1. The problem statement, all variables and given/known data Horseshoe bats use the Doppler Effect to determine their location. A horseshoe bat flies toward a wall at a speed of 15 m/s while emitting a sound of frequency 35 kHz. What is the beat frequency between the emission frequency and the echo? Assume that the air temperature is 20C. 2. Relevant equations Doppler effect f0=[(1-v0/v)/(1-vs/v)]fs Beat Frequency fbeat = Δf 3. The attempt at a solution f0=[(1-(-343m/s)/(15m/s))/(1-(343m/s)/(15m/s))]*35kHz f0=-38.20121951kHz fbeat = Δf = fs-f0 fbeat = 35kHz-(-38.20121951kHz) fbeat = 73.2kHz I'm not sure that I have solved this correctly. It seems strange to me that the echo has more than double the frequency of the original sound. edit; or should I say, that the beat frequency is more than double that of the original sound.