Propagating wave spherical spreading (or geometric spreading) question The intensity (power per unit area) of a spherical wave falls off as $1/4\pi r^2$.
My question: Is that equation correct? Does this mean the wave amplitude falls off as $1/2\pi^{1/2} r$ ?
I understand the $1/r$ and the $1/2$ but have not seen the $1/\pi^{1/2} $ used.  The sources I have seen only say that the amplitude is inversely proportional to $r$.
See
http://resource.isvr.soton.ac.uk/spcg/tutorial/tutorial/Tutorial_files/Web-basics-pointsources.htm
and
Amplitude decrease during geometrical spreading of a seismic pulse
 A: A spherical wave is a solution of the wave equation in the form:
$$
u(r,t) = \frac{A}{r}e^{i(kr\pm\omega t)}
$$
Its intensity is given by:
$$
I = \lvert u(r,t) \rvert^2 = \frac{A^2}{r^2}
$$
As you can see the intensity defined in this way decreases as $\frac{1}{r^2}$ and the amplitude as $\frac{1}{r}$. These are mathematical quantities. The inverse square law states doesn't consider the intensity as the square of the absolute value. Instead considers the net energy radiation $P$ over a closed surface that contains the source. P must conserve
$$
P = \int I \,d\textbf{A} 
$$
Being the sound pressure constant having supposed spherical waves
$$
P = \lvert I \rvert A = \lvert I \rvert 4\pi r^2
$$
We can invert this to obtain the inverse square law:
$$
I = \frac{P}{4\pi r^2}
$$
So generally speaking the amplitude of a sound wave is related to the solution of the wave equation, for sound the variable in which we solve said equation is the sound pressure. For spherical solution to this equation the amplitude obeys the $\frac{1}{r}$ rule. The intensity is not directly related to this quantity in general, so it follows for the inverse square law $\frac{1}{4\pi r^2}$
A: We relate the intensity at distance r to the power
$I=\frac{P}{4\pi r^2}$,
and the intensity to the amplitude at distance r (pressure difference in spherical sound waves)
$\Delta p=\sqrt{2\rho v}\sqrt{I}$.
Which put together gives
$\Delta p=\sqrt{2\rho v}\sqrt{\frac{P}{4\pi r^2}}=\frac{\sqrt{\rho v P}}{2\sqrt{\pi}r}$.
So we managed to retrieve the $1/\pi^{1/2}$ factor you were looking for.
$\Delta p$ is the pressure difference vel amplitude; $\rho$ is the density of the medium (air); $v$ is the propagation velocity; $I$ is the intensity; $P$ is the total power.
here is how to relate intensity to amplitude https://openstax.org/books/physics/pages/14-2-sound-intensity-and-sound-level
When we have an expression proportional to another expression, there isn't generally much point including constants like $\pi$. We say the intensity is inversely proportional to the distance squared and the amplitude is inversely proportional to the distance. Anything that is independent of $r$ can be attached according to preferences and that's fine.
So why do we keep seeing $1/4\pi r^2$, rather than $1/r^2?$ It's not relevant as far as proportions are concerned. We may add $4\pi$ to emphasize that it will appear in the "complete" expression.
