I am finding out the relationship between intensity of sound and speed of sound. To do that, I am using this equation: $$I =\frac{\Delta P^2_\text{max}}{2 \rho v}$$
where $\Delta P_\text{max}$ is pressure amplitude, $I$ is intensity of sound, $\rho$ is density of medium, and $v$ is wave speed.
I got this equation from this website, Physics.info, my only problem is that I cannot figure out how to measure the pressure amplitude. How do I do that?
First of all, this question is more or less about the engineering and that would need another SE site. But since the answer is not complicated:
If you consider sound in "common, general meaning" (i.e. something that could be percieved by an ear) then the simpliest pressure meter is the microphone. OK, there are microphones sensitive to air pressure or air velocity, but from recorded signal you can always read the pressure variations. Common way to do that is to record a normalised signal 94 dB at 1 kHz which corresponds to 1 Pa.
From pressure variations to $P_{max}$ there is a simple way using RMS (root mean square, consult this), i.e. for a sine pressure wave of amplitude $A$: $P_{max}=\frac{\sqrt{2}}{2}A$.
Lets denote Intensity from (I).Then we know $$I=\frac{1}{2}\rho vA^2\omega^2$$ Hence the "A" here is amplitude and "v" is velocity of wave and $\omega$ is the angular velocity of wave. So we know that $\Delta P_{max}=\beta Ak$ From this we could say $$A=\Delta P_{max}\frac{\beta}{Ak}$$(bulk modulus of the material is $\beta$ So we get $$I=\frac{1}{2}\rho v \left(\frac{(\Delta P_{max})^2}{\beta^2 k^2}\right)\omega^2$$ For Pressure amplitude consider Imagine a half vertically filled cylinder having moved the Liquid a $\Delta x$ So $\Delta V=A_1\Delta x$(Initial Volume ) and futhermore $\Delta y$ distance by it now next change $\Delta V=A_1\Delta y$(volume change). The equation $$\beta=\frac{\Delta P}{\left(-\frac{\Delta V}{V}\right)}$$ $$\Rightarrow\beta=\frac{\Delta P}{\left(-\frac{\Delta y}{\Delta x}\right)}$$ $$\Rightarrow\Delta P=(-\beta)\left(\frac{\Delta y}{\Delta x}\right)\tag1$$ we know by wave equation $$y=A\sin(\omega t-kx)$$ from here $$\frac{\Delta y}{\Delta x}=-kA\cos(\omega t-kx)$$ Putting this in equation 1 $$\Rightarrow\Delta P=(\beta Ak)(\cos(\omega t-kx))$$ from here $$(\omega t-kx)=0$$ Then $$\Delta P_{max}=\beta Ak$$
