Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 23.0 m/s. Both screech, one emitting a frequency of 2600 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 that each eagle receives, we need to use the formula for the Doppler effect. The Doppler effect describes the change in frequency of a wave (in this case, sound) due to relative motion between the source of the wave and the observer.
The formula for the observed frequency (f') is:
f' = f * (v + V) / (v - V)
Where:
f' is the observed frequency,
f is the emitted frequency,
v is the speed of sound,
V is the velocity of the observer relative to the medium (in this case, the speed of the eagle).
Let's calculate the frequencies that each eagle receives:
For the first eagle:
- Emitted frequency (f) = 2600 Hz
- Speed of sound (v) = 330 m/s
- Velocity of the first eagle (V) = -15.0 m/s (negative because it is moving towards the observer)
Using the formula:
f' = 2600 * (330 - 15) / (330 + 15)
f' ≈ 2555 Hz
Therefore, the first eagle receives a frequency of approximately 2555 Hz.
For the second eagle:
- Emitted frequency (f) = 3800 Hz
- Speed of sound (v) = 330 m/s
- Velocity of the second eagle (V) = 23.0 m/s
Using the formula:
f' = 3800 * (330 + 23) / (330 - 23)
f' ≈ 4138 Hz
Therefore, the second eagle receives a frequency of approximately 4138 Hz.
So, the frequencies that the eagles receive are approximately 2555 Hz and 4138 Hz.