A bat flying at 6.00 m/s emits a chirp at 40.0 kHz. If this sound pulse is reflected by an insect moving away from the bat at 2.00 m/s, what is the frequency of the echo received by the bat? Assume the air temperature is 25.0 ˚C.
Would I need two different equations for this? Have no clue how to do, please help!
Yes, you have to use Doppler shift equations twice; first to get the frequency received and reflected by the insect in its own frame of reference,l then to consider the insect a moving source of that frequency, and calculate the still higher frequency received by the bat.
Since the speeds are small compared to the speed of sound, you can get an accurate approximation by assuming the insect is still and the bat moves toward it at 4 m/s, resulting in an upwards Doppler shift of
delta f = fo * 2 * (4/340)
4/340 is the relative Doppler shift, fo the emitted frequency, and the factor of two accounts for the two Doppler shifts, sending and receiving.
To solve this problem, you can use the Doppler effect equation, which relates the observed frequency of a moving sound source to the frequency emitted by the source and the relative velocity between the source and the observer. The equation for the observed frequency is:
f' = f * ((v + v_observer) / (v + v_source))
Where:
f' = observed frequency
f = emitted frequency
v = speed of sound in air (assumed to be 343 m/s at 25.0 ˚C)
v_observer = velocity of the observer
v_source = velocity of the source
In this case, the bat is the source of the sound, and the insect is the observer.
First, let's determine the velocities:
v_observer = 2.00 m/s (since the insect is moving away from the bat)
v_source = -6.00 m/s (since the bat is moving towards the insect, the velocity is negative)
Now, substitute the known values into the equation:
f' = 40.0 kHz * ((343 + 2.00) / (343 - 6.00))
Simplifying,
f' = 40.0 kHz * (345.0 / 337.0)
f' ≈ 40.9 kHz
Therefore, the frequency of the echo received by the bat is approximately 40.9 kHz.