Physics Challenge Question 24: Solutions
Part 1
The speed of sound is 340 m/s. Since it takes the sound 0.1 seconds round
trip, it must have taken the sound 0.05 seconds for it to get from the bat to
the bug.
x
v
t
x  v  t  340 ms  0.05 s  17 m
The bug was 17 meters away.
Part 2
If its ears are only 2 cm apart, that means it must be able to distinguish
sounds with a time resolution t.
x
v
t
x 0.02 m
t 

 5.88  10 5 s  0.0000588 s  58.8 s
m
v
340 s
So the bat will hear a sound pulse about 0.00006 seconds apart from one ear
to the next if the sound is coming from its side!
Part 3
The bat is not moving.
When the sound reflects off the bug, it gets Doppler shifted because of its
movement. Since the echo is of a higher frequency (higher pitch), the bug
must be moving towards the bat.
Part 4
Since the bug is moving towards the bat, the bug will hear a higher
frequency than the original sound, since the sound is Doppler shifted. This
situation corresponds to a moving observer. We can find the frequency the
bug hears (f’) as
 v  vbug 

f   f  
 v 
Here, f is the original frequency (the 40 kHz), v is the speed of sound, and
vbug is the speed of the bug. However, f’ is not the same frequency the bat
hears!
The sound is now reflected back by the bug. It now acts as a moving source,
reflecting the sound with the frequency f’. The final frequency heard by
the bat (let’s call it f’’) is then
 v 
 v  vbug 
 v  vbug   v 
  f 

  f 


f   f   

 v  vv 
vv 
vv 
bug 
bug 
bug 

 


We can now solve for vbug: (The following is just a bunch of algebra.)
 v  vbug 

f   f  
vv 
bug 

f 
 v  vbug   v  vbug
f
f 
f 
v 
 vbug  v  vbug
f
f
f 
f 
 v  v  vbug 
 vbug
f
f
 f  

f  

  vbug
 1  v  1 
f 
 f


 f  

 1
f

 v
vbug 
 f  

 1
 f

 40.4 kHz 

 1
40.0 kHz 

vbug 
 340 ms  1.69 ms
 40.4 kHz 

 1
 40.0 kHz 
A Doppler shift of 400 Hz corresponds to a speed of just 1.69 m/s.