Calculating Wavelengths of Soundwaves: v = λ*f

  • Context: High School 
  • Thread starter Thread starter Denver Dang
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SUMMARY

The discussion centers on the calculation of soundwave wavelengths using the formula v = λ*f, where v represents the speed of sound, λ is the wavelength, and f is the frequency. A wavelength of 34.3 meters is calculated for a 10 Hz soundwave, while a wavelength of 1.715 cm is calculated for a 20,000 Hz soundwave. The participants explore the implications of these wavelengths on wall thickness required to block sound, considering factors such as the speed of sound in different materials. It is established that the speed of sound varies with the medium, affecting the necessary wall thickness for soundproofing.

PREREQUISITES
  • Understanding of the formula v = λ*f for soundwaves
  • Knowledge of sound frequencies and their corresponding wavelengths
  • Familiarity with the speed of sound in various materials
  • Basic principles of soundproofing and acoustic barriers
NEXT STEPS
  • Research the speed of sound in different materials and its impact on soundproofing
  • Learn about acoustic properties of materials used in sound barriers
  • Explore advanced soundwave behavior in various environments
  • Investigate practical applications of soundwave calculations in construction and engineering
USEFUL FOR

Acoustic engineers, soundproofing specialists, and anyone involved in audio engineering or construction requiring soundwave management will benefit from this discussion.

Denver Dang
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A little question I hope you can help me with.
If you got the low frequency bass notes and some high notes as well.
Then you have the formula:
v = [tex]\lambda[/tex]*f

A quick calculation gives you a wavelength of 34,3 m for af 10 Hz soundwave and 1.715 cm for a 20.000 Hz soundwave.

Does that mean that I would need a 34,4 m wall to stop the waves of a 10 Hz soundwave, if we imagine that wall being infinitetly tall and wide, and then only 1.716 cm wall to stop the waves from a 20.000 Hz soundwave ?

Or is there some other factors that play along ?


Regards
 
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Presumably you worked out the wavelength in air.

Have you thought that the speed of sound might change in the wall?
 
I've thought of that yes. But do you have to calculate the wall-thickness depending on the material that the wall is made of ?
So if the wall is made of material that doubles the speed of sound, compared to air, the wavelength doubles and you need a thicker wall ?
Or is it a mixture of both ? :S
 

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