[Phys-L] Re: Acoustics question about popped balloons.

[jsd at AV8N.COM (John Denker)]

John SOHL wrote:

>  My thinking says that the air rapidly leaving the
> popped bag or balloon is crashing into the surrounding air which then
> causes a shock wave

Be careful about terminology:  a _shock_ (also called a shock wave)
is a very special, specific phenomenon.  It always has a supersonic
propagation velocity.

It is very unlikely that a paper bag produces puts any significant
amount of energy into shocks.  Rather, it's just a snappy noise.
You can call it a snap, or a bang, or a pop ... but you shouldn't
call it a shock.

> which in turn propagates outward causing the sound
> that we hear. Specifically, our ears are detecting the shock wave and
> not any momentary increase in pressure caused by the released air.

Sound _is_ a pressure wave.  You can't detect sound without detecting
pressure.  It's impossible by definition.

Much of the point of the balloon, paper bag, or cymbal is to _concentrate_
the energy.  If you wave a drumstick back and forth, it doesn't make
much noise, partly because the frequency is too low to be heard,
and also because the coupling between stick and sound-field is poor.
Similarly fluffing air in and out of a bag doesn't make much sound.
But if the drumstick suddenly smacks into a cymbal, or if the bag
suddenly releases its internal overpressure, you get an outgoing
sound-pulse.  The pulse (initially at least) has a sharply-rising
leading edge.

> It seems unrealistic to me that the air pressure would increase and
> propagate such that you can easily hear a popped balloon even if you are
> 100 meters away. Yet, this seems perfectly reasonable if you are hearing
> a shock wave caused by colliding air.

What's special about colliding air?  All air pressure (including air
pressing against itself, and air pressing against eardrums) is entirely
due to "colliding" air.

As to sound being audible 100 meters away, why is that surprising?

> A similar question is that of thunder. Again, I assume that what we are
> hearing is the heated air expanding outward colliding with "still" air
> in the surrounding region.

Actually thunder does start out as a shock.  (A bit of dynamite also
suffices to launch a shock.)  But when the shock propagates out from
the source a ways, in cylindrical geometry (from a lightning strike)
or spherical geometry (from a small bit of explosive), then the
shock gets dispersed i.e. "diluted" to the point where it becomes an
ordinary sound wave.

> This also has the effect of leaving a lower
> pressure region (because of the heat) that the air then rushes back into
> almost instantly. When this air rushes back it collides and more shock
> waves are produced. The effect being a "ka-boom" sound having a longer
> or multiple impluse thunder clap vs. the single "pop" you get from a
> balloon.

Actually it's somewhat trickier than that.
  1) Thunder up close, and explosives up close, do not go ka-boom.  They
go SNAP!  The SNAP! noise is unmistakable and unforgettable.  The front
of the shock has a verrry steep rise, and your ears detect this.

  2) Because the wave equation in cylindrical-polar and/or spherical-polar
coordinates is dispersive (even when the medium itself is not dispersive),
an outgoing sound wave (not a shock) that started out with a very sharp
leading edge will disperse into a lot of wiggles.  It isn't as sharp,
and it doesn't sound as sharp.

You can figure out why this has to happen by direct application of the
conservation laws, as I wrote last week in another thread, to wit:


As a not-very-elementary application of conservation laws, to
illustrate the power thereof, consider the following:  at the
center of a large region of air, we have a balloon.  Using
some complex arrangement of clockworks, we use the balloon
to launch a spherically-symmetric outgoing sound wave.  The
details are arranged so that at time t = t1, the wave has a
square profile, i.e. it looks like this:

.               ____                   ____
.              |    |       (B)       |    |
.    __________|    |_________________|    |__________
.                <<                     >>

where "B" marks the position of the balloon.  The curve is
a graph of pressure versus position.  Assume the small-amplitude
limit, so everything is linear.

Prove that this wave cannot maintain its shape.  Specifically,
show that the wavefunction cannot be written as
     phi(r, t) = f(r - c t)
where f() is a function of one variable ... even though the
system is linear, and plane waves in the same medium can perfectly
well be written as
     Phi(x, t) = F(x - c t)

That is to say, prove that the wave equation in polar coordinates
is necessarily dispersive, event though the corresponding wave equation
in rectangular coordinates is not dispersive.

Sketch roughly what the wave must look like at some time t >> t1.

HUGE HINT:  There are two conservation laws in play here:  conservation
of energy, and conservation of air molecules.

HUGE HINT:  Scaling laws are very high on my list of important ideas,
right up there with the conservation laws.  Derive a scaling law for
amplitude-versus-r using conservation of energy, then derive another
scaling law for the same thing (amplitude-versus-r) using conservation
of air.  Two different scaling results for the same thing?!?  What
does that tell you?