030 m trough give $0.040 - 0.030 = 0.010$ m on the crest side. After the waves have passed through each other each continues unchanged.

A stationary (standing) wave forms when two progressive waves of the same frequency (and similar amplitude) travel in opposite directions and superpose. This usually happens when a wave is reflected back on itself, for example at a fixed end of a string or at a metal plate.

• Nodes are points where the two waves are always in antiphase: the displacement is always zero.
• Antinodes are points where they are always in phase: the amplitude is a maximum.
• Adjacent nodes (or adjacent antinodes) are half a wavelength apart: $\lambda/2$.
• Between two adjacent nodes all points oscillate in phase, with different amplitudes (largest at the antinode). Points on either side of a node are in antiphase.
• Unlike a progressive wave, no energy is transferred along the wave and the pattern does not move.

A string of length $l$ fixed at both ends must have a node at each end. Only certain wavelengths fit: $\lambda = \dfrac{2l}{n}$ for $n = 1, 2, 3, \ldots$ These are described as harmonics (the first, second, third harmonic). The first harmonic has one antinode, so half a wavelength fits along the string: $\lambda = 2l$.

For the first harmonic the frequency is

$f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$

where $T$ is the tension (N) and $\mu$ is the mass per unit length (kg m⁻¹). The wave speed on the string is $v = \sqrt{T/\mu}$. The $n$th harmonic has frequency $nf$.

Worked example. $l = 0.65$ m, $T = 80$ N, $\mu = 5.0\times10^{-3}$ kg m⁻¹: $f = \dfrac{1}{1.30}\sqrt{\dfrac{80}{5.0\times10^{-3}}} = 97$ Hz. The third harmonic is 291 Hz. Quadrupling $T$ doubles $f$.

Stationary waves are produced when a wave reflects from a barrier.

• Microwaves: a transmitter faces a metal plate. A small detector moved between them finds a series of minima (nodes) and maxima (antinodes). Successive nodes are $\lambda/2$ apart, so $\lambda$ is twice the spacing and $f = c/\lambda$.
• Sound: a loudspeaker facing a hard reflecting surface (or sound in a tube) gives nodes where the microphone signal is zero and antinodes where it is loudest, again $\lambda/2$ apart.

Worked example. Minima 1.4 cm apart: $\lambda = 2.8$ cm and $f = 3.00\times10^{8}/0.028 = 1.1\times10^{10}$ Hz.

To find how the first-harmonic frequency $f$ varies with $l$, $T$ and $\mu$, change one variable at a time.

• Vary $l$ between two bridges, keeping $T$ (a fixed hanging mass) and $\mu$ (the same wire) constant. Adjust the signal generator until there is a single large antinode and record $f$. Plot $f$ against $1/l$: a straight line through the origin shows $f \propto 1/l$.
• Vary $T$ using different masses ($T = mg$); plot $f$ against $\sqrt{T}$.
• Vary $\mu$ by using wires of different thickness; plot $f$ against $1/\sqrt{\mu}$.
• The gradient of the first graph is $\tfrac{1}{2}\sqrt{T/\mu}$. Measure $l$ with a metre rule, repeat and average, and wear eye protection in case the wire snaps.