University:
Boston University
Course:
CAS PY 211 | General Physics I
Academic year:

2023
Views:

177

Pages:

2
Author:

Tara Guerrero

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 µs 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 ′′ ⋅ vbug ⋅ v − v = 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   ⋅ 340 ms = 1.69 ms vbug =  40.4 kHz   + 1  40.0 kHz  A Doppler shift of 400 Hz corresponds to a speed of just 1.69 m/s.