A sound source emits frequency 520 Hz while moving right at 20 m s⁻¹ through still air. Sound travels at 330 m s⁻¹. A stationary observer is directly ahead of the source. Find the wavelength ahead, the wavelength behind, and the frequency heard by the observer ahead. Explain using the distance the source travels between successive wavefront emissions. Assume steady subsonic motion along the line joining source and observer.

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Accepted Answer

Ahead wavelength 0.5962 m; behind wavelength 0.6731 m; observed frequency ahead 553.5 Hz. The time between emitted wavefronts is the source period T = 1/f = 0.001923 s. During this time each sound wavefront advances cT through the stationary air. The source itself advances uT to the right before emitting the next wavefront. Ahead of it the separation is reduced to (c − u)T, whereas behind it the separation increases to (c + u)T. Thus λ_ahead = (330 − 20)/520 = 0.5962 m and λ_behind = (330 + 20)/520 = 0.6731 m. The stationary observer receives wavefronts travelling at 330 m s⁻¹, so f_observed = c/λ_ahead = 553.5 Hz. This exceeds the emitted frequency because the approaching fronts are closer together. The source speed does not add to the propagation speed of sound in the stationary medium. Setting u = 0 recovers equal wavelengths c/f in both directions and no Doppler shift, checking the model.

Source motion and compressed wavefronts · set 4

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