Interference of Waves, Sound Diminution

  • Thread starter alingy1
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  • #1
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A manufacturing firm has hired your company, Acoustical Consulting, to help with a problem. Their employees are complaining about the annoying hum from a piece of machinery. Using a frequency meter, you quickly determine that the machine emits a rather loud sound at 1400 Hz. After investigating, you tell the owner that you cannot solve the problem entirely, but you can at least improve the situation by eliminating reflections of this sound from the walls. You propose to do this by installing mesh screens in front of the walls. A portion of the sound will reflect from the mesh; the rest will pass through the mesh and reflect from the wall.How far should the mesh be placed in front of the wall for this scheme to work?

I have the answer right in front of me, along with the steps. But the solution manual just mentions that Δø0 (the phase difference) is simply 0. Can you check my reasoning of why it is 0? The sound wave gets into the mesh (-pi reversal for the reflected wave) and then reflects on the wall (-pi reversal again for the reflected wave on the wall). So, when sound waves go from denser to less dense areas, the waves don't get inverted. They do get inverted when they from less dense to dense.

So, because the waves that are travelling away from the wall are inverted, they have 0 phase difference? Right?

If anyone wants to solve the problem just for kicks, the answer is 7.15 cm.
 

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  • #2
NascentOxygen
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That's the line of thinking I would have used.

I'd like to see (i.e., hear) the scheme in action. Where can I glimpse a mesh that reflects a significant percentage of incident sound?
 
  • #3
PhysicsLearning
use the speed of sound to solve for wavelength. Then because technically the wall and the air is a close-open tube, you can say 4L = wavelength to solve for L.
 

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