Final answer:
The frequency of the reflected sound wave received by the bat, which is flying towards a wall at 7.7 m/s and emits a 53.6 kHz ultrasound burst, is approximately 54.8 kHz. This is calculated using the Doppler effect equation.
Step-by-step explanation:
The question relates to the Doppler effect, which is a change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, a bat is emitting an ultrasound burst and detecting the frequency of the reflected pulse from a wall. We can use the Doppler effect equation for the frequency heard by the bat when it is moving towards a stationary object. The formula is f' = f * (v + v0) / (v - vs), where f' is the received frequency, f is the source frequency, v is the speed of sound, v0 is the speed of the observer (bat), and vs is the speed of the source (0 since the wall is stationary).
Given the ultrasound burst frequency f = 53.6 kHz, the bat's speed v0 = 7.7 m/s, and the speed of sound v at 20°C is 343 m/s typically, the calculation for the received frequency f' would be:
f' = 53.6 kHz * (343 m/s + 7.7 m/s) / (343 m/s)
Plugging in the values we get:
f' ≈ 53.6 kHz * (350.7 m/s) / (343 m/s)
f' ≈ 53.6 kHz * 1.0225
f' ≈ 54.8 kHz
Therefore, the frequency of the reflected sound wave received by the bat is approximately 54.8 kHz.