A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 38.9 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.015 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

40085 Hz

Explanation:

We are given; Sound frequency emmision of bat;f = 38.9 kHz = 38900 Hz

Bat is moving at 0.015 times the speed of sound in air.

Speed of sound in air = 343 m/s

The formula for waves reflected off the wall is calculated from Doppler equation as:

f' = f(v + v_d)/(v - v_s)

Where;

f is the frequency = 38900 Hz

f' is the detected frequency,

v_d is the velocity of the detector = 0.015 × 343 = 5.145

v_s is the velocity of the source = 0.015 × 343 = 5.145 m/s

v is the velocity of the sound = 343 m/s

Thus;

f' = 38900(343 + 5.145)/(343 - 5.145)

f' ≈ 40085 Hz

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