# Horsehoe Bats, Insects, Speed and the Doppler Effect

• TFM
In summary, horseshoe bats use a unique method of hunting by emitting sounds from their nostrils and using the "horseshoe" shape to focus the sound in a beam. They listen to the frequency of the sound reflected from their prey to determine the prey's speed. By using the Doppler effect equations and given values of frequencies and speeds, the speed of the insect can be calculated. In this conversation, the speed of the insect is found to be 0.63 m/s.
TFM
[SOLVED] Horsehoe Bats, Insects, Speed... and the Doppler Effect

## Homework Statement

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, 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 $$v_b_a_t$$ emits sound of frequency $$f_b_a_t$$ ; the sound it hears reflected from an insect flying toward it has a higher frequency $$f_r_e_f_c$$ .

If the bat emits a sound at a frequency of 80.8 kHz and hears it reflected at a frequency of 83.1 kHz while traveling at a speed of 4.2 m/s, calculate the speed of the insect.
Use 344 m/s for the speed of sound in air. Express your answer using two significant figures.

## Homework Equations

Doppler Shift Equations:

$$f_I = (\frac{v + v_I}{v + v_b_a_t})f_b_a_t$$

$$f_r_e_f = (\frac{v + v_b_a_t}{v + v_I})f_I$$

## The Attempt at a Solution

the answer I got by using these two equations was 10.45, got via:

firstly, I inserted the values into the top equation, to get $$f_I$$ in terms of $$v_I$$ :

$$f_I = (\frac{v + v_I}{v + v_b_a_t})f_b_a_t$$

$$f_I = (\frac{344 + v_I}{344 + (-4.2)})80.8$$

$$f_I = (\frac{344 + v_I}{339.8})80.8$$

$$f_I = (\frac{344}{339.8} + \frac{v_I}{339.8})80.8$$

$$f_I = (81.7987 + \frac{80.8v_I}{339.8})$$

$$f_I = (81.7987 + 0.2672v_I)$$

I then rearranged the second equation:

$$f_r_e_f = (\frac{v + v_b_a_t}{v + v_I})f_I$$

$$83.1 = (\frac{344 + 4.2}{344 + v_I})f_I$$

$$83.1 = (\frac{348.2}{344 + v_I})f_I$$

$$83.1*(344 + v_I) = 348.2f_I$$

$$28586.4 + 83.1v_I) = 348.2f_I$$

Now inserting $$f_I$$:

$$28586.4 + 83.1v_I) = 348.2((81.7987 + 0.2672v_I))$$

$$28586.4 + 83.1v_I) = ((28482.31 + 93.045v_I))$$

$$104.0909 = 9.94v_I$$

Giving the Insects speed as 10.46

Any ideas where I have gone wrong?

TFM

Last edited:
Any ideas where I may have gone wrong?

TFM

It would help a lot if you would tell what each parameter means. I assume that "v" is the speed of sound. You appear to have two equations:
$$f_I= \frac{v+ v_i}{v- v_{bat}}f_{bat}$$
and
$$f_{ref}= \frac{v+ v_{bat}}{v+ v_I}f_I$$
If you just go ahead replace $f_I$ in the second equation by the first you get
$$f_{ref}=\frac{v+ v_{bat}}{v- v{bat}}f_{bat}$$
which has NO dependence on $v_I$!

Perhaps both $v_{bat}$ and $v_I$ should be positive in one equation and negative in the other.

Sorry, that is rather sill of me.

$$v$$ speed of sound
$$v_I$$ Speed of Insect
$$v_b_a_t$$ speed of bat
$$f_b_a_t$$ frequency emitted by bat
$$f_I$$ frequency heard by Insect
$$f_r_e_f$$ frequency reflect back and heard by bat

as fir the speed, in the book, it has a car and a wall, and the speed of the car for the emitted wave is negative, the speed when the wave is headed back is positive. the speed of the wall is zero, so I just took the insect speed parts from the relevant equations for moving source/moving listener.

Does this help?

TFM

I never realized that the latex had taken over the actual question. The question is:

If the bat emits a sound at a frequency of 80.8 kHz and hears it reflected at a frequency of 83.1 kHz while traveling at a speed of 4.2 m/s, calculate the speed of the insect.
Use 344 m/s for the speed of sound in air. Express your answer using two significant figures.

TFM

Using:

$$f_I = (\frac{v+v_I}{v-v_b_a_t})f_b_a_t$$

and

(This should be f reflected = ...) => $$f_r_e_f = (\frac{v+v_b_a_t}{v-v_I})f_I$$

and rearranging similarly to what I did in the first post, I now get a small answer of:
0.6274 (?)

Does this look right?

TFM

It is the right answer. The book gives another equation, whch returns the same value, and it is correct!

Thanks HallsofIvy,

TFM

Last edited:

## 1. What are horseshoe bats?

Horseshoe bats, also known as Rhinolophidae, are a family of bats that are characterized by their horseshoe-shaped nose and large ears. They are found in Asia, Europe, Africa, and Australia and are known for their echolocation abilities.

## 2. How do horseshoe bats use echolocation to hunt insects?

Horseshoe bats emit high-frequency sounds and then listen for the echoes that bounce off of objects, including insects. They use this information to locate and capture their prey with precision.

## 3. What role do insects play in the diet of horseshoe bats?

Insects make up the majority of the diet of horseshoe bats. These bats are especially fond of moths, beetles, and other flying insects that they can catch using their echolocation abilities.

## 4. How fast can horseshoe bats fly?

Horseshoe bats are relatively slow flyers, with an average speed of 10-20 miles per hour. However, they are incredibly agile and can make quick turns and maneuvers while hunting for insects.

## 5. What is the Doppler Effect and how does it relate to horseshoe bats and their echolocation abilities?

The Doppler Effect is a phenomenon where the frequency of sound or light waves appears to change when the source of the waves is moving relative to the observer. This is important for horseshoe bats because they use this effect to determine the distance and speed of their insect prey based on the changes in frequency of their echolocation calls.

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