Doppler effect bat

  • Thread starter bastige
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
81
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At first the frequency received by the prey is:

[tex] \nu' = \nu (\frac{v-v_{L}}{v-v_{S}}) [/tex]

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:

[tex] \nu'' = \nu (\frac{v+v_{S}}{v+v_{L}}) [/tex]

putting the two equation together we get:

[tex] \nu'' = \nu (\frac{v-v_{L}}{v-v_{S}}). (\frac{v+v_{S}}{v+v_{L}}) [/tex]

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

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