Separate observer motion from source motion

Move through an existing wave pattern

First set source velocity to zero. Rightward fronts are 0.85 m apart in the air frame. Move the observer left at −40 m/s: the fronts approach at 340 − (−40) = 380 m/s relative to the observer. Dividing by spacing gives 380/0.85 = 447.058824 Hz. Moving the observer did not compress the already emitted pattern in the air frame.

Compare that capture with source +40 m/s and observer zero. The closing speed is 40 m/s in both cases, but the frequencies are 447.058824 and 453.333333 Hz. Explain the discrepancy using the spacing and encounter-speed columns. Closing speed alone omits how each object moves relative to the air.

Keep one signed convention

Write vₒ for observer velocity relative to air, positive rightward, in metres per second. The wave is traveling right at speed c. Its speed relative to the observer is c − vₒ. Positive observer velocity reduces encounters; negative velocity increases them. For source frequency fₛ and source velocity vₛ, combine encounter speed with rightward spacing:

fₒ = (c − vₒ)/λ = fₛ(c − vₒ)/(c − vₛ).

Here fₒ is received frequency in hertz and λ = (c − vₛ)/fₛ is spacing in metres. Read the equation as “wave distance traveled relative to the observer each second, divided by metres between fronts.” The velocity ratio is dimensionless, so multiplying by the emitted frequency gives hertz. Do not choose a separate approach sign after substituting signed velocities; that would count direction twice.

At source zero and observer +40 m/s, the encounter speed is 300 m/s, giving 352.941176 Hz. At source +40 and observer +40, the encounter speed is also 300 m/s but spacing is 0.75 m: the received frequency returns to 400 Hz. Both objects move, yet there is no shift. Their common motion changes spacing and encounters in compensating proportions.

Verify an encounter using positions

Use the arrival table as an independent geometry check. A crest emitted at time tₑ leaves from source position vₛtₑ. At arrival time tₐ its position is vₛtₑ + c(tₐ − tₑ). The observer is then at 4 + vₒtₐ metres. An arrival requires these two positions to agree. All times in this calculation are seconds; convert the table’s milliseconds first.

For source zero and observer −40 m/s, the first arrival is at 4/380 = 0.010526316 s. The wave position is 3.578947 m; the observer position is 4 − 40 × 0.010526316 = 3.578947 m. The next arrival is 2.236842 ms later, consistent with 447.058824 Hz. The observer moves between arrivals: treating its initial 4 m position as fixed would give the wrong answer.

Transfer and explain the model boundary

Predict the received frequency for source +60 m/s and observer +20 m/s at the model’s 400 Hz emission frequency. Is the shift equal to the source +40 / observer zero case, whose closing speed is the same?

Check the moving-observer transfer

Rightward spacing is (340 − 60)/400 = 0.70 m. Encounter speed is 340 − 20 = 320 m/s. The received frequency is 457.142857 Hz, different from 453.333333 Hz despite the same 40 m/s closing speed.

The source starts left of the observer in every selected scenario. The formula and arrival events apply before they pass, with a rightgoing wave and stationary uniform air. After passing, the propagation direction reaching the observer changes and this formula cannot simply be reused. The investigation does not describe oblique motion or Doppler shifts of light. A microphone recording also involves calibration, ambient noise and frequency estimation that this exact calculation does not assess.

Return to source motion and front spacing if the denominator needs another explanation. OpenStax: the Doppler effect provides further classical sound cases.