# What is the speed of the bat gaining on its prey in m/s?

• bastige
In summary, a bat is chasing a flying insect by emitting a 47 kHz chirp and receiving back an echo at 47.73kHz. By using the equation for observed frequency, the velocity of the insect was determined to be 5.3363 m/s. The bat's velocity was found to be 1.96 m/s faster than the insect.
bastige

## Homework Statement

Assume: Take the speed of sound in air to be 341 m/s.
A bat, moving at 6.3 m/s, is chasing a flying insect.
The bat emits a 47 kHz chirp and receives back an echo at 47.73kHz.
At what speed is the bat gaining on its prey? Answer in units of m/s.

## Homework Equations

f=initial frequency
fo=observed frequency
x= velocity of insect
f(observed)=f(init){(v+v[observer])/(v-v[source])}

## The Attempt at a Solution

I determined the bat is the source, but because the chirp bounds off the insect and returns to the bat, I considered the insect to be the "source" in my equation:
Using the above equation:

f/fo=(343+Vbat)/(343+x)
46000/47730=(341+6.3)/(341+x)
.9637=347.3/(341+x)
341+x=346.3363
x=5.3363
then i take the velocity of the bat:
5.9-3.939=1.96

At first the frequency received by the prey is:

$$\nu' = \nu (\frac{v-v_{L}}{v-v_{S}})$$

Now, as you said correctly, the wave is reflected back and the the source and listener interchange to give us this i.e. the bat becomes the listener and the prey the source of the reflected sound wave:

$$\nu'' = \nu (\frac{v+v_{S}}{v+v_{L}})$$

putting the two equation together we get:

$$\nu'' = \nu (\frac{v-v_{L}}{v-v_{S}}). (\frac{v+v_{S}}{v+v_{L}})$$

Substituting the values, find the velocity of the prey and then find the relative velocity of the prey w.r.t the bat.

m/s

Based on the given information, the speed of the bat gaining on its prey is approximately 1.96 m/s. This is calculated by taking the difference between the bat's initial velocity of 6.3 m/s and the calculated velocity of the insect, which is 5.3363 m/s. This difference of 1.96 m/s represents the speed at which the bat is gaining on its prey. However, it should be noted that this calculation assumes the insect is moving in a straight line and does not take into account any changes in direction or acceleration of either the bat or the insect.

## 1. What is the Doppler effect bat?

The Doppler effect bat is a phenomenon that occurs when a bat emits sound waves at a certain frequency while flying. The frequency of the sound waves changes as the bat approaches or moves away from an observer, resulting in a shift in the perceived pitch of the sound.

## 2. How does the Doppler effect bat work?

The Doppler effect bat works by emitting sound waves at a specific frequency. As the bat moves towards an observer, the waves are compressed, resulting in a higher pitch. On the other hand, as the bat moves away, the waves are stretched, resulting in a lower pitch.

## 3. What causes the Doppler effect bat?

The Doppler effect bat is caused by the relative motion between the bat and the observer. As the bat and the observer move towards or away from each other, the frequency of the sound waves changes, resulting in a perceived change in pitch.

## 4. How is the Doppler effect bat used in science?

The Doppler effect bat is used in various scientific fields, such as astronomy, meteorology, and medical imaging. In astronomy, it is used to study the movement of celestial bodies. In meteorology, it is used to track the movement of storms. In medical imaging, it is used to measure blood flow and diagnose cardiovascular diseases.

## 5. Is the Doppler effect bat only applicable to bats?

No, the Doppler effect can be observed in various situations, such as the sound of a passing vehicle, the changing pitch of a siren, or the redshift of distant galaxies. It is a fundamental principle in physics that applies to any moving object emitting waves.

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