A sound source emits frequency 680 Hz while moving at 42 m/s directly towards a stationary observer in still air. Sound speed is 340 m/s. (a) Calculate the wavelength in front of the moving source. (b) Calculate the observed frequency before and after it passes and recedes directly. Use the classical moving-source model and neglect acceleration during each observation. (c) Explain whether the sound speed changes.

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Accepted Answer

Front wavelength 0.4382 m; approaching frequency 775.8 Hz; receding frequency 605.2 Hz; sound speed stays 340 m/s. The source emits consecutive wavefronts one period 1/680 s apart. During this interval sound travels 340/f while the source moves 42/f, so front wavelength is (340 − 42)/680 = 0.4382 m. The observer receives wavefronts moving through air at 340 m/s. Hence fapproach = 340/λfront = 680 × 340/(340 − 42) = 775.8 Hz. Behind the source, the separation increases to (340 + 42)/680. Thus frecede = 680 × 340/(340 + 42) = 605.2 Hz, smaller than the emitted frequency. The source motion changes wavefront spacing, not the propagation speed set by the still-air medium. The approaching frequency rises because more compressed wavefronts arrive each second; after passage the expanded spacing makes arrivals less frequent. Source and observer formulas are not interchangeable.

A moving sound source 10

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