Re: [Phys-L] sound intensity problem

["John Clement" <clement@hal-pc.org>]

It would seem that 1750 miles is unreasonable.  If a thunderstorm is that
far away you won't hear it.  Indeed if it is only a few miles away you only
hear the boom and not the initial crackle.  The sound from a thunderstorm is
much more powerful than a megaphone and it is in the air where objects won't
block the sound.  So the attenuation of high frequencies is substantial. You
can get an order of magnitude change in  sound level by just the
inefficiencies in the megaphone.

I really doubt the lensing is much of a factor, but backgound noise can be a
factor, especially in a stadium.  I would believe a mile of so, but not that
large a figure.

John M. Clement
Houston, TX

> The answers from the two Johns are useful, but the bottom 
> line is that in theory the distance is NOT unreasonable.  A 
> way to think about it is to imagine outer space filled with 
> wind free air (OK--no longer outer
> space....but) and now do the experiment.  The key here is the 
> large ratio between the initial sound intensity and the 
> intensity at the 
> threshold of hearing.    Even at inverse square fall-offs, it still 
> requires a very large spherical surface to reduce the 
> intensity by 10-12 orders of magnitude.
>
> R.W.Tarara
>
>
> On 4/5/2013 2:00 PM, Anthony Lapinski wrote:
> > I was considering this sound problem.. For a 100-W 
> megaphone, how far 
> > would you have to be so that it is barely audible (0 dB)?
> >
> > I = P/A
> >
> > Io = P/4pir2
> >
> > With Io = 10-12 W/m2, the result is 2821 km = 1750 mi
> >
> > This seems unreasonably far! I would think that the sound 
> level would 
> > drop off much closer, as we typically experience. Or am I 
> > missing/miscalculating something?
> >
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>
> --
> Richard Tarara
> Professor of Physics
> Saint Mary's College
>
> free Physics instructional software
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