= Solution
Choose complex amplitudes proportional to $e^{i\omega t}$ and define $\widetilde\sigma=G^*(\omega)\widetilde\epsilon$. The <Maxwell model> gives
$$
i\omega\widetilde\epsilon=\left(\frac{i\omega}{G_0}+\frac1\eta\right)\widetilde\sigma,
\qquad G^*=G_0\frac{i\omega\tau}{1+i\omega\tau}.
$$
Rationalizing the denominator yields the <storage modulus> and <loss modulus>:
$$
\boxed{G'=G_0\frac{(\omega\tau)^2}{1+(\omega\tau)^2},\qquad
G''=G_0\frac{\omega\tau}{1+(\omega\tau)^2}.}
$$
For $\omega\tau\ll1$, $G''\simeq\eta\omega$ dominates and $G'\simeq G_0\omega^2\tau^2$ is small: the material has <time> to flow. For $\omega\tau\gg1$, $G'\to G_0$ and $G''\simeq G_0/(\omega\tau)$: the <dashpot> scarcely deforms during a cycle and the response is mainly elastic. At $\omega\tau=1$, $G'=G''=G_0/2$, and $G''$ is maximal. The <stress> leads <strain> by $\delta=\arctan(G''/G')$, ranging from nearly $\pi/2$ to nearly zero.
For a fixed <strain> amplitude $\epsilon_a$, dissipation per cycle is $\pi G''\epsilon_a^2$. A peak in $G''$ therefore concerns that normalization. At fixed <stress> amplitude $\sigma_a$, it is instead $\pi\sigma_a^2/(\eta\omega)$, so it has no finite-frequency maximum in this ideal model. Reversing the Fourier <time> convention reverses the imaginary-part sign, while the physical dissipated <energy> remains positive.
\Image[/past-exam-of-the-soft-condensed-matter-course-of-the-university-of-cambridge/2012/paper-1-maxwell-response.png]
{title=Maxwell storage and loss}
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