Visualizing superposition and standing waves

The superposition principle states that the displacement of the medium at every point equals the sum of the displacements that each wave would produce alone. If two waves of equal frequency and amplitude travel toward each other, their sum carries no energy along the axis and forms a standing wave $y(x,t)=2A\sin(kx)\cos(\omega t)$. Points where $\sin(kx)=0$ never move at all — these are nodes, and the distance between neighbouring nodes is $\lambda/2$. Points where $|\sin(kx)|=1$ oscillate with the maximum amplitude $2A$ — these are antinodes. The boundaries set the spectrum of natural frequencies: for a string fixed at both ends, $L=n\lambda/2$ and $f_n=\dfrac{nv}{2L}$. Exactly such patterns were studied by Chladni (figures on plates, 1787), Melde (a vibrating string, 1859) and Kundt (dust in a tube, 1866).

Two waves of equal amplitude $A=0.020$ m and wavelength $\lambda=0.80$ m travel toward each other: $y_1=A\sin(kx-\omega t)$ and $y_2=A\sin(kx+\omega t)$. Find the displacement at $x=0.30$ m at time $t=0$ and decide whether this point is a node or an antinode.

In Kundt's experiment, dust in a glass tube of air forms heaps; the distance between neighbouring heaps is $0.170$ m. The speed of sound in air is $v=340$ m/s and the tube length is $L=0.510$ m. Find the wavelength, the frequency and the harmonic number.

Melde's experiment: a string of length $L=1.50$ m between a vibrator and a pulley has linear density $\mu=2.0$ g/m. The vibrator runs at $f=120$ Hz. Find the tension $T$ at which exactly $n=4$ loops fit on the string, and give the coordinates of the nodes and antinodes.

This practice runs in a simulation rather than with a real high-voltage vibrator, so it is safe.

1. Open PhET Wave on a String and choose Oscillate mode with a Fixed End.
2. Set the amplitude to $0.5$ cm and the frequency to $1.5$ Hz, then turn on Slow Motion and Show Rulers.
3. Pause the simulation at a moment of maximum displacement and record the coordinates of the motionless points (nodes) and the points of largest swing (antinodes).
4. Repeat for frequencies $1.0$, $2.0$ and $2.5$ Hz and fill in a table: $f$, number of loops $n$, $\lambda=2L/n$, $v=f\lambda$.
5. Compare the speed you obtain with the value set in the simulation and with the formula $v=f\lambda$; explain why the nodes stay still even though the waves travel.
6. As a check, note that the distance between neighbouring nodes is always $\lambda/2$ — measure it with the ruler and compare with the calculation.

Show how adding two counter-propagating travelling waves produces motionless nodes and oscillating antinodes, and relate the node spacing to the wavelength.

A three-panel diagram on a common $x$-axis.

• Top panel: the wave $y_1=A\sin(kx-\omega t)$ travelling to the right (arrow).
• Middle panel: the wave $y_2=A\sin(kx+\omega t)$ travelling to the left (arrow).
• Bottom panel: their sum $y=2A\sin(kx)\cos(\omega t)$ at the instant of maximum swing; nodes are marked with dots N and joined by a dashed line, antinodes are marked A, and arrows show the spacing $\lambda/2$ between neighbouring nodes.

The graph also shows the envelope $2A|\sin(kx)|$ as thin dotted lines.

In a school gym a rope of length $L = 4.0\,\text{m}$ is fixed at both ends and shaken at $f = 2.5\,\text{Hz}$. The rope settles into a pattern with $n = 2$ stationary loops: three places do not move at all ($x = 0$, $x = 2.0\,\text{m}$, $x = 4.0\,\text{m}$), and two places swing with the largest amplitude $\pm 0.30\,\text{m}$. For this mode $\lambda = 2L/n = 4.0\,\text{m}$, so the wave speed along the rope is $v = f\lambda = 2.5 \times 4.0 = 10\,\text{m/s}$.

Each loop is the sum of the wave travelling to the right and the wave reflected from the fixed end. At the three node positions the two displacements are always equal in magnitude and opposite in sign, so their sum is zero at every instant — that is the definition of a node. Halfway between neighbouring nodes the two waves are in phase, so the amplitude doubles to $2A$; the whole-number-of-half-wavelengths condition $L = n\lambda/2$, i.e. $\lambda_n = 2L/n$, selects the frequencies that build a steady pattern instead of a travelling wave.

A microwave oven drives its cavity at $f = 2.45\,\text{GHz}$. With $c = 3.00\times10^{8}\,\text{m/s}$ the wavelength in air is $\lambda = \frac{c}{f} = \frac{3.00\times10^{8}}{2.45\times10^{9}} = 0.122\,\text{m} \approx 12.2\,\text{cm}.$ The metal walls reflect the wave, so the electric field forms a standing pattern; the zones that heat food most strongly (electric-field antinodes) are separated by $\lambda/2 \approx 6.1\,\text{cm}$. Across a $25\,\text{cm}$ wide cavity that is about $25/6.1 \approx 4$ hot bands, which is why the turntable keeps food from sitting in a cold band.

Two identical waves travelling in opposite directions are added here: one launched by the magnetron, one reflected by the metal wall. The sum at a point fixed in space oscillates with a constant amplitude $2A\cos(2\pi x/\lambda)$ — zero right at the wall (a node: $E = 0$ on a conductor surface) and maximum at a distance $\lambda/2$ from it. The turntable averages the heating over many positions of the food, so the node/antinode map is the main reason an oven is fitted with a rotating plate.

Noise-cancelling headphones listen to the ambient sound and emit a mirror copy of it. Suppose a fan hums at $f = 150\,\text{Hz}$ with sound-pressure amplitude $0.20\,\text{Pa}$. In air at $20^{\circ}\text{C}$ ($v = 343\,\text{m/s}$) the wavelength is $\lambda = v/f = 343/150 = 2.29\,\text{m}$, and one period lasts $T = 1/f = 6.7\,\text{ms}$. To cancel the hum the headphone signal must be delayed by half a period, $\Delta t = T/2 = 3.3\,\text{ms}$, which corresponds to a phase shift $\Delta\varphi = \pi$.

This is superposition in time rather than in space: at the eardrum the two pressure waves add algebraically. With equal amplitudes and $\Delta\varphi = \pi$ the resultant pressure is $0\,\text{Pa}$ (complete destructive interference), whereas with $\Delta\varphi = 0$ it would be $0.40\,\text{Pa}$. Real headphones reach $10$–$20\,\text{dB}$ of reduction because amplitude matching and the half-period delay are only approximate.