--- title: "Standing waves on drum skins, on plates and in a liquid-filled channel" description: "Add two opposite sound waves into a standing wave, find a microchannel’s resonance and nodes, and see what a node means on a drum, a plate and in a liquid." canonical_url: "https://sandboxphysics.com/stories/acoustics-ultrasound/standing-waves-and-resonators/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/acoustics-ultrasound/standing-waves-and-resonators/" last_modified: "2026-10-29" --- [Acoustics and ultrasound](https://sandboxphysics.com/stories/acoustics-ultrasound/) L2 · Chapter 3 # Standing waves on drum skins, on plates and in a liquid-filled channel Sandbox Physics · 2026-10-29 Shut a sound wave between two hard walls and it bounces back and forth; only the right frequencies build up. The standing wave that forms has nodes that never move. A drum skin, a metal plate and a liquid-filled channel can all hold standing waves, but what vibrates differs, and so does what a node means. The acoustic trap of the next two chapters relies on the pressure nodes in a liquid channel. ## Before you begin Pressure, particle velocity and energy density from Chapter 1, and the sum-to-product identities for sines and cosines. Bessel functions appear once and need not be computed. - Write the pressure and velocity fields of a half-wave resonance between hard walls and find the resonance frequency and node positions. - Tell apart what a node means on a membrane, a plate and in a liquid pressure field, and say where particles and sand collect. ## Two opposite waves make a standing wave Add two plane waves of equal amplitude travelling in opposite directions and the travelling parts cancel. What is left keeps a fixed shape in space and only rises and falls in time. The pressure is always zero where $\cos(ky)=0$; these points are pressure nodes. It swings most where $\cos(ky)=\pm1$, the pressure antinodes. The velocity follows from the equation of motion. It has the shape $\sin(ky)$ and lags the pressure by a quarter period. So a pressure node is a velocity antinode, a quarter wavelength away from the nearest pressure antinode. In a standing wave, a “node” always needs the name of the quantity. $$ p_1=p_a\cos(ky)\cos(\omega t),\qquad v_1=\frac{p_a}{\rho_0c_0}\sin(ky)\sin(\omega t) $$ The origin is at the left side wall and $k=2\pi/\lambda$. This is Eq. 1 of Barnkob and co-workers with the origin moved from the channel centre to the wall. Half-wave resonance across a channel with hard side walls. Teal: pressure amplitude, largest at the walls and zero at the centre. Amber: velocity amplitude, zero at the walls and largest at the centre. Beads with $\Phi>0$ collect on the pressure node. Original schematic. Source: R. Barnkob · M. Settnes and H. Bruus ## Resonance between hard walls Liquid cannot flow through a hard wall, so the normal velocity there is zero: $\sin(ky)=0$. Both side walls, $y=0$ and $y=w$, must satisfy this, so the channel width has to be a whole number of half wavelengths. The silicon–glass chip of Barnkob and co-workers (2012) has a channel $w=377\,\mu\mathrm{m}$ wide and $157\,\mu\mathrm{m}$ high. With ideal hard walls the $n=1$ resonance would be at 1.985 MHz; they drove the chip at 1.940 MHz, about 2 % lower. A real chip is not an ideal hard-walled box, which is why experiments sweep the frequency to find the resonance. $$ w=n\frac{\lambda}{2},\qquad f_n=\frac{nc_0}{2w},\qquad y_{\mathrm{node}}=\Big(j+\tfrac12\Big)\frac{w}{n},\; j=0,\dots,n-1 $$ The last expression gives the pressure nodes. For $n=1$ there is a single node in the middle of the channel. Source: R. Barnkob · M. Settnes and H. Bruus ## Drum, plate and liquid: three standing waves, three kinds of node Two other experiments on this site show standing waves, but something different vibrates in each. A drum skin is a stretched membrane: what moves is its sideways displacement, and tension pulls it back. The natural frequencies of a round drum come from zeros of Bessel functions, so the overtones are not whole multiples of the fundamental; the first overtone is about 1.59 times the fundamental. A Chladni plate is a metal plate with bending stiffness, and bending pulls it back. For a square plate with simply supported edges the natural frequencies grow as $m^2+n^2$, faster with mode number than a membrane’s. Sand is thrown off where the plate moves and settles on the lines that stay still. Those are displacement nodes. The liquid in a channel is a pressure field. What oscillates is the scalar pressure, and the liquid moves back and forth along the direction of travel: a longitudinal wave. From the next chapter on, beads stiffer and denser than water are pushed to the pressure nodes, which is exactly where the liquid moves fastest. Both “collect at a node”, but Chladni sand stops where the displacement is zero, while beads in an acoustic trap stop where the pressure is zero and the velocity is largest. The mechanical lattice experiment is yet another case: discrete masses and springs, where the frequency stops being proportional to the wavenumber once the wavelength approaches the spacing. The continuum acoustics in this course holds only for wavelengths much longer than any microscopic structure. Source: R. Barnkob · M. Settnes and H. Bruus ## Worked example: pressure nodes in a 377 µm channel Take a sound speed of 1497 m/s. For $n=1$, $f_1=1497/(2\times377\times10^{-6})\,\mathrm{Hz}=1.985\,\mathrm{MHz}$, the wavelength is $\lambda=2w=754\,\mu\mathrm{m}$ and the pressure node sits at $y=188.5\,\mu\mathrm{m}$. For $n=2$ the frequency doubles to 3.97 MHz and the nodes are at 94.3 µm and 282.8 µm. Take an energy density $E_{\mathrm{ac}}=20\,\mathrm{J/m^3}$. From $E_{\mathrm{ac}}=\kappa_0p_a^2/4$, $p_a=\sqrt{4E_{\mathrm{ac}}/\kappa_0}=423\,\mathrm{kPa}$. At the pressure node the liquid’s velocity amplitude is $v_a=p_a/(\rho_0c_0)=0.283\,\mathrm{m/s}$ and its displacement amplitude $v_a/\omega=22.7\,\mathrm{nm}$. That displacement is more than two hundred times smaller than a 5 µm bead. The bead is not washed to the node by the liquid: in one period it only rocks back and forth by tens of nanometres. What moves it to the node is a second-order radiation force that survives time averaging, the subject of Chapter 5. Source: R. Barnkob · M. Settnes and H. Bruus ## Try it in the Lab ### [Count the nodes in the channel](https://sandboxphysics.com/labs/acoustic-radiation-trap/) 1. In the acoustic trap, set “Half-wavelengths across the channel” to 1, then 2 and 3. Count the lines the beads form and note where they sit. 2. Check each line against $y_{\mathrm{node}}=(j+\tfrac12)w/n$ and explain why no line touches a side wall. ### [Compare the nodal lines of a Chladni plate](https://sandboxphysics.com/labs/chladni-plate/) 1. Open the “Four lobes” preset. How many regions do the nodal lines cut the plate into, and how do neighbouring regions move relative to each other? 2. Now pick “Drive on a node”. Why does the pattern stay still when the driver sits on its nodal line? Compare with the zero-velocity condition at the channel walls. ### [Listen for a drum’s overtones](https://sandboxphysics.com/labs/drum-modes/) 1. Choose “Center strike · all modes” and read the selected-to-fundamental frequency ratios. Confirm they are not whole numbers. 2. Compare them with the 1 : 2 : 3 ratios of the channel resonances and explain the difference by the round, two-dimensional boundary. ## Exercises with explained answers 1. For the same channel at $n=3$, what are the resonance frequency and the three pressure-node positions? Show the answer $f_3=3\times1.985\,\mathrm{MHz}=5.96\,\mathrm{MHz}$. The nodes are at $w/6$, $w/2$ and $5w/6$: 62.8 µm, 188.5 µm and 314.2 µm. 2. Replace the water with a liquid of higher sound speed and keep the width. How do the $n=1$ resonance frequency and the node position change? Show the answer The frequency $f_1=c_0/(2w)$ rises in proportion to the sound speed. The node position depends only on the geometry and stays in the middle. That is why an experiment must sweep again to find the resonance, while the node position needs no recalculation. 3. Is the average acoustic energy zero at a pressure node? Show the answer No. The potential energy is zero there, but the liquid moves fastest, and the time-averaged kinetic energy density is $\rho_0v_a^2/4$. Averaged over the whole standing wave, kinetic and potential energy are equal, and together give $E_{\mathrm{ac}}=\kappa_0p_a^2/4$. [Previous: How much of a sound wave reflects at an interface and how much passes through](https://sandboxphysics.com/stories/acoustics-ultrasound/reflection-at-an-interface/) [Next: Why a point comes out as a blob when echo times become an image](https://sandboxphysics.com/stories/acoustics-ultrasound/pulse-echo-imaging/) ## Sources and what we recomputed 1. [R. Barnkob, P. Augustsson, T. Laurell and H. Bruus, Acoustic radiation- and streaming-induced microparticle velocities determined by micro-PIV in an ultrasound symmetry plane, arXiv:1208.6534v1; Physical Review E 86, 056307 (2012)](https://arxiv.org/abs/1208.6534v1) Version 1, 31 August 2012, read in full. Eq. 1 (standing-wave field), Eq. 2 (radiation force), Eqs. 5–7 (streaming scale and Rayleigh streaming), Eqs. 9–10 (geometry and thermoviscous corrections), Eqs. 12–14 (velocity amplitude, velocity ratio, critical diameter), Eqs. 16–18 and Table I (wall drag), Table II (material data), Tables IV–V and Figs. 8–9 (measured energy densities, streaming coefficients and velocity ratios). We recompute the theory values; the measured data are quoted, not reproduced. 2. [M. Settnes and H. Bruus, Forces acting on a small particle in an acoustical field in a viscous fluid, arXiv:1110.6037v1; Physical Review E 85, 016327 (2012)](https://arxiv.org/abs/1110.6037v1) Version 1, 27 October 2011, read in full. Eq. 1 (viscous penetration depth), Eqs. 4–5 (first-order pressure and compressibility), Eqs. 48a–e (Gor’kov potential with the viscous dipole coefficient) and Table I (Pyrex and cell parameters).