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For the normalized damped pendulum, $E={\dot{\theta }}^2/2+1-\cos \theta$ obeys $\dot{E}=-c{\dot{\theta }}^2\le 0$. Its mechanical energy is constant only along an equilibrium trajectory, and strictly decreases over every positive-length interval on a nonstationary trajectory. At an individual turning point its derivative vanishes. This excludes nonconstant periodic trajectories despite oscillatory approach to a stable focus.

Differentiate the mechanical energy and use the damped pendulum equation:The energy dissipation of a damped pendulum makes $E$ nonincreasing, rather than strictly decreasing at every instant: its derivative vanishes at turning points. It is constant along an equilibrium trajectory. Along every nonstationary trajectory it strictly decreases over any nonzero time interval, because a vanishing integral of $c{\dot{\theta }}^2$ over an interval would force an equilibrium there and hence everywhere by uniqueness.

A point mass constrained to a fixed-length massless arm under uniform gravitational acceleration is a simple pendulum. With no drag its angular equation is $\ddot{\theta }+(g/\ell )\sin \theta =0$. The small-angle approximation yields simple harmonic motion, whereas the full restoring force is nonlinear. Adding drag gives a damped pendulum.