A bat,flying at 7.00 m/s toward a wall,emits a chirp at 50.0 kHz.If the wall reflects this sound pulse,what is the frequency of the echo received by the bat? (v_sound = 340 m/s) A)52.1 kHz B)51.1 kHz C)50.0 kHz D)49.0 kHz E)48.6 kHz

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Quizplus · Accepted Answer

The frequency of the echo received by the bat is increased due to the Doppler effect, as the bat is moving towards the wall. The frequency of the echo can be calculated using the formula for the Doppler effect when the source is moving towards a stationary reflector (wall) and then back to the moving detector (bat). The formula for the frequency received $f'$ is given by $f' = f \times \frac{v + v_0}{v - v_s}$, where $f$ is the emitted frequency (50.0 kHz), $v$ is the speed of sound (340 m/s), $v_0$ is the speed of the observer (bat) towards the source (0 m/s, since the wall is stationary and reflects the sound back), and $v_s$ is the speed of the source (bat) towards the observer (7.00 m/s). Substituting the values, we get $f' = 50.0 \times \frac{340 + 7}{340 - 7} = 50.0 \times \frac{347}{333} \approx 52.1 kHz$.

A Bat,flying at 7