Calculating Wavelengths of Soundwaves: v = λ*f

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The discussion centers on calculating soundwave wavelengths using the formula v = λ*f, with specific examples of 10 Hz and 20,000 Hz soundwaves yielding wavelengths of 34.3 m and 1.715 cm, respectively. The question arises whether a wall must match these wavelengths to effectively block the sound, assuming the wall is infinitely tall and wide. Participants note that the speed of sound varies in different materials, which affects the wavelength and necessary wall thickness for soundproofing. It is suggested that a wall's effectiveness depends on both its thickness and the material's sound speed properties. Understanding these factors is crucial for effective sound insulation.
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 = \lambda*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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