# Wave Derivation: Intensity and Pressure Amplitude Relationship

1. Sep 8, 2007

### PFStudent

1. The problem statement, all variables and given/known data

My professor mentioned that,

$${I} \propto {\Delta{{A}_{p}}^{2}}$$

Where,

$I$ $\equiv$ Intensity measured in units of $\frac{[W]}{[{m}^{2}]}$

$\Delta{{A}_{p}}$ $\equiv$ pressure amplitude measured in units of $[Pa]$

What I want to do is actually derive the relationship for $I$ and $\Delta{A_{p}}^{2}$, I did it below but I am not sure if it is right...

Can anyone check?

2. Relevant equations

$${s(x, t)} = {A_{s}}cos({kx} \mp {\omega}{t} + {\phi}),{{.}}{{.}}{{.}}{{.}}{{.}}[\pm{{.}}x{{.}}propagation]$$

$${\Delta{p(x, t)}} = \Delta{A_{p}}sin({kx} \mp {\omega}{t} + {\phi}),{{.}}{{.}}{{.}}{{.}}{{.}}[\pm{{.}}x{{.}}propagation]$$

$${\Delta{A_{p}}} = {v}{\rho}{\omega}{A_{s}}$$

$$I = \frac{1}{2}{\rho}{v}{\omega}^{2}{{A_{s}}^{2}}$$

3. The attempt at a solution

Using,

$$I = \frac{1}{2}{\rho}{v}{\omega}^{2}{{A_{s}}^{2}}$$

and

$${\Delta{A_{p}}} = {v}{\rho}{\omega}{A_{s}}$$

So rearranging, I get the following,

$${\Delta{{A_{p}}^{2}}} = {2}{\rho}{v}{I}$$

Is that right?

Any help is appreciated.

Thanks,

-PFStudent