So in the case of a point source of a sound wave in a sphere, I'm wondering why the Intensity decreases with $r^2$ because if i setup the formula the Power always increases with $r^2$. So these two factors should cancel eachother out no?

So to start $I = \frac{P}{A}$ with $A=4\pi r^2$

And $E_{mech}=\frac{1}{2}m\omega^2\Delta s^2$ with $\Delta s$ being the amplitude.

we substitute m with $m=\rho V_{sphere} $

$m=\rho \frac{4}{3}\pi \Delta x^3 $ using $\Delta x$ as the radius.

plugging everything in the equation for $P=\frac{\Delta E}{\Delta t}$ we get: $P=\frac{\frac{1}{2}\rho \frac{4}{3}\pi \Delta x^3\omega^2\Delta s^2$}{\Delta t}$ in this we substitute $v_{x} = \frac{\Delta x}{\Delta t}$ and get: $P=\frac{1}{2}\rho \frac{4}{3}\pi \Delta x^2 v_{x}\omega^2\Delta s^2$}$

If we substitute this in the equation for I and remember that $r=\Delta x$ we get: $I=\frac{\frac{2}{3}\rho \pi \Delta x^2 v_{x}\omega^2\Delta s^2}{4 \pi \Delta x^2}$

$I=\frac{1}{6}\rho v_{x}\omega^2\Delta s^2$

from which we can clearly see that the intensity does not decline with distance. But ofcourse I know that in reality this is clearly the case. So my question is, what am I doing wrong? or what factor am i forgetting?


In most of these sorts of theories the amplitude decays like $1/r$. For example in quantum electrodynamics, the quantum amplitude for a photon to travel through space to a point a distance $r$ away is $$\psi(r) = \frac1r e^{2\pi i r/\lambda},$$where $\lambda$ is the photon's wavelength. The thing on the left is the magnitude of the amplitude; it diminishing by $1/r$ allows the intensity $|\psi|^2$ to fall off like $1/r^2$; the thing on the right is basically a rotation matrix by angle $2\pi~r/\lambda$ and it does all of the wavy stuff that light as an electromagnetic wave does.

  • $\begingroup$ Yeah I tought it would have something to do with the wave not being a perfect Harmonic but a deminishing one. thanks for confirming $\endgroup$ – Dries Van Eyck May 23 '19 at 13:02

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