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Wave Derivation: Intensity and Pressure Amplitude Relationship

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    My professor mentioned that,

    [tex]
    {I} \propto {\Delta{{A}_{p}}^{2}}
    [/tex]

    Where,

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

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

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

    Can anyone check?

    2. Relevant equations

    [tex]
    {s(x, t)} = {A_{s}}cos({kx} \mp {\omega}{t} + {\phi}),{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}[\pm{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}propagation]
    [/tex]

    [tex]
    {\Delta{p(x, t)}} = \Delta{A_{p}}sin({kx} \mp {\omega}{t} + {\phi}),{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}[\pm{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}propagation]
    [/tex]

    [tex]
    {\Delta{A_{p}}} = {v}{\rho}{\omega}{A_{s}}
    [/tex]

    [tex]
    I = \frac{1}{2}{\rho}{v}{\omega}^{2}{{A_{s}}^{2}}
    [/tex]

    3. The attempt at a solution

    Using,

    [tex]
    I = \frac{1}{2}{\rho}{v}{\omega}^{2}{{A_{s}}^{2}}
    [/tex]

    and

    [tex]
    {\Delta{A_{p}}} = {v}{\rho}{\omega}{A_{s}}
    [/tex]

    So rearranging, I get the following,

    [tex]
    {\Delta{{A_{p}}^{2}}} = {2}{\rho}{v}{I}
    [/tex]

    Is that right?

    Any help is appreciated.

    Thanks,

    -PFStudent
     
  2. jcsd
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