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Doppler Effect (Approaching Observer)

f′=f(v+vo)vf' = \frac{f (v + v_o)}{v}

Worked example: 500 Hz, observer at 34.3 m/s (v = 343) → f' = 550 Hz — press Try an example to run it live, then adjust anything.

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Grade 11Grade 11 Physics

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Doppler Effect (Approaching Observer) explained

fvvof′

When you move toward a stationary source, the waves themselves are unchanged — you simply run into them more often. Cycling at 10 m/s toward a 500 Hz factory whistle in 343 m/s air, you intercept 500 × (343 + 10)/343 ≈ 515 Hz, a shift of 15 Hz. Unlike the moving-source case there is no runaway denominator: even sprinting at the speed of sound toward the whistle would only double the frequency, because the waves keep their spacing and you just sweep through them faster.

That asymmetry — moving source squeezes the wavelength, moving observer changes only the encounter rate — is a favorite exam trap, and the two formulas give slightly different answers for the same relative speed. At everyday speeds the difference is tiny: both predict about a 3% shift at 10 m/s. Bats live by this physics, correcting their echolocation chirps for their own flight speed so the returning echo lands in the narrow frequency band where their ears are sharpest.

Doppler Effect (Approaching Observer) formula

f′=f(v+vo)vf' = \frac{f (v + v_o)}{v}
Where
  • f′f'= Observed frequency (Hz)
  • ff= Source frequency (Hz)
  • vv= Speed of sound (m/s)
  • vov_o= Observer speed (m/s)

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