Question

An eagle flying at 31 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 16 m/s. (Assume the speed of sound is 343 m's.) (a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird? Hz (b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird? Hz

Answer 1

(a) The frequency the blackbird hears is 461.15 Hz.

(b) The frequency the blackbird hears after the eagle passes the blackbird is 422.35 Hz.

Given that, velocity of the eagle v₁ = 31 m/s

Frequency of the eagle F₁ = 440 Hz

Velocity of the blackbird v₂ = 16 m/s

Speed of sound v = 343 m/s

(a) The equation for frequency is given as F = f₀ (v-v₂)/(v-v₁)

F = 440 * (343-16)/(343-31) = 440 * 327/312 = 461.15 Hz

(b) The equation for frequency is given as F = f₀ (v+v₂)/(v+v₁)

F = 440 * (343+16)/(343+31) = 440 *359/374 = 422.35 Hz

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