Question

if the bat emits a sound at a frequency of 80.8 khz and hears it reflected at a frequency of 83.2 khz while traveling at a speed of 3.8 m/s , calculate the speed of the insect.

Answer 1

This is a classic example of the Doppler effect, where the frequency of a wave changes due to the relative motion of the source and the observer. In this case, the bat is the source and the insect is the observer.

The Doppler equation for frequency is:

f' = f * (v + u) / (v - u)

where f is the frequency emitted by the source (bat), f' is the frequency heard by the observer (insect), v is the speed of sound in air, and u is the speed of the observer (insect) relative to the medium (air) in which the wave is traveling.

We can rearrange the Doppler equation to solve for the speed of the insect:

u = (v * (f' / f) - v - f' * u / f) / (f' / f - 1)

Plugging in the given values, we get:

u = (343 m/s * (83.2 kHz / 80.8 kHz) - 343 m/s - 83.2 kHz * u / 80.8 kHz) / (83.2 kHz / 80.8 kHz - 1)

Simplifying the equation:

u = 3.13 m/s

Therefore, the speed of the insect is approximately 3.13 m/s.

Step-by-step explanation:

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