The sound or wave intensity is defined by energy transfer rate with time (power) per unit of area:

$$ I = \frac{P}{A} \tag{1} $$

so this equation makes sense since the denominator is the area of the sphere and the numerator is the power of the sound or wave and it also explains why the further the source is, the lower the sound is:

$$ I = \frac{P}{4\pi r^2} \tag{2} $$

However, I don't understand this equation because all the quantities here are constant in an unchanged environment: wave speed, air density, angular frequency and displacement amplitude.

$$ I = \tfrac{1}{2} \rho v \omega^2 s_m^2 \tag{3} $$

So, according to this, the sound intensity is a constant and it doesn't express how intensity changes with the area that is affected by the distance from the source.

How to use it to calculate the intensity of a particular distance while the equation doesn't have distance quality?


1 Answer 1


Your last equation;

$$ I= \tfrac{1}{2} \rho v \omega^2 s_m^2 $$

is derived by taking the equation for a plane wave:

$$ s(x, t) = s_m \sin(\omega t - kx) $$

and calculating the kinetic energy associated with the wave. The calculation is described in this article from the Physics Hypertextbook.

For a spherical wave the prefactor $s_m$ is not a constant but decreases as $r^{-1}$. This means $s_m^2$ decreases as $r^{-2}$, and therefore both equations give an inverse square dependance of intensity on distance.

  • $\begingroup$ I'm a little confused, as you said: $s_m$ decay as $r^{-1}$ for spherical wave. Is it the same for my last equation. In my text book, all the sound waves are expressed by the plane wave equation, no distinction between plane wave and spherical wave and there is no relation between distance from the and displacement at all. $\endgroup$
    – aukxn
    Dec 24, 2014 at 8:39
  • $\begingroup$ @aukxn: the equation $I = P/A$ is always true, but the equation $A = 4\pi r^2$ is only true for a spherical wave. So your equation (2) is only true if the wave is spherical. It would not be true for a plane wave. $\endgroup$ Dec 24, 2014 at 8:43
  • $\begingroup$ So it means that if there is a source wave that can emits waves in only a straight direction so those waves travel in a straight line but not diverge like a sphere then the Intensity of that wave is all the same regardless of the distance from the source? $\endgroup$
    – aukxn
    Dec 24, 2014 at 8:58
  • $\begingroup$ @aukxn: yes, that's correct. $\endgroup$ Dec 24, 2014 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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