A bat is moving in still air toward an insect at constant speed v_{b} = 7 m/s. The insect is moving away from the bat at constant speed v_{i}. The bat emits ultrasounds of frequency υ_{0}. If the returning echo has a pitch 2% lower than the original sound, find the speed of the insect.
Givens: speed of sound v = 340 m/s.
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Solution:
Some species of bat use echolocation to determine both the position and speed of their prey. In order to do so, they detect the difference in pitch between the sound they emit and its returning echo. This difference in pitch (frequency) is due to the Doppler effect.
In the figure below is shown the given physical situation. The bat, the insect and the wavefronts are moving in the same direction.
In order to find the frequency perceived by the bat as a function of both its speed and that of the insect, we are going to split the problem in two parts. First, we will determine the frequency υ_{i} of the sound wave that reaches the insect and then we will use this frequency as the initial frequency of the returning echo that will reach the bat.
Calculation of υ_{i}:
The general expression for Doppler shift is:
Where:

 v: is the speed of the waves with respect to the medium (air in this case).
 v_{m}: is the speed of the medium with respect to Earth.
 v_{S}: is the speed of the source emitting the waves.
 v_{R}: is the speed of the observer (or listener).
 υ_{0}: is the frequency of the waves emitted by the source.
In this first part of the problem the bat plays the role of the source and the insect is the “receiver”, as you can see in the figure below. The bat emits waves of frequency υ_{0} and the frequency perceived by the insect is υ_{i}:
In this problem we are assuming still air (v_{m} = 0); therefore, υ_{i} is given by:
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In the second part of the problem the wave is reflected by the insect and the echo travels toward the bat, as you can see in the figure below:
Now the insect plays the role of the source and the bat is the receiver. The initial frequency of the waves reflected by the insect is υ_{i} and the frequency perceived by the bat is υ.
Now we have to determine what signs we need to use for the various speeds in the equation. We have to change the sign of both the source and the receiver speed. The receiver (bat) is moving toward the wavefronts and therefore its speed goes in the numerator and has positive sign. The insect (receiver) moves in the opposite direction to the wavefronts, and therefore its speed goes in the denominator with positive sign. After inserting all this information, the expression for the Doppler effect will be:
Finally, substituting υ_{i} into the previous formula we get the frequency υ perceived by the bat as a function of the initial frequency υ_{0}:
Isolating and substituting the numerical values we determine the speed of the insect v_{i}:
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