# Confusion with time symmetry - pendulum example

If the classic nonlinear pendulum with friction equation is reviewed as example, in Wolfram-Alpha it can be seen it have decaying solutions as expected: $$\ddot{x}+2\cdot 0.021\,\dot{x}+0.2\sin(x)=0, x(0)=\frac{\pi}{2}, \dot{x}(0)=0 \tag{Eq. 1}\label{Eq. 1}$$

But following the analysis for T-Symmetry, under the transformation $$\hat{t} \to - t$$ the position and acceleration remains the same, but the velocity profile change in sign: this applied to \eqref{Eq. 1} shows it change under this transformation, so it not fulfill T-Symmetry.

If instead the standard Drag force $$F_{\text{drag}}\propto (\dot{x})^2$$ is used as is shown here for the equation: $$\ddot{x}+0.021(\dot{x})^2+0.2\sin(x)=0, x(0)=\frac{\pi}{2}, \dot{x}(0)=0 \tag{Eq. 2}\label{Eq. 2}$$ their solution aren't showing the expected decay one can see on the experimental pendulums. This is commonly solve using an ansatz for the drag force $$F_{\text{drag}}\propto \dot{x}|\dot{x}|$$, which as can be seen here for the equation: $$\ddot{x}+0.021\dot{x}|\dot{x}|+0.2\sin(x)=0, x(0)=\frac{\pi}{2}, \dot{x}(0)=0 \tag{Eq. 3}\label{Eq. 3}$$ their solution are indeed having the expected decaying behavior for a pendulum with friction.

Since \eqref{Eq. 2} fulfills T-Symmetry but fails to proper model the pendulum with friction, but both \eqref{Eq. 1} and \eqref{Eq. 3} don't fulfill T-Symmetry but modeled properly the amplitudes, its look like for having a decaying behavior it was required to choose a differential equation that is not fulfilling the time-reversal-symmetry.

Since commonly is said that physics laws must fulfill T-Symmetry being the only counterexample the 2nd Law of Thermodynamics for the Entropy, I want to understand:

• What I am doing wrong here?, since the only solutions that shows having the experimentally observed decay are the differential equations which are not fulfilling T-Symmetry.
• Does this imply the kinetic energy must be $$K=\frac{1}{2}mv|v|$$ instead of $$K=\frac{1}{2}mv^2$$?