Explain how source motion changes wavefront spacing

Follow successive emissions

Set both velocities to zero. The source emits 400 cycles per second, so consecutive crests leave 1/400 s = 2.5 ms apart. In that interval a crest travels 340 × 0.0025 = 0.85 m through the stationary air. A stationary observer encounters these equally spaced crests at 400 Hz.

Now set source velocity to +40 m/s and keep the observer stationary. During one emission period the previous crest advances 0.85 m, but the source advances 0.10 m before emitting the next crest. Their rightward spacing is therefore 0.75 m. To the left the spacing is 0.95 m. Inspect both sides of the snapshot before reading the frequency ledger. The source has not increased its emission rate.

Derive a spacing, then count encounters

Let c be sound speed in metres per second, fₛ the emitted frequency in cycles per second, and vₛ source velocity relative to air in metres per second. Rightward velocity is positive. During the period Tₛ = 1/fₛ, the old crest moves cTₛ and the source moves vₛTₛ. Rightward wavelength is therefore λ = (c − vₛ)/fₛ. Here λ is in metres and is positive because the source remains subsonic.

For the stationary observer, received frequency is c/λ. At +40 m/s this is 340/0.75 = 453.333333 Hz. At −40 m/s the rightward spacing is 0.95 m and the received frequency is 357.894737 Hz. Compare these cases on the same 0–800 Hz bar scale. Equal approach and recession speeds do not produce equal-sized positive and negative shifts.

Read the two kinds of time evidence

The snapshot is at time zero, with the source at x = 0 m and observer at x = 4 m. Each displayed front was emitted earlier at a different source position. Only six recent fronts on each side are shown; missing older fronts are not evidence that sound has not yet reached the observer.

The arrival table instead follows two future crests: one emitted now and one 2.5 ms later. At source +40 m/s and observer zero, their arrivals are 11.764706 ms and 13.970588 ms. The difference is 2.205882 ms, the received period. The first crest’s travel time is 11.764706 ms; it is not the time between arrivals. Record both values and explain which one determines pitch-related frequency.

Transfer without a slider preset

A source emits 500 Hz in air with c = 340 m/s. It moves right at 20 m/s toward a stationary observer to its right. Predict rightward spacing and received frequency before opening the answer.

These are steady, one-dimensional, subsonic scenarios before the source passes the observer. They do not model loudness, acceleration, wind, sonic booms or a recording. Next, move the observer to change encounter speed separately from spacing.

OpenStax: the Doppler effect provides the classical sound framework. Calculations and diagrams here use the explicitly signed air-frame convention above.