Stress and Strain lag for viscous materials As stated here,

In purely viscous materials, strain lags stress by a 90 degree phase.

How can we derive this statement? Is it experimental? If not, which is its proof and what is the physical cause of this lag between stress and strain?
 A: Viscoelastic material behavior can be modeled by a combination of springs and dashpots. A springs represents the elastic behavior of the material. A dashpot represents the viscous behavior. The dashpot acts as a damper which resists motion via viscous friction. The resulting force is proportional to the velocity, but acts in the opposite direction, slowing the motion and absorbing energy. In the dashpot the displacement lags the force. The springs act to resist displacement by storing elastic potential energy.
A perfectly viscous material can be modeled as an ideal dashpot. All energy is dissipated as viscous friction heating. See Figure below.
In the equation below $x(t)$ can be viewed as displacement as a function of time, and $\frac{dx(t)}{dt}$ the displacement rate, due to the applied force $F(t)$. $C$ is the damping coefficient due to viscous friction. Note that if $F(t)$ is a sine function then then $x(t)$ is a cosine function.  Displacement and force are analogous to strain and stress. Stress and strain are therefore 90 deg out of phase.
Physically, if you apply a force $F$ suddenly to the dashpot, it resists motion. But it will gradually give way to the applied force. The strain lags the stress.
Hope this helps.

A: First, for comparison, let's consider a purely elastic material. Its constitutive equation is Hooke's Law, $$\sigma(t)=E\varepsilon(t),$$
where $\sigma$ is the stress, $t$ is time, $E$ is Young's modulus, and $\varepsilon$ is the strain.
When you pull on an elastic material, it stretches immediately and stops. If you pull less hard, it recovers somewhat. If you let go, it returns to its original dimensions. If you push (i.e., pull in the opposite direction), you get deformation in the opposite direction. Make sure these relations are familiar.
In the context of sinusoidal phase lag, all this is compatible with a load of $\sigma(t)\propto\sin(\omega t)$ (where $\omega$ is the angular frequency), a deformation of $\varepsilon(t)\propto\sin(\omega t)$, and a phase lag of zero.
OK, now only to a purely viscous material. Its constitutive equation is $$\sigma(t)=\mu\dot\varepsilon(t),$$
where $\mu$ is the viscosity; note that we're now working with the time derivative of the strain.
What this means is that, unlike the purely elastic solid, if you pull on this material, it begins to move continuously. If you pull less, it only moves less quickly. If you let go, it stops in its new position. The only way to recover the original dimensions is to push (i.e., pull in the opposite direction).
For sinusoidal oscillation, we thus obtain the greatest deformation rate at the peak load and the maximum deformation when the load is zero. (Note how different this behavior is from that of the elastic solid, where we saw the maximum deformation at the maximum load and a deformation of zero at a load of zero.)
All this is compatible with a deformation of $\varepsilon(t)\propto -\cos(\omega t)$ in response to a load of $\sigma(t)\propto\sin(\omega t)$. Since $-\cos(\omega t)=\sin\left(\omega t-\frac{\pi}{2}\right)$, we obtain a phase lag of $\pi/2$ or 90 degrees, as subtracting something from a sinusoidal argument is equivalent to pushing the waveform to the right.
