1
$\begingroup$

In my medical physics book it says that Intensity is 1/2 a2/pc=a2/2z Where a is the amplitude, p medium of density, c velocity of that wave and offcourse Z is the impedance. When I google it it just comes up that intensity of sound waves are proportional to the pressure of a area. Could someone pleas explain me this equation if that is the case?

$\endgroup$
1
$\begingroup$

Sound waves are pressure waves. We measure it as a logarithmic ratio of intensity. Sound intensity is a useful parameter to measure because it's related to the energy incident on a surface which can be easy to measure. Sound intensity is proportional to pressure squared. When calculating decibels we would have to handle that like so: \begin{equation} I = p^2/Z \end{equation} for some constant of proportionality k \begin{equation} 10log(I_1/I_0) = 10log(Zp_1^2/Zp_0^2) =20log(p_1/p_0) \end{equation} where p1 is the wave pressure and p2 is the ambient or reference pressure.

The equation you have quoted is simply one that equates the impedance to \begin{equation} c\rho = Z \end{equation} If you wish to understand where that comes from google the derivation of the acoustic wave equation. Really it's just a property of the medium that determines the relation ship between pressure particle velocity and energy.

$\endgroup$

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.