Post by: scientizscht on 12/06/2018 13:36:01
When a sound wave travels in the atmosphere, there are areas of high and low pressure formed.
What is the difference in that pressure for eg 100dB? Does this difference attenuates as we stand farther from the sound source?
Also, is there any movement of air mass due to the sound wave? Even if that's a little?
Thanks
Post by: Colin2B on 12/06/2018 17:47:13
So to find the actual pressure you would need to use:
Where p0=2x10-5 Pa
Lp=100
P is rms pressure and it is pressure swing either side of ambient - there are compressions & rarefactions so pressure difference between these two is 2p.
In a free field - no reflections - p decreases at 6dB for every doubling of distance so p varies as 1/r not 1/r2.
The air molecules vibrate rather than move so there is no net movement of air.
Post by: scientizscht on 12/06/2018 18:27:59
The sound pressure of 100dB is not an absolute measure against local air pressure, but a measure against a reference auditory threshold of 2x10-5 Pa.
So to find the actual pressure you would need to use:
Where p0=2x10-5 Pa
Lp=100
P is rms pressure and it is pressure swing either side of ambient - there are compressions & rarefactions so pressure difference between these two is 2p.
In a free field - no reflections - p decreases at 6dB for every doubling of distance so p varies as 1/r not 1/r2.
The air molecules vibrate rather than move so there is no net movement of air.
Thanks but how much the max and min pressure would be in a 100dB wave in atmosphere?
Post by: evan_au on 12/06/2018 22:31:29
Post by: Colin2B on 12/06/2018 22:44:00
Presumably, if the pressure difference is too great, you end up with a shock wave?Yes, but a shockwave tends to be a transient rather than a continuous sound. Part of the destructive effect is the difference between the compression on rarefaction, objects can be pressurised by the overpressure and explosivly decompressed by the following partial vacuum. Houses are good at resisting overpressure but not so good when decompressed. Similarly a shockwave from a bomb can pressurise sewers and when the decompression comes the manhole covers will fly up making people think the bomb was in the sewer.
Post by: Bored chemist on 12/06/2018 23:07:04
If I got the arithmetic right that equation gives 2 Pa as the peak air pressure change due to a 100dB sound.The sound pressure of 100dB is not an absolute measure against local air pressure, but a measure against a reference auditory threshold of 2x10-5 Pa.
So to find the actual pressure you would need to use:
Where p0=2x10-5 Pa
Lp=100
P is rms pressure and it is pressure swing either side of ambient - there are compressions & rarefactions so pressure difference between these two is 2p.
In a free field - no reflections - p decreases at 6dB for every doubling of distance so p varies as 1/r not 1/r2.
The air molecules vibrate rather than move so there is no net movement of air.
Thanks but how much the max and min pressure would be in a 100dB wave in atmosphere?
Normal atmospheric pressure is about 100,000Pa
Post by: Colin2B on 13/06/2018 00:18:20
If I got the arithmetic right that equation gives 2 Pa as the peak air pressure change due to a 100dB sound.Spot on.
Normal atmospheric pressure is about 100,000Pa
Sound waves require very little pressure change, or on the otherhand we might say atmospheric pressure is really big - lots of air in the column.
Whichever way you look at it the ears are very sensitive, 15min is maximum permissible exposure to 100dB SPL to avoid hearing damage.
Post by: evan_au on 13/06/2018 11:52:06
Quote from: Colin2B
a shockwave tends to be a transient rather than a continuous soundI vaguely recall seeing a Mythbusters episode where they used a special piece of test equipment to measure the sound pressure of an explosion.
They punctured some of them rated for 50 pounds per square inch, located at various distances from the explosion.
Normal atmospheric pressure is 15 pounds per square inch, so they achieved more than 3 x normal atmospheric pressure.
Post by: Colin2B on 13/06/2018 14:16:00
They punctured some of them rated for 50 pounds per square inch, located at various distances from the explosion.And they weren’t even using nucs!
Normal atmospheric pressure is 15 pounds per square inch, so they achieved more than 3 x normal atmospheric pressure.
Post by: scientizscht on 13/06/2018 15:07:16
They punctured some of them rated for 50 pounds per square inch, located at various distances from the explosion.And they weren’t even using nucs!
Normal atmospheric pressure is 15 pounds per square inch, so they achieved more than 3 x normal atmospheric pressure.
Oh I see thanks.
But there still is movement of air molecules, right? So that areas of high and low pressure are created.
Is the amount of air moved little?
Post by: evan_au on 13/06/2018 22:34:05
Quote from: scientizscht
But there still is movement of air molecules, right?I recently listened to an interview about the life of Enrico Fermi.
He was actively involved in the design and test of nuclear weapons (nukes) during WW2.
At the first atomic bomb test, he took a handful of paper strips, and released them at the right time. He found that they landed about 2 meters from where he released them, allowing him to instantly estimate the yield of the weapon.
After all the instrument readings were analysed, he turned out to be pretty right.
So I guess with nukes, there is a net movement of air, over a few seconds - possibly due to the heating of a cubic kilometer of air. But for ordinary sounds, there is no net movement.
Listen: https://www.scientificamerican.com/podcast/episode/enrico-fermi-the-last-man-who-knew-everything/
Post by: Colin2B on 13/06/2018 23:25:01
At the first atomic bomb test, he took a handful of paper strips, and released them at the right time. He found that they landed about 2 meters from where he released them, allowing him to instantly estimate the yield of the weapon.Interestingly, the generalised method he used to make the estimate of yeild from the paper movement is known as a Fermi estimate or Fermi problem, a form of ‘back of envelope’ calculation based on a number of likely assumptions.