# Calculating Doppler Effect: Bat Approaching Wall at 10 m/s with 80 kHz Squeak

• maccha
In summary, the frequency of the sound the bat hears in its echo will be the same as the frequency it emitted, regardless of whether the bat is the moving source or observer. This is because the frequency of a reflected sound wave is determined by the properties of the reflecting surface and not by the motion of the source or observer.
maccha
A bat is approaching a wall. If it is traveling at 10 m/s and squeaks at 80 kHz, what frequency does it hear in its echo?

f=fo(1 +/- vo/v)
f=fo (1/(1 +/- vs/v))

So I got the right answer by initially making the wall the stationary "observer" and the bat the moving source, and finding the frequency the wall would "hear". Then I made the bat the moving observer and the wall the stationary source- for the frequency emitted by the wall, I used the frequency I found in the first step. I just want to confirm that this makes sense because I feel like I only got the answer by trial and error-so would the wall in fact reflect the sound at the frequency it "heard" from the bat and why?

Yes, this makes sense. The wall will reflect the sound at the frequency it heard from the bat because the frequency of a reflected sound wave is determined by the properties of the reflecting surface. Since the wall is not a source of sound, it will simply reflect whatever frequency is incident upon it back to the bat.

Your approach to solving this problem is correct. The Doppler effect describes the change in frequency of a wave due to the relative motion of the source and the observer. In this case, the bat is both the source and the observer, and the wall is the stationary object that reflects the sound wave. When the bat is moving towards the wall, the frequency of the sound wave it emits will appear higher to the wall, and when the bat is moving away from the wall, the frequency will appear lower. This is because the sound waves are being compressed or stretched as they travel towards or away from the wall. The wall then reflects the sound wave back to the bat, and the bat perceives this reflected sound wave as an echo. Since the wall is stationary, the frequency of the reflected sound wave will be the same as the frequency that the wall "heard" from the bat. This is why you were able to use the frequency you found in the first step to calculate the frequency of the echo.

## 1. What is the Doppler Effect?

The Doppler Effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.

## 2. How does the Doppler Effect work?

The Doppler Effect works by compressing the waves in front of a moving object and stretching them behind the object, causing a change in frequency that is perceived by an observer.

## 3. What causes the Doppler Effect?

The Doppler Effect is caused by the relative motion between a wave source and an observer. This can be observed in various phenomena such as the sound of a siren from a moving ambulance or the color shift of stars in the sky.

## 4. How is the Doppler Effect used in science?

The Doppler Effect is used in various fields of science, such as astronomy, meteorology, and physics. It is used to determine the speed and direction of objects, study the behavior of waves, and observe the movement of celestial bodies.

## 5. What are some real-world applications of the Doppler Effect?

The Doppler Effect has many practical applications, including in radar and sonar technology, medical imaging, and weather forecasting. It is also used in speed measurement devices such as radar guns and speed cameras.

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