Physics / Kundt's Tube Standing Waves
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EXPERIMENT 42 · PHYSICS LABORATORY

Kundt's Tube Standing Waves theory

Understand kundt's tube standing waves: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
d = \left(\frac{v_{\text{air}}}{v_{\text{rod}}}\right) L_{\text{rod}} \iff y = m x + b \iff v_{\text{rod}} = \frac{v_{\text{air}}}{m}, \quad E = \rho_{\text{rod}} v_{\text{rod}}^2

Keep the physical quantities, graph variables and units consistent.

01Acoustic Wave Interference and Longitudinal Standing Waves

When two identical longitudinal acoustic waves propagate in opposite directions within a confined tube, their linear superposition produces an acoustic standing wave. Let the two counter-propagating acoustic pressure perturbations be represented by p₁(x, t) = P₀ sin(kx − ωt) and p₂(x, t) = P₀ sin(kx + ωt). Applying trigonometric product-to-sum identities, the resultant acoustic pressure field is p(x, t) = p₁ + p₂ = 2 P₀ sin(kx) cos(ωt). Simultaneously, the acoustic particle displacement ξ(x, t) follows the spatial gradient of pressure: ξ(x, t) = 2 ξ₀ cos(kx) sin(ωt), where ξ₀ = P₀ / (ρ₀ v ω). Crucially, acoustic displacement and acoustic pressure are in space-time quadrature (a quarter-wave spatial phase shift of π/2 radians). At locations where kx = nπ (x = n λ/2), the displacement amplitude reaches a maximum (2 ξ₀), defining displacement antinodes where particle velocity u = ∂ξ/∂t swings violently. At these same positions, sin(kx) = 0, meaning acoustic pressure variations vanish (pressure nodes). Conversely, where kx = (2n + 1)π/2 (x = (2n + 1) λ/4), particle displacement is strictly zero (displacement nodes), while acoustic pressure fluctuations reach their peak amplitude (2 P₀, pressure antinodes). At the rigid reflector piston, the solid boundary enforces zero particle displacement (u = 0, displacement node) and complete acoustic reflection with zero phase shift in pressure.

p(x, t) = 2 P_0 \sin(kx) \cos(\omega t), \qquad \xi(x, t) = 2 \xi_0 \cos(kx) \sin(\omega t), \qquad u = \frac{\partial \xi}{\partial t}
In a Kundt's tube, do the dust heaps form at displacement nodes or displacement antinodes, and why?

Dust heaps accumulate at displacement nodes. At displacement nodes, acoustic particle velocity is zero, so dust particles remain undisturbed and settle quietly. In contrast, at displacement antinodes, acoustic particle velocity is maximal, generating violent air agitation and radiation pressure that sweep the dust away into striations.

02Longitudinal Elastic Vibrations in Clamped Solid Rods

Longitudinal wave propagation through a solid elastic rod of uniform cross-section A, density ρ, and Young's modulus E is governed by the 1D wave equation: ∂²u/∂t² = v_rod² (∂²u/∂x²), where the phase velocity is given by v_rod = √(E / ρ). In Kundt's experiment, the rod is clamped rigidly at its exact midpoint (x = 0), leaving both ends (x = −L_rod/2 and x = +L_rod/2) entirely free. The rigid clamp enforces a zero-displacement boundary condition at the center: u(0, t) = 0 (displacement node). At the free ends, the longitudinal mechanical stress must vanish: σ = E (∂u/∂x) = 0, which requires the spatial derivative to vanish, creating displacement antinodes (strain nodes). The spatial mode function for anti-symmetric displacement is u_n(x) = C sin(k_n x). Enforcing the free-end condition cos(k_n L_rod / 2) = 0 yields k_n L_rod / 2 = (2n − 1) π / 2, or k_n = (2n − 1) π / L_rod for n = 1, 2, 3, ... For the fundamental longitudinal mode (n = 1), k₁ = π / L_rod, which corresponds to an acoustic wavelength of λ_rod = 2π / k₁ = 2 L_rod. Thus, the rod vibrates in half-wave resonance with its fundamental natural frequency given by f₁ = v_rod / λ_rod = v_rod / (2 L_rod). Clamping exactly at the center suppresses all even harmonics and supports only odd harmonic overtones f_n = (2n − 1) f₁.

v_{\text{rod}} = \sqrt{\frac{E}{\rho}}, \qquad \lambda_{\text{rod}} = 2 L_{\text{rod}}, \qquad f_1 = \frac{v_{\text{rod}}}{2 L_{\text{rod}}} = \frac{1}{2 L_{\text{rod}}} \sqrt{\frac{E}{\rho}}
Why must the metal rod in Kundt's tube be clamped at its exact center rather than at one-quarter of its length to produce the fundamental tone?

The fundamental longitudinal standing wave in a free-free bar has a single displacement node at the exact midpoint (x = L/2) and displacement antinodes at both free ends (λ = 2L). Clamping at the exact midpoint enforces this displacement node without damping the fundamental mode. Clamping at L/4 would suppress the fundamental and force higher harmonic modes.

03Kundt's Tube Acoustic Coupling and Powder Striation Dynamics

The operational brilliance of Kundt's tube lies in using a lightweight indicator powder to visualize microscopic acoustic wave motion. A small, light cork or cardboard disc mounted on the free end of the vibrating rod projects into the glass tube mouth, acting as an acoustic piston that oscillates back and forth at frequency f. When the air column length is tuned to resonance, stationary standing waves are established with acoustic particle velocity u(x, t) = u_max sin(kx) sin(ωt). The aerodynamic behavior of the fine lycopodium powder or cork dust is governed by acoustic radiation pressure, Bernoulli forces, and Rayleigh acoustic streaming. At displacement antinodes (sin(kx) = ±1), particle velocity oscillates with large amplitudes (often exceeding 1 to 2 m/s). This high velocity creates intense localized aerodynamic drag and Bernoulli suction, which violently lifts the powder off the tube floor and forms characteristic transverse striations (fine parallel ridges perpendicular to the tube axis). At displacement nodes (sin(kx) = 0), the air is completely quiescent: no acoustic forces act on the dust particles, allowing them to remain at rest and accumulate into well-defined, sharp heaps. The spatial distance between any two consecutive heaps corresponds to the distance between adjacent displacement nodes, which is exactly half of the acoustic wavelength in the air: d = λ_air / 2.

d = \frac{\lambda_{\text{air}}}{2} \iff \lambda_{\text{air}} = 2 d, \qquad u_0 = \omega \xi_0 = \frac{P_0}{\rho_0 v_{\text{air}}}
How is the distance between two consecutive dust heaps related to the wavelength of sound in air?

The distance between consecutive dust heaps equals half of the acoustic wavelength in air: d = λ_air / 2. Therefore, the full wavelength of sound in air is twice the heap spacing: λ_air = 2d.

04Comparative Velocity Ratio and Invariant Frequency Equivalence

Because the vibrating rod acts as the acoustic driver for the air column, the frequency of the sound waves propagating through the air inside the tube is strictly identical to the mechanical vibration frequency of the rod: f_air = f_rod. Using the universal wave relationship v = f λ for both media, we have: f_air = v_air / λ_air and f_rod = v_rod / λ_rod. Equating these two expressions yields: v_air / λ_air = v_rod / λ_rod. Substituting the fundamental standing wave conditions λ_rod = 2 L_rod and λ_air = 2 d gives: v_air / (2 d) = v_rod / (2 L_rod), which simplifies directly to the classical Kundt velocity relation: v_rod / v_air = L_rod / d. Rearranging for the dust node spacing d yields: d = (v_air / v_rod) L_rod. This powerful relationship demonstrates that the ratio of sound velocities in the solid rod and the gaseous medium equals the ratio of the sounding rod length to the nodal dust spacing. If the speed of sound in air v_air is known from Laplace's adiabatic formula v_air = √(γ R T / M) (v_air = 331.3 √(1 + t / 273.15) m/s, yielding 343.20 m/s at 20.0 °C), the sound velocity in the metal rod can be determined with remarkable accuracy without requiring an electronic frequency counter.

\frac{v_{\text{rod}}}{v_{\text{air}}} = \frac{\lambda_{\text{rod}}}{\lambda_{\text{air}}} = \frac{L_{\text{rod}}}{d} \implies d = \left(\frac{v_{\text{air}}}{v_{\text{rod}}}\right) L_{\text{rod}}, \qquad v_{\text{rod}} = v_{\text{air}} \left(\frac{L_{\text{rod}}}{d}\right)
If a brass rod of length 100 cm clamped at its center produces dust heaps spaced 9.81 cm apart in air where sound speed is 343.2 m/s, what is the speed of sound in the rod?

Using v_rod = v_air × (L_rod / d) = 343.2 × (1.000 / 0.0981) ≈ 3500 m/s (or 3.50 km/s).

05Determination of Dynamic Young's Modulus and Continuum Elasticity

Kundt's tube provides an elegant method to measure the dynamic Young's modulus E of metals and alloys. From continuum mechanics, the phase velocity of longitudinal waves in an elastic rod whose lateral dimensions are small compared to the wavelength (r << λ) is given by v_rod = √(E / ρ_rod), where ρ_rod is the mass density. Squaring both sides yields: E = ρ_rod v_rod². Combining this with the Kundt velocity ratio gives: E = ρ_rod [v_air (L_rod / d)]². Measuring elastic moduli acoustically offers substantial advantages over static tensile testing (such as hanging weights from a wire). Static tensile tests are vulnerable to mechanical creep, microscopic plastic deformation, thermal expansion drifts, and support compliance. In contrast, acoustic vibrations operate at high audio frequencies (1 kHz to 5 kHz) under infinitesimal strain amplitudes (< 10⁻⁵) and purely adiabatic conditions, measuring the true reversible intermolecular bond stiffness of the crystal lattice. For cold-rolled cartridge brass with density ρ = 8500 kg/m³ and measured wave velocity v_rod = 3500 m/s, the dynamic Young's modulus is E = 8500 × (3500)² = 1.041 × 10¹¹ Pa = 104.1 GPa, in excellent agreement with standard ultrasonic pulse-echo benchmarks.

E = \rho_{\text{rod}} v_{\text{rod}}^2 = \rho_{\text{rod}} \left[ v_{\text{air}} \left(\frac{L_{\text{rod}}}{d}\right) \right]^2, \qquad \sigma_{xx} = E \varepsilon_{xx}
How can the dynamic Young's modulus of the brass rod be calculated from the measured speed of sound in the rod?

From the continuum wave equation in a thin elastic bar, v_rod = √(E / ρ_rod). Squaring both sides yields E = ρ_rod v_rod². Given brass density ρ ≈ 8500 kg/m³ and v_rod ≈ 3500 m/s, E ≈ 8500 × 3500² ≈ 1.04 × 10¹¹ Pa = 104 GPa.

06Linear Regression Formulation and OLS Parameter Isolation

To determine the longitudinal wave velocity and Young's modulus with rigorous metrological precision, experimentalists vary the sounding rod length L_rod across a calibrated series (e.g. 80.0 to 130.0 cm) and measure the corresponding mean dust node spacing d. From the theoretical relation d = (v_air / v_rod) L_rod, defining x = L_rod (in metres) as the independent variable and y = d (in metres) as the dependent variable yields the standard linear regression model: y = m x + b. Here, the regression slope m represents the dimensionless acoustic velocity ratio: m = v_air / v_rod. Inverting this relation isolates the longitudinal sound speed in the rod directly: v_rod = v_air / m. The intercept b provides a crucial experimental diagnostic: theoretically b = 0, so any statistically significant non-zero intercept detects end-coupling disc offsets, finite clamp width artifacts, or progressive visual reading bias. Ordinary least-squares (OLS) regression minimizes the sum of squared residuals, averaging out random errors in identifying dust heap crests and yielding unbiased slope and standard error estimates.

d = m L_{\text{rod}} + b \implies m = \frac{v_{\text{air}}}{v_{\text{rod}}} \iff v_{\text{rod}} = \frac{v_{\text{air}}}{m}, \qquad \delta v_{\text{rod}} = v_{\text{air}} \frac{\delta m}{m^2}
When plotting dust node spacing d on the vertical axis against rod length L_rod on the horizontal axis, what does the slope m represent?

The slope m = d / L_rod represents the velocity ratio v_air / v_rod. Dividing the known speed of sound in air v_air by the regression slope m directly gives the speed of sound in the metal rod: v_rod = v_air / m.

07Gas Thermometry, Ratio of Specific Heats (γ), and Modern Applications

In addition to investigating solids, Kundt's tube is a classic apparatus for studying the thermodynamics of gases. By sealing the tube and flushing it with test gases (such as Helium, Argon, or Carbon Dioxide), the ratio of specific heats γ = C_p / C_v can be determined. According to Laplace's equation, sound velocity in an ideal gas is v_gas = √(γ R T / M), where R is the universal gas constant, T is absolute temperature, and M is molar mass. Combining this with Kundt's formula gives: d_gas = (v_gas / v_rod) L_rod = (√(γ R T / M) / v_rod) L_rod. Comparing the nodal spacing d_gas in a test gas to d_air in air under identical rod excitation isolates the gas sound velocity: v_gas = v_air (d_gas / d_air). From v_gas, the heat capacity ratio is calculated as γ = (M v_gas²) / (R T). This enabled historical physicists to prove that noble gases like Argon and Helium are monatomic (γ ≈ 1.67, 3 translational degrees of freedom), while air is diatomic (γ ≈ 1.40, 5 degrees of freedom). In modern technology, the standing wave principles of Kundt's tube underpin acoustic levitation of pharmaceutical micro-droplets in containerless processing, thermoacoustic heat pumps and cryo-coolers, and ultrasonic non-destructive testing of structural pipelines.

v_{\text{gas}} = \sqrt{\frac{\gamma R T}{M}} \implies \gamma = \frac{M v_{\text{gas}}^2}{R T}, \qquad \frac{d_{\text{gas}}}{d_{\text{air}}} = \frac{v_{\text{gas}}}{v_{\text{air}}} = \sqrt{\frac{\gamma_{\text{gas}} M_{\text{air}}}{\gamma_{\text{air}} M_{\text{gas}}}}
If the air in Kundt's tube is replaced by carbon dioxide (CO₂, sound speed ≈ 267 m/s) while exciting the same rod, how does the spacing between dust heaps change?

Since d = (v_gas / v_rod) L_rod, the node spacing is directly proportional to the gas sound speed. Because sound travels slower in dense CO₂ (267 m/s) than in air (343.2 m/s), the wavelength is shorter and the dust heaps become closer together by a factor of 267 / 343.2 ≈ 0.78 (7.65 cm instead of 9.81 cm).

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Connect the equations to a measurement.