Discover the Frequency of an Echo Using the Doppler Effect - Solved!

In summary, the conversation discusses the Doppler Effect and how it applies to a bat emitting a sound pulse at a certain frequency and reflecting off a wall. The correct equations and steps for finding the frequency of the echo received by the bat are provided, and the concept of sound waves being inverted when they hit a wall is mentioned. The correct answer is found through a series of equations and calculations.
  • #1
Dreams2Knight
8
0
[SOLVED] Doppler Effect?

Homework Statement



A bat flying at 4.0 m/s emits a chirp at 20 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat?

Homework Equations



Not sure. Do I use the f(1)=f*[(v+/-observer velocity)/(v+/-source velocity)]

The Attempt at a Solution



Not sure where to start. I seem to remember something about sound waves being returned inverted if they hit a wall. Does this factor in?
 
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  • #2
Hi Dreams2Knight,

As a first step, if the bat emits at a frequency of 20 kHz, what frequency does the wall receive?

After you have that, the reflection is handled by treating the wall as a source. What do you get?
 
  • #3
alphysicist said:
Hi Dreams2Knight,

As a first step, if the bat emits at a frequency of 20 kHz, what frequency does the wall receive?

After you have that, the reflection is handled by treating the wall as a source. What do you get?

Hello,

Thank you for replying.

Source: Bat
Observer: Wall
For the first equation I have f(1)=20[345/(345-4)] which is 20.23460411 kHz


Source: Wall
Observer: Bat
For the second equation I have f(1)=20[(345+4)/345] which is 20.23188406 kHz

Are my equations correct? If so, does the second equation give me my answer, or do I have to do something else?
 
  • #4
The bat emits sound at a frequency at 20 kHz, but the wall does not. After reflection from the wall, what frequency does the sound have? That number is the source frequency f on the right hand side. What do you get?
 
  • #5
alphysicist said:
The bat emits sound at a frequency at 20 kHz, but the wall does not. After reflection from the wall, what frequency does the sound have? That number is the source frequency f on the right hand side. What do you get?

Ok, so if I understand correctly, I need to set my equation up like this:

20.234=f(2)*[(345+4)/345]. This gives me 20.00209.

I confused myself at this point because when I put the numbers into my calculator, I accidently subtracted 4 from 345 instead of adding. Doing that gave me the right answer of 20.46 kHz. I was confused because I thought if the source was still, you used V+V(observor) on top, yet subtracting gave me the right answer.

Then I remembered my teacher mentioning sound waves being inverted when they hit a wall. So, I took the reciprocal of (345+4)/345 and divided 20.234 by that and also got the right answer.

So, did I have my equation set up correctly as 20.234=f(2)*[(345+4)/345]? Was taking the reciprocal of (345+4)/345 also the correct thing to do?
 
  • #6
No, not quite. It's giving close to the right answer, but I think the way it should be set up is:

step 1: frequency received by wall:
[tex]
f = 20 \frac{345}{345 -4} = 20.2346
[/tex]

step 2: frequency received by bat:


[tex]
f = (20.2346) \frac{345 + 4}{345} = 20.469
[/tex]


Your other ways get close to the answer becuase, for example 1+x is close to 1/(1-x) when x is small compared to one.
 
  • #7
alphysicist said:
No, not quite. It's giving close to the right answer, but I think the way it should be set up is:

step 1: frequency received by wall:
[tex]
f = 20 \frac{345}{345 -4} = 20.2346
[/tex]

step 2: frequency received by bat:


[tex]
f = (20.2346) \frac{345 + 4}{345} = 20.469
[/tex]


Your other ways get close to the answer becuase, for example 1+x is close to 1/(1-x) when x is small compared to one.


Ah, I see my mistake now. Thanks so much for your help. I really appreciate it.
 

1. What is the Doppler Effect?

The Doppler Effect is a phenomenon in which the perceived frequency of a wave (such as sound or light) changes when the source of the wave is moving relative to the observer.

2. How does the Doppler Effect work?

The Doppler Effect works by compressing or stretching the wavelength of a wave as the source and observer move closer or farther apart. This results in a change in the perceived frequency of the wave.

3. What causes the Doppler Effect?

The Doppler Effect is caused by the relative motion between the source of a wave and the observer. For example, as a car with a siren moves towards you, the sound waves are compressed, resulting in a higher perceived frequency.

4. What is the difference between the Doppler Effect and the Doppler shift?

The Doppler Effect and Doppler shift are often used interchangeably, but technically the Doppler shift refers to the change in frequency of a wave due to the Doppler Effect, while the Doppler Effect encompasses the broader concept of the change in perceived frequency due to relative motion.

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

The Doppler Effect has many practical applications, such as in weather radar to detect the movement of precipitation, in medical imaging techniques like ultrasound to measure blood flow, and in astronomy to study the motion of celestial objects.

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