Question

Please provide as much detail as possible.

Some bat species eat up to 1,000 mosquitoes in an hour. A bat, flying with a speed of 4.35 m/s, is in pursuit of a mosquito. While in motion, the bat produces a 39.70 kHz sound and hears the sound which is reflected off the mosquito at a frequency of 40.40 kHz. (Assume the speed of sound in air is v = 343 m/s.) (a) How fast is the mosquito flying (in m/s)? apply the Doppler equation twice, first to find an expression for the frequency as observed by the mosquito, and then for the echo frequency. For the first equation, vs = +vbat mosquitochirp and-observed by mosquitor and for the second equation,observed by mosquitor Vo oat sYmosquitor and -echo Be sure you understand the signs for Enter a number each speed. m/s

Answer 1

Let V: speed of mosquito, V_bat = 4.35 m/s frequency heard by mosquito: F^1 = f times (V_sound- V/v_sound- V_bat) further, frequency of echo heard by bat: f^11 = f^1 times (V_sound + V_bat/V_souns + V) = f times (V_sound- V) (V_sound + V_bat)/(V_sound- V_bat) (V_sound + V_bat) 40.4 = 39.7 times [(343 times V) (343 times 4.35)/(343- 4.35) (343 + v)] 0.992 times 343 - v/343 + V rightdoublearrow = 1.35 m/s