Finding frequency of moving sources with beats

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SUMMARY

The discussion centers on calculating the frequency of train whistles based on the observed beat frequency of 6.00 Hz while two trains approach a stationary observer at equal speeds of 11 m/s. Using the speed of sound at 331 m/s, the derived frequency of the whistles is calculated to be 90.2 Hz. However, participants noted that this frequency seems implausibly low for train whistles, suggesting that the beat frequency should be higher for realistic scenarios. The equations used include the beat frequency formula and Doppler effect equations for moving sources.

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  • Understanding of the Doppler effect and its equations
  • Familiarity with beat frequency concepts
  • Basic algebra for solving equations
  • Knowledge of sound speed in air (331 m/s)
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  • Learn about the Doppler effect in detail, focusing on moving sources and observers
  • Explore advanced sound wave properties and their applications in real-world scenarios
  • Study beat frequencies and their significance in acoustics
  • Investigate realistic sound frequencies used in train whistles and their design
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Students studying physics, particularly those focusing on wave mechanics and sound, as well as educators looking for practical examples of the Doppler effect and beat frequencies.

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Homework Statement



Jane waits on a railroad platform while two trains approach from the same direction at equal speeds of 11 m/s. Both trains are blowing their whistles (which have the same frequency), and one train is some distance behind the other. After the first train passes Jane but before the second train passes her, she hears beats of frequency 6.00 Hz. What is the frequency of the train whistles? (Use 331 m/s as the speed of sound.)

Homework Equations



1) fb = | f1 - f2 |

where fb is the frequency of the beats.

2) f1 = f * (v/ (v-vs) )

3) f2 = f * (v/ (v+vs))

where v is the speed of sound
and vs is the speed of the source

equation 2 is for the source moving towards Jane, and equation 3 is for the source moving away from Jane.

The Attempt at a Solution



fb = | f1 - f2 |

-- Sub in equation 2 and 3 --

fb = | f * ( v / (v-vs) ) - f * ( v / (v+vs) )

-- Solve for f --

fb = | f * (2v*vs) / (v^2 - vs^2) |

-- Term in absolute value always positive since v >> vs --

f = fb [ (2v*vs) / (v^2 - vs^2 ) ]^(-1)

f = 90.2Hz

This just doesn't seem like a good answer. There's no way any train would use such a low pitch whistle, I can't even hear it on my computer at max volume. Are there any mistakes in my work? Is there perhaps a different way to solve this?

P.S. I will work on making the equations more tidy in the future, perhaps learning how to use LATEX or something along those lines. Thanks for your patience.
 
Last edited:
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Your solution is correct, good job! The data were not realistic, the beat frequency should have been higher.
 
Thanks.
 

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