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two eagles fly directly toward one another, the first at 19.8 m/s and the second at 14.6 m/s. the first eagle screeches, emitting a frequency of 2,609 hz. what frequencies (in hz) does the second eagle receive if the speed of sound is 340 m/s?

User Cognophile
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When two eagles fly directly toward each other, the second eagle receives a frequency of 2955 Hz.

Step-by-step explanation:

When two eagles fly directly toward each other, their relative velocity is the sum of their individual velocities. In this case, the relative velocity is (19.8 m/s + 14.6 m/s) = 34.4 m/s.

The Doppler effect formula for frequency is given by:

f' = f(v + vd) / (v - vs)

Where f' is the observed frequency, f is the emitted frequency, v is the speed of sound, vd is the velocity of the detector, and vs is the velocity of the source.

Plugging in the given values, we get:

f' = 2609 Hz * (340 m/s + 14.6 m/s) / (340 m/s - 19.8 m/s) = 2955 Hz

Therefore, the second eagle receives a frequency of 2955 Hz.

User Hisham Ahamad
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