Doppler Effect?

  • #1
[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?
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
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
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
alphysicist
Homework Helper
2,238
1
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
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
alphysicist
Homework Helper
2,238
1
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
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.
 

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