Question

In each of the four scenarios to the right, a large bat lets out a short burst of ultrasonic sound, which the smaller bat hears a moment later. If the large bat flies at 3.420 m/s and the small bat flies at 14.60 m/s, rank the frequency that the smaller bat detects in the four scenarios, from highest to lowest. Assume that the speed of sound is 343.0 m/s. In scenario C, some of the large bat's signal reflects off of the small bat and returns to the large bat, warning it of the smaller bat's presence. If the initial signal has a frequency of 45.50 kHz, what return frequency will the large bat detect? Calculate the final frequency to four significant figures.

Answer 1

Answer 2

The detected frequency ?$f$ of a sound wave depends upon the motions of both the source of the sound and the detector. The detected frequency can be written in terms of the emitted frequency ?0$f_0$ , the source velocity ?s$v_{\text{s}}$ , and the detector velocity ?d$v_{\text{d}}$ .

$$f=f_0\left(\frac{V\pm v_{\text{d}}}{V\pm v_{\text{s}}}\right)$$

The variable ?$V$ denotes the propagation speed of the sound wave. The appropriate sign in each case depends on the direction of motion of the detector or source. If one moves toward the other, choose the sign that shifts the frequency higher. Conversely, if one moves away from the other, choose the sign that shifts the frequency lower. In all four scenarios, the large bat is the source of the sound, whereas the small bat is the detector.

In Scenario A, the source is moving toward the detector, so a negative sign is applied to the source velocity in the denominator to shift the frequency higher. The detector is moving away from the source, so a negative sign is applied to the detector velocity in the numerator to shift the frequency lower. The result is

$$\begin{array}{rl}{f}_{\text{A}}& =f_0\left(\frac{V-v_{\text{d}}}{V-v_{\text{s}}}\right)\\ \\ & =f_0\left(\frac{\left(343.0\text{ m}/\text{s}\right)-\left(15.60\text{ m}/\text{s}\right)}{\left(343.0\text{ m}/\text{s}\right)-\left(4.850\text{ m}/\text{s}\right)}\right)\\ \\ & =0.9682f_0\end{array}$$

In Scenario B, the source and the detector are moving away from each other. Both movements shift the frequency lower, which calls for a negative sign in the numerator and a positive sign in the denominator.

$$\begin{array}{rl}{f}_{\text{B}}& =f_0\left(\frac{V-v_{\text{d}}}{V+v_{\text{s}}}\right)\\ \\ & =f_0\left(\frac{\left(343.0\text{ m}/\text{s}\right)-\left(15.60\text{ m}/\text{s}\right)}{\left(343.0\text{ m}/\text{s}\right)+\left(4.850\text{ m}/\text{s}\right)}\right)\\ \\ & =0.9412f_0\end{array}$$

In Scenario C, the source and the detector are moving toward each other. Both movements shift the frequency higher, which calls for a positive sign in the numerator and a negative sign in the denominator.

$$\begin{array}{rl}{f}_{\text{C}}& =f_0\left(\frac{V+v_{\text{d}}}{V-v_{\text{s}}}\right)\\ \\ & =f_0\left(\frac{\left(343.0\text{ m}/\text{s}\right)+\left(15.60\text{ m}/\text{s}\right)}{\left(343.0\text{ m}/\text{s}\right)-\left(4.850\text{ m}/\text{s}\right)}\right)\\ \\ & =1.060f_0\end{array}$$

In Scenario D, the source is moving away from the detector, which necessitates a positive sign in the denominator to shift the frequency lower. However, the detector is moving toward the source, which necessitates a positive sign in the numerator to shift the frequency higher.

$$\begin{array}{rl}{f}_{\text{D}}& =f_0\left(\frac{V+v_{\text{d}}}{V+v_{\text{s}}}\right)\\ \\ & =f_0\left(\frac{\left(343.0\text{ m}/\text{s}\right)+\left(15.60\text{ m}/\text{s}\right)}{\left(343.0\text{ m}/\text{s}\right)+\left(4.850\text{ m}/\text{s}\right)}\right)\\ \\ & =1.031f_0\end{array}$$

So, the order of the detected frequencies is C > D > A > B.

If the emitted sound wave in Scenario C is reflected off the small bat's body and returns to the large bat's ears, the frequency is Doppler shifted again. The frequency received by the small bat, as found previously, is

$$\begin{array}{rl}{f}_{\text{C}}& =f_0\left(\frac{V+v_{\text{d}}}{V-v_{\text{s}}}\right)\\ \\ & =f_0\left(\frac{V+v_{\text{small bat}}}{V-v_{\text{large bat}}}\right)\end{array}$$

Now, treat the small bat as the source of a reflected sound wave of frequency ?C$f_{\text{C}}$ and the large bat as the detector. Because both bats are still moving toward each other, the signs used on the detector and source velocities remain unchanged. However, the detector and source roles have reversed, so the detector and source velocities are interchanged. The final frequency that the large bat detects is

?C2=?C(?+?d?−?s)=?C(?+?large bat?−?small bat)=?0(?+?small bat?−?large bat)(?+?large bat?−?small bat)=(50.80 kHz)((343.0 m/s)+(15.60 m/s)(343.0 m/s)−(4.850 m/s))((343.0 m/s)+(4.850 m/s)(343.0 m/s)−(15.60 m/s))=57.24 kHz