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I am looking at total internal reflection for an acoustic wave, defined in terms of its pressure such that

$$p = p_1 \,exp\left[-i\{\omega t-\vec{k} \cdot\vec x\}\right]$$

Using the definition of mean-square pressure, given by

$$ (\Delta p)^2 = \frac{1}{T}\int_0^T [Re(p)]^2 dt$$

I am required to thus show that

$$(\Delta p)^2 = \frac{1}{2}|p_1|^2$$

Given that $p_1$ is complex, I have observed that if

$$p_1 = Re(p_1) + i\,Im(p_1)$$

and further if

$$\omega t-\vec{k} \cdot\vec x = \varphi$$

then

$$Re(p) = Re(p_1)cos(\varphi) + Im(p_1)sin(\varphi)$$

So then I see that

$$(\Delta p)^2 = \frac{1}{T}\int_0^T \left[Re(p_1)^2cos^2(\varphi) + Im(p_1)^2 sin^2(\varphi) + 2Re(p_1)Im(p_1)cos(\varphi)(\sin(\varphi)\right]dt$$

I am slightly lost at this point. Do I need to equate $Re(p_1)$ with $Im(p_1)$? But how would such a move be legal?

Victoria
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  • What is preventing you from evaluating the integral term by term? You know that $\phi=\omega t-K$, where $K$ is a constant (in time), so just evaluate each piece. $p_1$ is constant. – Alex R. Mar 28 '15 at 04:14
  • Okay, I've done this and found that: $$ \frac{1}{T}\left[ \frac{Re(p_1)^2}{2\omega}\left(\varphi + \cos(\varphi)\sin(\varphi)\right) + \frac{Im(p_1)^2}{2\omega}\left(\varphi - \cos(\varphi)\sin(\varphi)\right) - \frac{Re(p_1)Im(p_1)}{\omega}\cos^2(\varphi)\right]_0^T$$

    Am I any closer? Looks overly messy.

     – Victoria Mar 28 '15 at 04:56

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