Time Average of a function $f(\phi(t),t)$, $\phi(t)$ is only varying very slowly with time.

Question

I am a bit stumped on a task. Namely I have the function: \begin{align} f(\phi(t),t) = \cos{(\omega (t-t_0))} \, \cos{\phi} + \cos^2{(\omega (t-t_0))}\,\sin{\phi}\,. \end{align} And my task is to calculate the time average. The only thing I know is that $\phi(t)$ is only varying very slowly with time. Personally, however, this means nothing to me. Maybe it's something I had 4 or 5 semesters ago, but it's not in my head anymore. I guess I just develop the function to $\phi$ using Taylor and then from the second order I can ignore the terms because $\frac{\text{d}^2\phi}{\text{d}t^2}$ is very small. But I am not sure. Can anyone here help me briefly?

Answer 1

I have now gone the following way. But this is wrong, because the solution should be given at the end in quantities of $\phi(t)$.

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Now, we assume that $\phi(t)$ is only varying very slowly with time. This means: \begin{align*} \phi(t) &= \phi(t_0) + \dot{\phi}(t_0) (t-t_0) + \underbrace{\frac{\ddot{\phi}(t_0)}{2}}_{=\,\text{very small}} (t-t_0)^2 + \ldots \\ &\approx \phi_0 + \omega_\phi (t-t_0) \end{align*} For the terms now applies: \begin{align*} \langle\cos{\phi(t)}\,\cos{\left(\omega(t-t_0)\right)}\rangle &= \frac{1}{2}\left\langle\cos{\left[(\omega_\phi + \omega)t -\omega t_0 -\omega_\phi t_0 + \phi_0\right]}\right\rangle \\ &\hspace{10pt} + \frac{1}{2}\left\langle\cos{\left[(\omega_\phi - \omega)t + \omega t_0 -\omega_\phi t_0 + \phi_0\right]}\right\rangle\\ &= \frac{1}{2}\left\{\begin{array}{cc} 0, & \omega_\phi \ne \omega \\ \cos{\left[\omega t_0 -\omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = \omega \end{array}\right. \\ \langle\cos^2{\left(\omega(t-t_0)\right)}\,\sin{\phi(t)}\rangle &= \frac{1}{4}\left\langle\sin{\left[(2\omega+\omega_\phi)t - 2\omega t_0 - \omega_\phi t_0 + \phi_0\right]}\right\rangle \\ &\hspace{10pt} -\frac{1}{4} \left\langle\sin{\left[(2\omega-\omega_\phi)t -2\omega t_0 + \omega_\phi t_0 - \phi_0\right]}\right\rangle + \frac{1}{2}\langle\sin\left[\phi(t)\right]\rangle \\ &= \frac{1}{4}\left\{\begin{array}{cc} 0, & \omega_\phi \ne \omega \\ \sin{\left[2\omega t_0 - \omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = 2\omega \end{array}\right. \end{align*} The final solution is now: \begin{align*} \langle f(\phi(t),t) \rangle = \left\{\begin{array}{ll} 0, & \omega_\phi \ne \omega, \omega_\phi \ne 2\omega \\ \frac{1}{2}\cos{\left[\omega t_0 -\omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = \omega \\ \frac{1}{4}\sin{\left[2\omega t_0 - \omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = 2\omega \end{array}\right. \end{align*}