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Doppler Effect Bat Problem

  • Thread starter lzh
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  • #1
lzh
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Assume: Take the speed of sound in air to be 343 m/s.
A bat, moving at 5.9 m/s, is chasing a flying insect.
The bat emits a 46 kHz chirp and receives back an echo at 46.26 kHz.
At what speed is the bat gaining on its prey? Answer in units of m/s.
I tried this:
f=initial frequency
fo=observed frequency
x= velocity of insect
fo/f=(343+Vbat)/(343+x)
but it doesn't seem to work:
46260/46000=(343+5.9)/(343+x)
1.0056=348.9/(343+x)
343+x=346.94
x=3.939
then i take the velocity of the bat:
5.9-3.939=1.96
this answer isn't right though, but I dont see why.
 
Last edited:

Answers and Replies

  • #2
960
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Is that the right equation to be using?

I'm getting a little confused myself, but since its an echo, I'm thinking there are two doppler shifts, one outgoing and one on the return leg, simply due do the bats speed, and then a third shift which is due to the insects motion. How did you arrive at the eqns you used?
 
  • #3
lzh
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well technically 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:
f(observed)=f(init){(v+v[observer])/(v-v[source])}
v is just the speed of the sound, which is just 343. And I just plugged in all my values and solved for v[source].
 
  • #4
960
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Good idea, and since the bat hears a blue shifted frequency, seems like you chose the right eqn, i get Vs=3.64, so the bat is making ground but at 5.9-3.64 m/s
 
  • #5
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Prettysure denver is right and there are 2 doppler shifts (one raising the frequency and one lowering it)

Try to draw it out on paper. Bat moving to the right shifts the frequency up. however the insect moving away shifts the frequency down.
 
  • #6
Doc Al
Mentor
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There are two doppler shifts, but in both shifts the source and observer are approaching each other (relatively speaking, since the bat is gaining ground on the insect).
 
  • #7
lzh
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how would you be able to consider two doppler shifts with one equation? We have always done single doppler shifts, and so this is pretty confusing. and denverdoc, did u get that answer with just my eqn? Or did you do some more to consider the two doppler shifts? Because i checked that answer and its not right.
 
  • #8
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how would you be able to consider two doppler shifts with one equation? We have always done single doppler shifts, and so this is pretty confusing. and denverdoc, did u get that answer with just my eqn? Or did you do some more to consider the two doppler shifts? Because i checked that answer and its not right.
No I thought you had a great approach to the problem interchanging Vs and Vo from the physical situation, and just used the eqn you posted. Not sure why the answers disagree, I did'nt double check, however.
 
  • #9
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using 2 doppler shifts i got ans as 5.9 - 4.93. m i right?
 
  • #10
lzh
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how did u get to 4.93? can u show me ur eqn?
 
  • #11
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f/fo=(343-x)(343+5.9)/(343+x)(343-5.9)
first consider bat as source & insect as observer
then exchang roles
 
  • #12
lzh
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thats quite interesting, and makes sense. But i really can't "afford" to check the answers again(since every time i enter an answer it deducts points from my webhomework), do you guys think the way this eqn is structured is right?
 
  • #13
960
0
this was the most analogous link I could find: http://www.rwc.uc.edu/koehler/biophys/9c2.html [Broken]

From a quick glance i believe that the above looks good. Damn glad there are not relativistic issues involved as well!!:bugeye:
 
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  • #14
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r u conviced with that "Izh" ?
 
  • #15
lzh
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i think so and i will try it
but is it f/fo?
because i had fo/f
 
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  • #16
960
0
let us know,lzh, if that was right. Good problem.
 
  • #17
lzh
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i got -4.177 as x, perhaps one of the signs is flipped in the eqn?
 
  • #18
lzh
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it didn't seem to work...
 
  • #19
960
0
I got 4.93 as well.
 
  • #20
lzh
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that didn't work...hmm i wonder what is wrong
 
  • #21
960
0
that didn't work...hmm i wonder what is wrong
well just to make sure 4.93 was the bugs speed, the closing velocity was 5.9-4.93, is that the answer that was wrong?
 
  • #22
lzh
111
0
yeah, it was
 
  • #23
960
0
if so maybe it's time to rethink things and see if we get a concordance. Lets just consider the situation where the bat is flying toward a brickwall; what does the bat hear? I would argue that since he is moving while producing his ping, the ping he hears as it rushes in front of him is still 14K, now on its return, the echo relative to the bat as perceived by the bat will be shifted by v/(v-vb) . I would also argue that if the brickwall were moving away at a velocity equal to that of the bat he would never get an echo. Now if the source were moving away at some intermediate velocity, this would attenuate the shift by a factor v/(v+vins-vb). If its moving toward the bat, then it would be v/(v-vins-vb) We can tell which of the two situations it is by comparing the brick wall shift with that actually recorded--
343/(343-5.9). The shift is 1.0175, much greater than the 1.0056 noted.

Hence the situation is where the bug is flying away from the bat. so solving for vins in the first case gives vins=343/1.0056-343+Vb=3.97 and so the bat closes in at 5.9-3.97.

this was the first answer I got, but discarded it as it seemed too simple.
I'm gonna kill myself if this was right after all.
 
  • #24
lzh
111
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here is a list of all the answers i tried:
date time answer
----------- -------- ---------------
Mar 24 2007 12:08 pm 1.9387
Mar 24 2007 12:16 pm 1.96096
Mar 24 2007 12:30 pm -1.96096
Mar 25 2007 11:05 am 2.26

the first one was close to ur answer ans and was wrong sadly, and it accepts anything near 1% of the actual answer.
 
  • #25
Doc Al
Mentor
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The formula that Sourabh N provided in post #11 looks right to me. It's just two Doppler shifts compounded. I haven't cranked through the numbers, but did you? (When denverdoc said he got 4.93 m/s as the bug's speed, was that for this formula? If so, that's one answer you didn't try: 5.9-4.93 = 0.97 m/s)
 

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