Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 21.0 m/s. Both screech, one emitting a frequency of 3200 Hz and the other one of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s?
To find the frequencies received by the two eagles, we need to use the concept of the Doppler effect. The Doppler effect describes the change in frequency of a wave due to the relative motion between the source of the wave and the observer.
In this case, both eagles are moving towards each other, so the sound waves they emit will be compressed, resulting in an increase in frequency. Let's calculate the frequency received by each eagle.
The formula for the Doppler effect is as follows:
f' = (v + v₀) / (v - vᵥ) * f₀
Where:
f' is the received frequency
v is the speed of sound in the medium (330 m/s in this case)
v₀ is the speed of the observer (the other eagle)
vᵥ is the speed of the source (the eagle emitting the sound wave)
f₀ is the emitted frequency
Let's solve for each eagle:
For the first eagle, using:
v₀ = 21.0 m/s (speed of the second eagle)
vᵥ = 15.0 m/s (speed of the first eagle)
f₀ = 3200 Hz (frequency emitted by the first eagle)
f'₁ = (330 + 21) / (330 - 15) * 3200
f'₁ = 351 / 315 * 3200
f'₁ = 3570.48 Hz
Therefore, the frequency received by the first eagle is approximately 3570.48 Hz.
For the second eagle, using:
v₀ = 15.0 m/s (speed of the first eagle)
vᵥ = 21.0 m/s (speed of the second eagle)
f₀ = 3800 Hz (frequency emitted by the second eagle)
f'₂ = (330 + 15) / (330 - 21) * 3800
f'₂ = 345 / 309 * 3800
f'₂ = 4250.16 Hz
Therefore, the frequency received by the second eagle is approximately 4250.16 Hz.
To summarize, the first eagle receives a frequency of approximately 3570.48 Hz, and the second eagle receives a frequency of approximately 4250.16 Hz.