An observer measures a 100 Hz Doppler shift as an ambulance goes by. At rest, the frequency of the ambulance's siren is 2,000 Hz. What is the speed of the ambulance?

8.6 m/s

Explanation:

The observer is stationary

The source is moving

Let v₀ be the speed of sound in the air

Let v be the speed of the ambulance

As the siren approaches

f₁ = 2000(v₀ / (v₀ - v))

As the siren departs

f₁' = 2000(v₀ / (v₀ + v))

                                                                         f₁ - f₁' = 100  

                     2000(v₀ / (v₀ - v)) - 2000(v₀ / (v₀ + v)) = 100

                                           v₀ / (v₀ - v) -  v₀ / (v₀ + v) = 100/2000

v₀(v₀ + v) / (v₀ - v)(v₀ + v) -  v₀(v₀ - v) / (v₀ - v)(v₀ + v) = 0.05

                              (v₀² + vv₀) -  (v₀² - vv₀) / (v₀² - v²) = 0.05

                                                          2vv₀ / (v₀² - v²) = 0.05

                                                                         2vv₀  = 0.05 (v₀² - v²)

                                           0.05v² + 2vv₀ - 0.05v₀² =  0

                                                         v² + 40vv₀ - v₀² = 0

quadratic formula positive answer

v = (-40v₀ + √((40v₀)² - 4(1)(v₀²))) / 2

v = (-40v₀ + √(1604v₀²)) / 2

v = (0.049968v₀) / 2

v = 0.02498439v₀

If we assume the speed of sound in air is 343 m/s

v = 8.56964... = 8.6  m/s

Approaching frequency heard is 2051 Hz

Departing frequency heard is 1951 Hz

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