Doppler Effect

Question

An ambulance siren emits sound at 660 Hz while the vehicle moves directly toward a stationary pedestrian at 30 m per second, with sound speed 330 m per second. What frequency does the pedestrian hear?

RankGuru · Accepted Answer

The Doppler effect changes the observed frequency when a source and observer move relative to each other. For a stationary observer and a source approaching at speed $v_s$, the apparent frequency is $f' = f\left(\frac{v}{v - v_s}ight)$, where $v$ is the speed of sound. The denominator shrinks because successive wavefronts are emitted from closer positions, compressing the wavelength. Substituting $f = 660	ext{ Hz}$, $v = 330	ext{ m/s}$, and $v_s = 30	ext{ m/s}$ gives $f' = 660\left(\frac{330}{330 - 30}ight) = 660 	imes \frac{330}{300} = 660 	imes 1.1 = 726	ext{ Hz}$. The value $600	ext{ Hz}$ wrongly adds $v_s$ in the denominator, which applies to a receding source. The value $660	ext{ Hz}$ ignores the motion entirely. The value $693	ext{ Hz}$ comes from putting $v_s$ in the numerator using the observer-motion formula by mistake. This is the NCERT Doppler formula for a moving source. As a plausibility check, an approaching source must raise the pitch, so the heard frequency must exceed 660 Hz, and the modest 10 percent increase to 726 Hz is consistent with a source speed that is one-eleventh of the sound speed.

Doppler Effect

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