0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.