BIO Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey’s speed. (The “horseshoe” that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed v_bat emits sound of frequency f_bat; the sound it hears reflected from an insect flying toward it has a higher frequency f_refl.(a) Show that the speed of the insect is

${\mathrm{v}}_{\mathrm{in}\mathrm{sect}}=\frac{\mathrm{v}\left({\mathrm{f}}_{\mathrm{refl}}\left(\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}\right)-{\mathrm{f}}_{\mathrm{bat}}\left(\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}\right)\right)}{{\mathrm{f}}_{\mathrm{refl}}\left(\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}\right)+{\mathrm{f}}_{\mathrm{bat}}\left(\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}\right)}$$cv$

where v is the speed of sound. (b) Iff_bat = 80.7 kHz, f_refl=83.5 kHz, andv_bat = 3.9 m>s, calculate the speed of the insect.

The frequency emitted from the bat ${\mathrm{f}}_{\mathrm{bat}}$, and the frequency detected by the bat after being reflected from the insect is f_refl ,In the first case, the bat will be the source of sound that moves towards the listener "the insect"

The formula that describes the frequency detected by the insect, is ${\mathrm{f}}_1={\mathrm{f}}_{\mathrm{s}}\left(\frac{\mathrm{v}\pm {\mathrm{v}}_{\mathrm{L}}}{\mathrm{v}\mp {\mathrm{v}}_{\mathrm{s}}}\right)$ where, v, the speed of sound in the air, VL = v_{in sect} the speed of the insect, vs = ${\mathbf{v}}_{\mathbf{bat}}$, the speed of the source "the bat", the frequency detected by the insect, fs = f_bat , the frequency emitted from the bat.

Therefore, equation can be written as role="math" localid="1668143929396" ${\mathrm{f}}_{\mathrm{L}}={\mathrm{f}}_{\mathrm{bat}}\left(\frac{\mathrm{v}\pm {\mathrm{v}}_{\mathrm{insect}}}{\mathrm{v}\mp {\mathrm{v}}_{\mathrm{bat}}}\right)$(1)

Now, after the sound reflects from the insect, it will travel towards the bat and the frequency detected by the bat in so the equation (1) can be written as

${\mathrm{f}}_{\mathrm{refl}}={\mathrm{f}}_{\mathrm{L}}\left(\frac{\mathrm{v}+{{\mathrm{v}}_{\mathrm{bat}}}_{}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{insect}}}\right)$(2)

Substitute ${\mathrm{f}}_{\mathrm{L}}$in equation 2 from 1

${\mathrm{f}}_{\mathrm{refl}}={\mathrm{f}}_{\mathrm{bat}}\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{insect}}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}\right)\left(\frac{\mathrm{v}+{{\mathrm{v}}_{\mathrm{bat}}}_{}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{insect}}}\right)$

$$\begin{gathered} \frac{{\mathrm{f}}_{\mathrm{refl}}}{{\mathrm{f}}_{\mathrm{bat}}}=\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{insect}}}{\mathrm{v}-{\mathrm{v}}_{{}_{\mathrm{insect}}}}\right)\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}\right) \\ \frac{{\mathrm{f}}_{\mathrm{refl}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)=\left(\frac{\mathrm{v}+{\mathrm{v}}_{{\mathrm{insect}}_{}}}{\mathrm{v}-{\mathrm{v}}_{{\mathrm{insect}}_{}}}\right) \end{gathered}$$

Let$\frac{{\mathrm{f}}_{\mathrm{refl}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)$ be x hence.

$$\begin{gathered} \mathrm{x}=\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{insect}}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{insect}}}\right) \\ \mathrm{xv}-{\mathrm{v}}_{\mathrm{insect}}\mathrm{x}=\mathrm{v}+{\mathrm{v}}_{\mathrm{insect}} \\ {\mathrm{v}}_{\mathrm{insect}}+{\mathrm{v}}_{\mathrm{insect}}\mathrm{x}=\mathrm{vx}-\mathrm{v} \\ {\mathrm{v}}_{\mathrm{insect}}(1+\mathrm{x})=\mathrm{v}(\mathrm{x}-1) \\ {\mathrm{v}}_{\mathrm{insect}}=\frac{\mathrm{v}(\mathrm{x}-1)}{(1+\mathrm{x})} \end{gathered}$$

Now let’s replace x with$\frac{{\mathrm{f}}_{\mathrm{refl}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)$

${\mathrm{v}}_{\mathrm{insect}}=\frac{\mathrm{v}\left(\frac{{\mathrm{f}}_{\mathrm{reff}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)-1\right)}{\frac{{\mathrm{f}}_{\mathrm{ref}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)+1}$

Note that $1=\frac{{\mathrm{f}}_{\mathrm{bat}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)$Hence,

${\mathrm{v}}_{\mathrm{insect}}=\frac{\mathrm{v}\left(\frac{{\mathrm{f}}_{\mathrm{ret}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)-\frac{{\mathrm{f}}_{\mathrm{bat}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)\right)}{\frac{{\mathrm{f}}_{\mathrm{ref}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)+\frac{{\mathrm{f}}_{\mathrm{bat}}}{{\mathrm{f}}_{\mathrm{bat}}}\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}}\right)}$

${\mathrm{v}}_{\mathrm{insect}}=\frac{\mathrm{v}\left({\mathrm{f}}_{\mathrm{ref}}\left(\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}\right)-{\mathrm{f}}_{\mathrm{bat}}\left(\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}\right)\right)}{{\mathrm{f}}_{\mathrm{refl}}\left(\mathrm{v}-{\mathrm{v}}_{\mathrm{bat}}\right)+{\mathrm{f}}_{\mathrm{bat}}\left(\mathrm{v}+{\mathrm{v}}_{\mathrm{bat}}\right)}$(1)

$$\begin{gathered} {\mathrm{v}}_{\mathrm{in}\mathrm{sect}}=344\times \frac{\left[\left(83.5\times {10}^3\right)\left(344\right)-\left(3.9\right)\right]-\left[\left(80.7\times {10}^3\right)\left(344\right)+\left(3.9\right)\right]}{\left[\left(83.5\times {10}^3\right)\left(344\right)-\left(3.9\right)\right]+\left[\left(80.7\times {10}^3\right)\left(344\right)+\left(3.9\right)\right]} \\ =1.97\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, 1.97m/s is the speed of the insect.