Sound, acoustics, musical instruments, and the Doppler effect

Start with an open air column

Imagine an ideal tube whose two ends are open to a stationary surrounding medium. Its effective length is 0.75 m. In our supplied model, sound travels at 300 m/s, so a disturbance takes 0.005 s to travel twice that length. The lowest standing mode repeats at 200 Hz. One hertz (Hz) means one complete cycle per second.

At either ideal open end, the pressure variation is zero. That does not say the air is motionless there: the same mode has its largest air-displacement amplitude at the ends. A node must name the quantity whose variation vanishes.

Units, standards, and physical constants prepares quantity notation; Algebra, functions, graphs, and logarithms prepares ratios and substitutions. Geometry, trigonometry, and coordinate systems prepares the sine and cosine profiles introduced below.

Start with the quantities and tables. Derivatives, Taylor series, and local approximation and The classical wave equation are optional published return paths for checking the differential equations. No later acoustics lesson is required.

Separate pressure, displacement and velocity

Let x be the undisturbed longitudinal coordinate in metres, positive to the right, and t be time in seconds. The equilibrium pressure is p₀ = 100000 Pa. The excess pressure p′ is the small difference from p₀; total pressure is p₀+p′. The signed displacement χ, in metres, describes the averaged movement of a small gas element along the tube, rather than the random motion of an individual molecule.

The gas-element velocity u has units m/s and is the time rate of χ: u = ∂χ/∂t. It is distinct from the stipulated wave speed c = 300 m/s. A displacement profile plotted upwards on a page still describes longitudinal movement along x.

Take equilibrium density ρ₀ = 1.2 kg/m³ and the small-signal stiffness B = ρ₀c² = 108000 Pa. In the one-dimensional linear, lossless model, p′ = −B ∂χ/∂x and ρ₀ ∂u/∂t = −∂p′/∂x. Compression gives a positive pressure excess; a pressure gradient changes the gas velocity.

These equations assume a continuum description, small disturbances and constant supplied medium parameters. The numbers are not a universal atmospheric specification. The model does not include heat loss, viscosity, flow through a real mouthpiece or radiation into the surrounding space.

Choose an allowed column mode

For 0 ≤ x ≤ L = 0.75 m and t ≥ 0, impose the ideal pressure-release conditions p′(0,t) = p′(L,t) = 0. A sine pressure profile fits them when kL = nπ, with positive integer n. Here k is wave number in rad/m, and the angular frequency is ω = ck in rad/s. The ordinary frequency is f = ω/(2π), in Hz.

For the lowest mode n = 1, k = 4π/3 rad/m, ω = 400π rad/s and f = 200 Hz. With pressure amplitude P = 0.3 Pa, choose the displacement amplitude S = P/(Bk) = 1/(480000π) m, about 6.63×10⁻⁷ m. Radians have dimension one in these relations.

The supplied standing mode is p′(x,t) = P sin(kx) cos(ωt) and χ(x,t) = S cos(kx) cos(ωt). Differentiation gives u(x,t) = −Sω cos(kx) sin(ωt). Substitution checks the two linear equations and both pressure-release ends.

The same ideal two-open-end column admits fₙ = nc/(2L). If instead the left end is closed to gas motion and the right end is pressure-release, the lowest allowed frequency is c/(4L), and only odd positive multiples occur in that ideal model.

These boundary conditions help explain how a resonant air column can enter a musical-instrument model. They do not determine a complete played note: real geometry, excitation, end corrections and losses can change the effective response. L here is an already stipulated effective length, not a measured bore length.

Compare two snapshot quantities

At t = 0, the pressure profile rises from zero at each end to P at the centre. The displacement profile runs from +S at the left end through zero at the centre to −S at the right end. The two normalized ordinates p′/P and χ/S are dimensionless; equal graph heights do not mean equal physical quantities.

The separate blue and violet panels use the same physical x scale. Their dotted horizontal guides mark the true normalized zero; the lower frame edges are not zero. Hollow circles identify selected pressure values, and hollow squares identify selected displacement values.

At this instant u is zero everywhere. A quarter-period later, at t = 0.00125 s, p′ and χ are zero everywhere, but u is generally nonzero and its largest magnitude is Sω = 1/1200 m/s. A zero displacement snapshot is not a proof of rest.

Count phases in the medium frame

Now use a separate example: a periodic emitter and a receiver move through a stationary, homogeneous, nondispersive medium with c = 300 m/s. The emitter produces an equivalent right-going phase every 1/f_S seconds, with f_S = 200 Hz. This emission rate is stipulated separately from the stationary-column resonance above.

Measure signed source and receiver velocities v_S and v_R in the medium frame, positive to the right. The receiver is ahead of the source during the selected observation window. Assume constant velocities with |v_S| < c and |v_R| < c, and follow each phase only after its emission.

During one emission period, the earlier phase advances c/f_S metres while the source advances v_S/f_S metres. The right-going spacing is therefore λ₊ = (c−v_S)/f_S. A receiver encounters that spacing at relative speed c−v_R, so Δa = (c−v_S)/(f_S(c−v_R)) and f_R = f_S(c−v_R)/(c−v_S).

Moving the source changes the spacing in the medium. Moving the receiver changes how quickly it encounters that spacing. Both contributions are needed; replacing them by an unqualified source–receiver closing speed loses this distinction.

Compare source and receiver motion

First keep the emitter stationary and let the receiver approach at v_R = −60 m/s. The phase spacing stays 1.5 m, but the encounter speed is 360 m/s. The received frequency is 240 Hz.

Instead let the source approach at v_S = +60 m/s and keep the receiver stationary. The spacing contracts to 1.2 m; the encounter speed is 300 m/s, giving 250 Hz. These cases have the same closing speed, 60 m/s, but different received frequencies.

In the main combined case v_S = +60 m/s and v_R = −30 m/s. The phase spacing is 1.2 m and the encounter speed is 330 m/s, so the arrival interval is 1/275 s and the received frequency is 275 Hz.

Read profiles and phase arrivals

The upper panels compare the stationary-column quantities already defined. The lower panel is a different position–time diagram for the combined Doppler case. Time increases upwards from 0 to 40 ms; position increases to the right. A sloping green line follows an emitted phase, rather than a moving gas element.

For the lower panel, x_S(t) = (60 m/s)t and x_R(t) = 4.8 m−(30 m/s)t. Phase n is emitted at eₙ = n/(200 Hz), from x_S(eₙ), for n = 0,…,5. For t ≥ eₙ its position is xₙ(t) = x_S(eₙ)+c(t−eₙ).

Solving xₙ(aₙ) = x_R(aₙ) gives aₙ = ((n+4)/275) s. Each arrival follows its emission; the differences between successive arrivals are exactly 1/275 s. Filled source circles mark emissions, and hollow receiver diamonds mark arrivals.

The plotted interval is 0 ≤ t ≤ 0.04 s. The receiver remains ahead: x_R−x_S = 4.8 m−(90 m/s)t is at least 1.2 m there. All six shown phases arrive within that interval. The green segments start at their own emissions; extending them backwards would invent pre-emission events.

These are exact mathematical phase events and snapshots from the supplied models, not measured arrivals or molecular paths. The tables give the same quantities without requiring the graph.

Keep the acoustic models in scope

Pressure, gas displacement and gas velocity remain different even when they share a frequency. In this column the pressure amplitude relative to equilibrium is 3×10⁻⁶, while the largest displacement gradient and the largest ratio |u|/c are both 1/360000. These small supplied values are consistent with the linear approximation; they are not measured accuracy guarantees.

The Doppler formula above is conditional on the declared one-dimensional, stationary-medium, subsonic geometry. A source at v_S = c is outside it. A zero denominator is not an admitted prediction of an infinite observed frequency; sonic or supersonic propagation needs a different treatment.

This lesson does not calculate loudness, perceived pitch thresholds, spherical radiation, a real instrument’s tuning or relativistic Doppler shifts. Return to the optional differential-equation lessons to check the field relations, or compare another explicitly stated boundary or velocity case before extending a conclusion.

Worked examples

Try it yourself

1. At x = 0 and t = 0 in the supplied open column, find p′ and χ. Is the end a pressure node or a displacement node?

   Show the answer and reasoning

   p′ = 0 Pa and χ = +S = 1/(480000π) m. It is a pressure node and a displacement antinode.

   sin(0) = 0 while cos(0) = 1. A boundary statement about pressure does not set displacement to zero.

2. Keep c = 300 m/s and two ideal pressure-release ends, but change the effective length to 1.5 m. What is the fundamental frequency?

   Show the answer and reasoning

   100 Hz.

   The wavelength of the lowest-frequency mode is 2L = 3 m. Dividing 300 m/s by 3 m gives 100 Hz; the old 200 Hz depended on the original effective length.

3. For a separate ideal lossless column of length 0.75 m and c = 300 m/s, impose u(0,t) = 0 at a closed end and p′(L,t) = 0 at an open end. Find its fundamental and allowed positive harmonic indices.

   Show the answer and reasoning

   The fundamental is 100 Hz; the indices are 1, 3, 5,… .

   A pressure profile proportional to cos(kx) has zero gas velocity at x = 0. cos(kL) = 0 requires kL = (2j+1)π/2 for integers j ≥ 0, giving f = (2j+1)c/(4L).

4. Keep a stationary 200 Hz source in the stationary c = 300 m/s medium. An ahead receiver moves rightwards at +30 m/s. Find its received frequency and arrival interval.

   Show the answer and reasoning

   f_R = 180 Hz and Δa = 1/180 s.

   Spacing remains 1.5 m. The receiver encounters phases at 300−30 = 270 m/s, so the interval is 1.5/270 s. Receding motion lowers the arrival rate.

5. In the same medium, let source and ahead receiver both move at +30 m/s, with f_S = 200 Hz. Find λ₊, Δa and f_R.

   Show the answer and reasoning

   λ₊ = 27/20 m, Δa = 1/200 s and f_R = 200 Hz.

   The source makes spacing (300−30)/200 = 27/20 m; the receiver encounters it at 270 m/s. Their ratio returns the emission period. Neither body is stationary in the medium.

6. May you insert v_S = 300 m/s and v_R = 0 into this model and report an infinite received frequency?

   Show the answer and reasoning

   No. v_S = c violates the strict subsonic assumption.

   The spacing derivation and admitted formula require |v_S| < c. The zero denominator signals that this case is outside the model; it does not establish an infinite measured frequency.

What to carry forward

• Name the acoustic quantity: pressure in Pa, gas displacement in m, gas velocity in m/s, or wave speed in m/s.
• A pressure node can coincide with a gas-displacement antinode; ideal boundary conditions select the allowed column modes.
• Sound Doppler frequency combines medium-frame phase spacing with the receiver’s encounter speed.
• A supplied ideal mode or phase-arrival formula needs its boundary, medium, amplitude and velocity assumptions.