Waves and Sounds - speed of a bat

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SUMMARY

The discussion focuses on calculating the speed at which a bat is gaining on its prey using the Doppler effect. A bat emits a chirp at 52 kHz and receives an echo at 52.75 kHz while moving at 4.5 m/s. The equation used is f1 = f ((velocity +/- velocity observer)/(velocity +/- velocity source)), with the speed of sound in air taken as 342 m/s. The user initially calculated the observer's velocity as approximately 0.367788 m/s but encountered issues with the final speed gain calculation.

PREREQUISITES
  • Understanding of the Doppler effect in sound waves
  • Familiarity with basic physics equations related to wave frequency
  • Knowledge of sound speed in air (342 m/s)
  • Ability to manipulate equations to solve for unknown variables
NEXT STEPS
  • Review the Doppler effect and its applications in sound wave scenarios
  • Practice solving problems involving frequency shifts due to relative motion
  • Explore the implications of sound speed variations in different mediums
  • Learn about the effects of temperature and pressure on sound speed in air
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Students studying physics, particularly those focusing on wave mechanics, as well as educators looking for practical examples of the Doppler effect in real-world scenarios.

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Homework Statement



A bat moving at 4.5 m/s is chasing a flying insect insect. The bat emits a 52 kHz chirp and receives back an echo at 52.75 kHz. At what speed is the bat gaining on its prey? Take the speed of sound in air to be 342 m/s.


Homework Equations




f1 = f ((velocity +/- velocity observer)/(velocity +/- velocity source))

The Attempt at a Solution



To start off, I converted 52 kHz to 52,000 Hz and 52.75 kHz

Then my equation looked like:

52750 = 52000 ((342 - vo)/(342-4.5)

I found vo to be .367788 and subtracted that from 4.5 to find the speed gained, but that doesn't seem to work. Can anyone help?
 
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